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Quantum entanglement

Quantum entanglement is a phenomenon where each particle constituting a composite system cannot be described by its own characteristic state. Instead, the particles must be expressed as a single composite state. One such entangled composite system is the decay of a spin-0 particle at rest into two spin-\frac{1}{2} particles with zero relative orbital angular momentum. According to the conservation of linear momentum, the particles move in opposite directions after decay. If we measure the spin of the particles by passing them through two Stern-Gerlach devices, we would get correlated results due to the conservation of spin momentum. For example, if one particle is measured to be spin up with respect to the z-axis (see diagram below), the other particle is always measured to be spin down with respect to the same axis, and vice versa.

The quantum mechanical explanation of the results is that the singlet state \vert\psi\rangle=\frac{1}{\sqrt{2}}[\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2] collapses into \vert\phi_1\rangle=\vert+z\rangle_1\otimes\vert-z\rangle_2 if the first measurement is spin up. If the first measurement is spin down, the composite state collapses into \vert\phi_2\rangle=\vert-z\rangle_1\otimes\vert+z\rangle_2. Whether the first measurement is spin up or spin down is purely random, as evident from the coefficient of the singlet state. In other words, quantum mechanics provides, via \vert\psi\rangle, a statistical distribution of possible outcomes of a measurement, in a way that the certainty that a system possesses a physical property is only determined after a measurement is made.

 

Question

How does the coefficient of the singlet state show that the collapse of the state is random?

Answer

The singlet state is normalised and is a linear combination of the states \vert\phi_1\rangle=\vert+z\rangle_1\otimes\vert-z\rangle_2 and \vert\phi_2\rangle=\vert-z\rangle_1\otimes\vert+z\rangle_2; i.e. \vert\psi\rangle=\sum_{i=1}^{2}c_i\vert\phi_i\rangle, where c_1=c_2=\frac{1}{\sqrt{2}}. Since \vert c_i\vert^{2} is interpreted as the probability that a measurement of a system will yield an eigenvalue associated with the eigenfunction \phi_i, the probability of obtaining an up-spin (or a down-spin) for the first measurement is 50%.

 

If we conduct the experiment a hundred times, with the first Stern-Gerlach device in the +z direction and the second device in the +x direction, we have (assuming a random distribution of spins after decay) the following results:

 

Particle 1 (measured in +z) Particle 2 (inferred in +z) Particle 2 (measured in +x)
EV_1 Population EV_{2,z} EV_2 Population EV_1EV_2 Average EV_1EV_2
+\frac{\hbar}{2} 50 -\frac{\hbar}{2} +\frac{\hbar}{2} 25 +\frac{\hbar^{2}}{4} \frac{(25\times\frac{\hbar^{2}}{4})+[25\times(-\frac{\hbar^{2}}{4})]}{50}=0
-\frac{\hbar}{2} 25 -\frac{\hbar^{2}}{4}
-\frac{\hbar}{2} 50 +\frac{\hbar}{2} +\frac{\hbar}{2} 25 -\frac{\hbar^{2}}{4} \frac{[25\times(-\frac{\hbar^{2}}{4})]+(25\times\frac{\hbar^{2}}{4})}{50}=0
-\frac{\hbar}{2} 25 +\frac{\hbar^{2}}{4}

 

In general, if the second Stern-Gerlach device is rotated at an angle \theta relative to the first device, the composite spin angular momentum operator associated with EV_1EV_2 (the product of EV_1 and EV_2) is:

\hat{S}_z^{\; (1)}\otimes\hat{S}_r^{\; (2)}=\left (\hat{S}_z^{\; (1)}\otimes I\right )\left (I\otimes\hat{S}_r^{\; (2)}\right )

Substitute eq174 and eq184 (where we let \phi=0^{\circ}) into the above equation,

\hat{S}_z^{\; (1)}\otimes\hat{S}_r^{\; (2)}=\frac{\hbar^{2}}{4}\begin{pmatrix} cos\theta &sin\theta &0 &0 \\ sin\theta &-cos\theta &0 &0 \\ 0 &0 &-cos\theta &-sin\theta \\ 0 &0 &-sin\theta &cos\theta \end{pmatrix}

The average value of EV_!EV_2 over multiple experiments is:

\small \langle\psi\vert\hat{S}_z^{\; (1)}\otimes\hat{S}_r^{\; (2)}\vert\psi\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 &1 &-1 &0 \end{pmatrix}\frac{\hbar^{2}}{4}\begin{pmatrix} cos\theta &sin\theta &0 &0 \\ sin\theta &-cos\theta &0 &0 \\ 0 &0 &-cos\theta &-sin\theta \\ 0 &0 &-sin\theta &cos\theta \end{pmatrix} \frac{1}{\sqrt{2}}\begin{pmatrix} 0\\1 \\ -1 \\ 0 \end{pmatrix}

\langle\psi\vert\hat{S}_z^{\; (1)}\otimes\hat{S}_r^{\; (2)}\vert\psi\rangle=-\frac{\hbar^{2}}{4}cos\theta\; \; \; \; \; \; \; \; 226

or in spin units of \frac{\hbar}{2}

\langle\psi\vert\hat{S}_z^{\; (1)}\otimes\hat{S}_r^{\; (2)}\vert\psi\rangle=-cos\theta\; \; \; \; \; \; \; \; 227

Finally, an interesting behaviour of an entangled system is that the particles constituting the system remain entangled if they are separated by a large distance, even at a distance greater than light can travel within the time between measurements. If so, and if the outcome of the first measurement is random, one may perceive that the two particles are able to influence each other instantaneously. This property of a composite system seems to be at odds with special relativity, which implies that nothing (including communication of any form) can travel faster than the speed of light. In opposition to the quantum mechanical interpretation of the results, Albert Einstein, along with Boris Podolsky and Nathan Rosen, proposed an alternate explanation called the Einstein-Podolsky-Rosen paradox.

 

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The singlet and triplet states

The singlet state describes a composite system with a coupled spin eigenstate of \small \vert j=0,m_j=0\rangle, while a triplet state refers to a composite system with coupled spin eigenstates of \small \vert 1,+1\rangle, \small \vert 1,0\rangle and \small \vert 1,-1\rangle.

Consider a system of two spin-\small \frac{1}{2} particles ( \small s_1=\frac{1}{2} and \small s_2=\frac{1}{2}). According to the Clebsch-Gordan series, the allowed angular momenta of the system are \small j=1,0, which implies that the system is characterised by the spin eigenstates \small \vert 1,+1\rangle, \small \vert 1,0\rangle, \small \vert 1,-1\rangle and \small \vert 0,0\rangle. The basis states forming these four spin eigenstates are given by eq193, with the general spin eigenstate of the system being a linear combination of these basis states (see eq195).

Let’s begin with the determination of the values of the coefficients for the spin eigenstates \small \vert 1,0\rangle and \small \vert 0,0\rangle, which must adhere to the condition: \small \hat{S}_z^{\; (T)}\vert\psi\rangle=0\hbar\vert\psi\rangle. Letting the operator \small \hat{S}_z^{\; (T)} act on eq195 and using eq193 and 197,

\hat{S}_z^{\; (T)}\vert\psi\rangle=c_1\hbar\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}-c_4\hbar\begin{pmatrix} 0\\0 \\ 0 \\ 1 \end{pmatrix}+0\hbar\left [c_2\begin{pmatrix} 0\\1 \\ 0 \\ 0 \end{pmatrix}+c_3\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ]

For the eigenvalue equation \hat{S}_z^{\; (T)}\vert\psi\rangle=0\hbar\vert\psi\rangle to be valid, c_1=c_4=0, i.e. one possible spin eigenstate when m_j=0 is \vert\psi\rangle=c_2\vert+z\rangle_1\otimes\vert-z\rangle_2+c_3\vert-z\rangle_1\otimes\vert+z\rangle_2, while the other possible spin eigenstate is \vert\psi\rangle=c_2\vert+z\rangle_1\otimes\vert-z\rangle_2-c_3\vert-z\rangle_1\otimes\vert+z\rangle_2. These two spin eigenstates must correspond to either \vert 1,0\rangle or \vert0,0\rangle. To distinguish them, we use eq203 and let the operator act on the two spin eigenstates to give:

\hat{{S}^{2}}^{(T)}\left [ c_2\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}\pm c_3\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ] =\hbar^{2}\left [ c_2\begin{pmatrix} 0\\1 \\1 \\0 \end{pmatrix}\pm c_3\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}\right ]

Since the two eigenstates are characterised by j=1 and j=0, the expected results are \hat{{S}^{2}}^{(T)}\vert\psi\rangle=j(j+1)\hbar^{2}\vert\psi\rangle=2\hbar^{2}\vert\psi\rangle and \hat{{S}^{2}}^{(T)}\vert\psi\rangle=j(j+1)\hbar^{2}\vert\psi\rangle=0\hbar^{2}\vert\psi\rangle respectively. Therefore, the above equation is valid only if c_2=c_3=c, such that:

\small \hat{{S}^{2}}^{(T)}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}+ c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ] =\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\1 \\0 \end{pmatrix}+ c\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}\right ]=2\hbar^{2}c\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}=2\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}+ c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ]

and

\small \hat{{S}^{2}}^{(T)}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}- c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ] =\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\1 \\0 \end{pmatrix}- c\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}\right ]=\hbar^{2}c\begin{pmatrix} 0\\0 \\ 0 \\ 0 \end{pmatrix}=0\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}- c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ]

Therefore, \small \vert1,0\rangle=c(\vert+z\rangle_1\otimes\vert-z\rangle_2+\vert-z\rangle_1\otimes\vert+z\rangle_2) and \small \vert0,0\rangle=c(\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2). The spin state \small \vert0,0\rangle as mentioned earlier is the singlet state, which after normalisation becomes

\small \vert0,0\rangle=\frac{1}{\sqrt{2}}(\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2)\; \; \; \; \; \; \; \; 220

 

Question

Show that the normalisation constant is \small \frac{1}{\sqrt{2}}.

Answer

\langle\psi\vert\psi\rangle=1\; \; \; \; \Rightarrow \; \; \; \; c^{*}\left [ \begin{pmatrix} 0\\1 \\ 0 \\ 0 \end{pmatrix}-\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix} \right ] ^{\dagger}c\left [ \begin{pmatrix} 0\\1 \\ 0 \\ 0 \end{pmatrix}-\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix} \right ]=1

\vert c\vert^{2}\begin{pmatrix} 0 &1 &-1 &0 \end{pmatrix}\begin{pmatrix} 0\\1 \\ -1 \\ 0 \end{pmatrix}=1\; \; \; \; \Rightarrow \; \; \; \; c=\frac{1}{\sqrt{2}}

 

Likewise, the state \vert1,0\rangle after normalisation is

\vert1,0\rangle=\frac{1}{\sqrt{2}}(\vert+z\rangle_1\otimes\vert-z\rangle_2+\vert-z\rangle_1\otimes\vert+z\rangle_2)\; \; \; \; \; \; \; \; 221

We have two more spin eigenstates of the system to consider: \vert1,+1\rangle and \vert1,-1\rangle. Since the allowed values of the total magnetic quantum number m_j are the sum of the allowed values of the two contributing magnetic quantum numbers (see eq207), \vert1,+1\rangle=\vert+z\rangle_1\otimes\vert+z\rangle_2 and \vert1,-1\rangle=\vert-z\rangle_1\otimes\vert-z\rangle_2., both of which are already normalised because they are basis states.

 

Question

Verify that \hat{{S}^{^{2}}}^{{T}}\vert1,\pm1\rangle=2\hbar^{2}\vert1,\pm1\rangle and \hat{S}_z^{\; {T}}\vert1,\pm1\rangle=\pm\hbar\vert1,\pm1\rangle.

Answer

Using eq203

\hat{{S}^{2}}^{(T)}\vert1,+1\rangle=\hbar^{2}\begin{pmatrix} 2 &0 &0 &0 \\ 0 &1 &1 &0 \\ 0& 1 & 1 & 0\\ 0& 0& 0& 2 \end{pmatrix}\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}=2\hbar^{2}\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}

Similarly, \hat{{S}^{2}}^{(T)}\vert1,-1\rangle=2\hbar^{2}\vert1,-1\rangle. Using eq197,

\hat{S}_z^{\; {T}}\vert1,+1\rangle=\hbar\begin{pmatrix} 1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 0& 0 & 0 & 0\\ 0& 0& 0& -1 \end{pmatrix}\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}=\hbar\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}

Similarly, \hat{S}_z^{\; (T)}\vert1,-1\rangle=-\hbar\vert1,-1\rangle.

 

Therefore, we have:

Triplet\: state=\left\{\begin{matrix} \vert1,+1\rangle&=\vert+z\rangle_1\otimes\vert+z\rangle_2 \\ \vert1,0\rangle &=\frac{1}{\sqrt{2}}(\vert+z\rangle_1\otimes\vert-z\rangle_2+\vert-z\rangle_1\otimes\vert+z\rangle_2)\\ \vert1,-1\rangle&=\vert-z\rangle_1\otimes\vert-z\rangle_2 \end{matrix}\right.

Singlet\: state=\vert0,0\rangle=\frac{1}{\sqrt{2}}[\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2]

or in the wave function notation:

Triplet\: state=\left\{\begin{matrix} \sigma_a&=\alpha(1)\alpha(2)\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 222 \\ \sigma_b &=\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)+\beta(1)\alpha(2)]\; \; \; \; \; \; 223\\ \sigma_c&=\beta(1)\beta(2)\; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 224 \end{matrix}\right.

Singlet\: state=\sigma_d=\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)-\beta(1)\alpha(2)]\; \; \; \; \; \; \; \; 225

The helium atom with the excited state configuration of 1s12s1 is commonly used to illustrate the singlet and triplet states. If the electrons have anti-parallel spins, the excited atom is characterised by the singlet state. If the electrons have parallel spins, the excited helium is in the triplet state.

 

 

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Composite systems – beyond spin operators

In the previous article, we showed that the total z-component spin angular momentum operator\hat{S}_z^{\; (T)}  is \hat{S}_z^{\; (T)}=\hat{S}_z^{\; (1)}\otimes I+I\otimes\hat{S}_z^{\; (2)}, which is a special case of the general form:

\hat{J}_i=\hat{M}_i^{\; (1)}\otimes I+I\otimes\hat{M}_i^{\; (2)}\; \; \; \; \; \; \; \; 205

\hat{J}_i (where i=x,y,z) is the total angular momentum component operator. \hat{M}_i^{\; (1)} and \hat{M}_i^{\; (2)} are component operators of \hat{M}^{ (1)} and \hat{M}^{ (2)} respectively. \hat{M}^{ (1)} and \hat{M}^{ (2)} are operators of two sources of angular momentum. For example, they may be:

1) the orbital angular momentum operator of particle 1 and orbital angular momentum operator of particle 2 respectively;
2) the spin angular momentum operator of particle 1 and spin angular momentum operator of particle 2 respectively;
3) the orbital angular momentum operator and spin angular momentum operator respectively of a particle;
4) the total angular momentum of particle 1 and the total angular momentum of particle 2.

Also mentioned in the previous article is that L_z=\sum_{i=1}^{n}l_{z,i} and S_z=\sum_{i=1}^{n}s_{z,i}. It is therefore easy to accept the validity of points 1) and 2). For point 3, the proposal that j_z=l_z+s_z may seem untenable. However, spin angular momentum, like orbital angular momentum, is a form of angular momentum. In fact, the total angular momentum \boldsymbol{\mathit{J}} of a system is defined as the vector sum \boldsymbol{\mathit{J}}=\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{S}}. If point 3) is valid, \hat{J}_i must satisfy the same commutation relations as described by eq99, eq100 and eq101.

 

Question

Show that \hat{J}_i satisfies the same commutation relations as described by eq99, eq100 and eq101.

Answer

\left [ \hat{J}_x,\hat{J}_y \right ]=\left [\hat{M}_x^{\; (1)}\otimes I+I\otimes\hat{M}_x^{\; (2)},\hat{M}_y^{\; (1)}\otimes I+I\otimes\hat{M}_y^{\; (2)}\right ]

Expanding the RHS of the above equation and noting that \hat{M}_i^{\; (1)} and \hat{M}_i^{\; (2)} commute because they act on different vector spaces, we have

\left [ \hat{J}_x,\hat{J}_y \right ]=\left [\hat{M}_x^{\; (1)}\otimes I,\hat{M}_y^{\; (1)}\otimes I\right ] +\left [I\otimes\hat{M}_x^{\; (2)},I\otimes\hat{M}_y^{\; (2)}\right ]

\left [ \hat{J}_x,\hat{J}_y \right ]=\left [\hat{M}_x^{\; (1)} ,\hat{M}_y^{\; (1)}\right ]\otimes I +I\otimes\left [\hat{M}_x^{\; (2)},\hat{M}_y^{\; (2)}\right ]

With reference to eq99, eq100, eq101, eq165, eq166 and eq167, \left [\hat{M}_i^{\; (l)} ,\hat{M}_j^{\; (l)}\right ]=i\hbar\epsilon_{ijk}\hat{M}_k^{\; l}, where l=1,2 and \epsilon_{ijk} is the Levi-Civita symbol. So,

\left [ \hat{J}_x,\hat{J}_y \right ]=i\hbar\hat{M}_z^{\; (1)} \otimes I +i\hbar I\otimes\hat{M}_z^{\; (2)}=i\hbar\hat{J}_z

Similarly, we have \left [ \hat{J}_y,\hat{J}_z \right ]=i\hbar\hat{J}_x and \left [ \hat{J}_z,\hat{J}_x \right ]=i\hbar\hat{J}_y.

 

Since the total angular momentum component operators satisfy the form of commutation relations as described by eq99, eq100 and eq101, the raising and lowering operators also apply to the total angular momentum operator \hat{J}. We would therefore expect

\hat{J}^2\psi=j(j+1)\hbar^2\psi\; \; \; \; and\; \; \; \; \; \hat{J}_z\psi=m_j\hbar\psi\; \; \; \; \; \; \; \; \; 205a

You’ll realise from the workings of the above Q&A that we can simplify the notation \hat{J}_z of eq205 as

\hat{J}_z=\hat{M}_{z,1}+\hat{M}_{z,2}\; \; \; \; \; \; \; \; 206

To show that \hat{J}_z commutes with \hat{M}_{z,i}, where \hat{M}_{z,i} can be either \hat{l}_{z,i} or \hat{s}_{z,i}, we have \left [ \hat{J}_z,\hat{l}_z \right ]=\left [ \hat{l}_z,\hat{l}_z \right ]+\left [ \hat{s}_z,\hat{l}_z \right ]=0 and \left [ \hat{J}_z,\hat{s}_z \right ]=\left [ \hat{l}_z,\hat{s}_z \right ]+\left [ \hat{s}_z,\hat{s}_z \right ]=0. Therefore, the eigenstate of \hat{J}_z is simultaneously the eigenstates of \hat{M}_{z,1} and \hat{M}_{z,2}. This implies that the eigenvalues of \hat{J}_z are the sum of the eigenvalues of \hat{M}_{z,1} and \hat{M}_{z,2}, i.e. m_j\hbar=m_1\hbar+m_2\hbar, or

m_j=m_1+m_2\; \; \; \; \; \; \; \; 207

In other words, the allowed values of the total magnetic quantum number m_j are the sum of the allowed values of the two contributing magnetic quantum numbers. As for the allowed values of the total angular momentum quantum number j, let’s further define the eigenvalues of \hat{M}_1^{\; 2} and \hat{M}_2^{\; 2} as M_1(M_1+1)\hbar^{2} and M_2(M_2+1)\hbar^{2} respectively. This allows us to work with the quantum numbers M_1 and M_2.

Now, the maximum value of m_j in eq207 is m_{j,max}=m_{1,max}+m_{2,max}. Since the maximum value of a magnetic quantum number is the angular momentum quantum number (i.e. m_{j,max}=j_{max} and m_{i,max}=M_{i}, where i=1,2), the highest value of j is

j_{max}=M_1+M_2\; \; \; \; \; \; \; \; 208

Furthermore, for a particular value of \boldsymbol{\mathit{j}} in the coupled representation, there are 2j+1 values of m_j and therefore 2j+1 states. So j_{max} has 2j_{max}+1=2M_1+2M_2+1 states. These states are \vert j_{max},m_{j,max}\rangle,\vert j_{max},m_{j,max}-1\rangle,\vert j_{max},m_{j,max}-2\rangle,\cdots. The states for the next lower value of j (denoted by j') are \vert j',m'_{j,max}\rangle,\vert j',m'_{j,max}-1\rangle,\vert j',m'_{j,max}-2\rangle,\cdots. The same logic applies for states all the way to the lowest value of j.

 

Question

Show that the total number of states in the uncoupled representation is (2M_1+1)(2M_2+1).

Answer

In eq193, the total number of states in the uncoupled representation \vert M_1,m_1;M_2,m_2\rangle is the number of ways to form Kronecker products of basis vectors from each vector space. Since there are 2M_1+1 basis vectors in the 1st vector space and 2M_2+1 basis vectors in the 2nd vector space,

total\: states\: in\: uncoupled\: representation=(2M_1+1)(2M_2+1)\; \; \; \; \; \; \; \; 209

 

To determine the lower values of j, we consider the lower values of m_{j,max}, the first being m_{j,max}-1=m'_{j,max}. There are two possible ways to obtain this value, with m_{1,max}-1 and m_{2,max}, or m_{1,max} and m_{2,max}-1.  Since each state is characterised by a unique value of m_j for a particular value of j, one of the two possibilities is accounted for by the state \vert j_{max},m_{j,max}-1\rangle. The remaining possibility must be due to \vert j',m'_{j,max}\rangle. Since m_{j,max}=j_{max}, we must have m'_{j,max}=j'. Furthermore, because m_{j,max}-1=m'_{j,max}, we have j'=m_{j,max}-1=j_{max}-1. The state \vert j',m'_{j,max}\rangle is therefore \vert j_{max}-1,m_{j,max}-1\rangle.

For m_{j,max}-2, there are three possible ways to obtain it. Again, one of the possible ways is accounted for by \vert j_{max},m_{j,max}-2\rangle and the second way by \vert j',m'_{j,max}-1\rangle=\vert j_{max}-1,m_{j,max}-2\rangle. The remaining possibility must be due to the state \vert j'',m''_{j,max}\rangle=\vert j'-1,m'_{j,max}-1\rangle=\vert j_{max}-2,m_{j,max}-2\rangle.

Therefore, the allowed values of  are

j=j_{max},j_{max}-1,j_{max}-2,\cdots,j_{min}

or

j=M_1+M_2,M_1+M_2-1,M_1+M_2-2,\cdots,j_{min}

To determine j_{min}, we note that the total number of states for the system can be written as \sum_{j=j_{min}}^{j_{max}}(2j+1) because there are 2j+1 states associated with each value of j. Since j_{min}\geq 0, we can further split the sum as:

total\: states=\sum_{j=0}^{j_{max}}(2j+1)-\sum_{j=0}^{j_{min}-1}(2j+1)\; \; \; \; \; \; \; \; 210

 

Question

Show that \sum_{j=0}^{n-1}(2j+1)=n^{2} and hence \sum_{j=0}^{x}(2j+1)=(x+1)^{2}.

Answer

\sum_{j=0}^{n-1}(2j+1)=1+\sum_{j=1}^{n-1}(2j+1)=1+2\sum_{j=1}^{n-1}j+\sum_{j=1}^{n-1}1\; \; \; \; \; \; \; \; 211

For the 2nd term on RHS of 2nd equality of eq211, \sum_{j=1}^{n-1}j=1+2+\cdots+(n-2)+(n-1), which if written in the reverse order becomes \sum_{j=1}^{n-1}j=(n-1)+(n-2)+\cdots+2+1. Adding the two sums, we have

2\sum_{j=1}^{n-1}j=n(n-1)\; \; \; \; \; \; \; \; 214

For the 3rd term on RHS of 2nd equality of in eq211

\sum_{j=1}^{n-1}1=\sum_{j=1}^{n-1}j^{0}=1+1+\cdots+1=n-1\; \; \; \; \; \; \; \; 215

Substitute eq214 and eq215 back in eq211, we have \sum_{j=0}^{n-1}(2j+1)=n^{2}. Let x=n-1, we have

\sum_{j=0}^{n-1}(2j+1)=(x+1)^{2}\; \; \; \; \; \; \; \; 216

 

Using eq216, where x=j_{max} for \sum_{j=0}^{j_{max}}(2j+1), and x=j_{min}-1 for \sum_{j=0}^{j_{min}-1}(2j+1), eq210 becomes

total\: states=(j_{max}+1)^{2}-j_{min}^{\; \;\; \; \; \; 2}=j_{max}^{\; \;\; \; \; \; 2}+2j_{max}+1-j_{min}^{\; \;\; \; \; \; 2}\; \; \; \; \; \; \; \; 217

The total number of states (energy levels) of a system must be independent of the chosen representation. Substituting eq209 in LHS of eq217 and eq208 in RHS of eq217 and simplifying,

j_{min}=\pm(M_1-M_2)\; \; \; \; \; \; \; \; 218

Eq218 is equivalent to j_{min}=\vert M_1-M_2\vert because j_{min}\geq 0, M_1\geq 0, M_2\geq 0, and M_2 may be a larger value than M_1. Therefore, for a given value of M_1 and a given value of M_2, the allowed values of the total angular momentum quantum number j are:

j=M_1+M_2,M_1+M_2-1,\cdots,\vert M_1-M_2\vert\; \; \; \; \; \; \; \; 219

which is called the Clebsch-Gordan series.

 

Question

Write all the eigenstates (in the form of \vert j,m_j\rangle) and basis states (in the form of \vert m_1,m_2\rangle) of a system with two sources of angular momentum, M_1=2 and M_2=1.

Answer

There are a total of 15 eigenstates and also 15 basis states. The allowed values of j are 3, 2 and 1. The eigenstates are \vert 3,3\rangle, \vert 3,2\rangle, \vert 3,1\rangle, \vert 3,0\rangle, \vert 3,-1\rangle, \vert 3,-2\rangle, \vert 3,-3\rangle, \vert 2,2\rangle, \vert 2,1\rangle, \vert 2,0\rangle, \vert 2,-1\rangle, \vert 2,-2\rangle, \vert 1,1\rangle, \vert 1,0\rangle and \vert 1,-1\rangle. The basis states are \vert 2,1\rangle, \vert 1,1\rangle, \vert 0,1\rangle, \vert -1,1\rangle, \vert -2,1\rangle, \vert 2,0\rangle, \vert 2,-1\rangle, \vert 1,0\rangle, \vert 1,-1\rangle, \vert 0,0\rangle, \vert 0,-1\rangle, \vert -1,0\rangle, \vert -1,-1\rangle, \vert -2,0\rangle and \vert -2,-1\rangle. Each spin eigenstate of the system is a linear combination of the 15 basis states.

 

What we have described so far pertains to a system with two sources of angular momentum. If the system has more than two sources of angular momentum, the Clebsch-Gordan series is applied repeatedly, i.e. a first series is written with M_1 and M_2, and then the Clebsch-Gordan procedure is again applied to each value of this series with M_3 to form a second resultant series, and the procedure is repeated until a final resultant series is developed with M_{final}. For example, a system with three sources of angular momentum, s_1=\frac{1}{2}, s_2=\frac{1}{2} and s_3=\frac{1}{2}, has the following allowed values of S:

1st series using s_1 and s_2, S=1,0

2nd and final series using 1,0 and s_3, S=\frac{3}{2},\frac{1}{2},\frac{1}{2}

For this system, there are 8 basis states, whose explicit forms can be expressed as follows:

\small \begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix}

\small \begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix}

 

Question

What are the allowed angular momenta of a system with three sources of angular momentum, \small l_1=1, \small l_2=1 and \small l_3=1, and how many basis states are there in total?

Answer

\small j=3,2,2,1,1,1,0. The total number of basis states is 27.

 

 

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Composite systems (quantum mechanics)

A composite system is one that has more than one part; for instance, a system of two spin-\frac{1}{2} particles. One possible way to denote a basis set for the eigenstates of the two spin-\frac{1}{2} particles system is:

\uparrow\uparrow\; \; \; \; \uparrow\downarrow\; \; \; \; \downarrow\uparrow\; \; \; \; \downarrow\downarrow

or in terms of the z-component of the spins:

\vert +z,+z\rangle\; \; \; \;\vert +z,-z\rangle\; \; \; \;\vert -z,+z\rangle\; \; \; \;\vert -z,-z\rangle

where the 1st term and 2nd term in the each basis state refer to the z-component spin states of the 1st particle and that of the 2nd particle, respectively.

The full notation of \vert +z,+z\rangle is \vert s=\frac{1}{2},m_s=\frac{1}{2};s=\frac{1}{2},m_s=\frac{1}{2}\rangle, which can be condensed to \vert m_s=\frac{1}{2},m_s=\frac{1}{2}\rangle, or simply \vert +z,+z\rangle.

To represent the 4 kets in column vector form, we borrow the notation of basis vectors that are used to derive the Pauli matrices and give the kets the following assignments:

\vert +z,+z\rangle=\begin{pmatrix} 1\\0 \\0 \\ 0 \end{pmatrix} \; \; \; \;\vert +z,-z\rangle=\begin{pmatrix} 0\\1 \\0 \\ 0 \end{pmatrix}

\vert -z,+z\rangle=\begin{pmatrix} 0\\0 \\1 \\ 0 \end{pmatrix} \; \; \; \;\vert -z,-z\rangle=\begin{pmatrix} 0\\0 \\0 \\ 1 \end{pmatrix} \; \; \; \; \; \; \; \; 193

If you’ve read the article on Kronecker product, you’ll realise that

\begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix}=\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix}=\begin{pmatrix} 0\\1 \\ 0 \\ 0 \end{pmatrix}

\; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix}=\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix}=\begin{pmatrix} 0\\0 \\ 0 \\ 1 \end{pmatrix}

Therefore,

\vert\pm z,\pm z\rangle=\vert\pm z\rangle_1\otimes\vert\pm z\rangle_2\; \; \; \; \; \; \; \; 194

Eq194 is called the uncoupled representation of the state of the system. The general state of the system, which is called the coupled representation, is a linear combination of the four basis states:

\small \vert\psi\rangle=c_1\vert +z\rangle_1\otimes\vert +z\rangle_2+c_2\vert +z\rangle_1\otimes\vert -z\rangle_2+c_3\vert -z\rangle_1\otimes\vert +z\rangle_2+c_4\vert -z\rangle_1\otimes\vert -z\rangle_2\; \; \; \; \; \; 195

where \small c_1, \small c_2, \small c_3 and \small c_4 are constants called Clebsch-Gordan coefficients.

Eq195 is often written in the general form:

\small \vert J,M_J\rangle=\sum_{m_{j1}+m_{j2}=M_J}C_{j_1,m_{j1};j_2,m_{j2}}^{JM_J}\vert j_1,m_{j1};j_2,m_{j2}\rangle

 

Question

Why is eq194 the uncoupled representation, while eq195 is the coupled representation? Why is \small m_{j1}+m_{j2}=M_J?

Answer

The \small z-component of the spins in the basis vectors \small \vert\pm z\rangle_1 and \small \vert\pm z\rangle_2 that form the state \small \vert\pm z,\pm z\rangle in eq194 are assumed to behave independently (non-interacting), whereas the state \small \vert\psi\rangle in eq195 takes into account the coupled interactions of the spins. For example, the coupled representation is used for the singlet and triplet states (interacting spins), while both the coupled and uncoupled representations are utilised in deriving atomic term symbols where spin-orbit coupling is neglected.

In terms of \small m_{j1}+m_{j2}=M_J, see the derivation of eq207 for explanation.

 

Consider the Hilbert spaces \small H_1 and \small H_2 that are spanned by the basis states \small \begin{pmatrix} 1\\0 \end{pmatrix}_1,\begin{pmatrix} 0\\1 \end{pmatrix}_1 and \small \begin{pmatrix} 1\\0 \end{pmatrix}_2,\begin{pmatrix} 0\\1 \end{pmatrix}_2 respectively. The Hilbert space of the composite system of the two spin-\small \frac{1}{2} particles is the Kronecker product of \small H_1 and \small H_2, i.e. \small H_1\otimes H_2. To construct a total spin angular momentum operator for \small H_1\otimes H_2, we have to consider the following:-

  1. According to the principle of the conservation of angular momentum of a system, the total \small z-component of the orbital angular momentum of a system is the sum of all \small z-components of the orbital angular momentum of particles constituting the system \small L_z=\sum_{i=1}^{n}l_{z,i}. The total \small z-component of spin angular momentum, which is also a form of angular momentum, is postulated to be the sum of all \small z-components of the spin angular momentum of particles constituting the system \small S_z=\sum_{i=1}^{n}s_{z,i}.
  2. The spin operator \small \hat{S}_z^{\; (1)} that acts on eigenstates in the Hilbert space \small H_1, and the spin operator \small \hat{S}_z^{\; (2)} that acts on eigenstates in the Hilbert space \small H_2, are 2×2 matrices.
  3. The eigenvectors in \small H_1\otimes H_2 are column vectors with 4 elements, and hence, the spin operator must be a 4×4 matrix.

Taking the above points in consideration, \small \hat{S}_z^{\; (1)} and \small \hat{S}_z^{\; (2)} acting on \small \vert\pm z\rangle_1\otimes\vert\pm z\rangle_2 in the Hilbert space \small H_1\otimes H_2 are defined as:

\small \left ( \hat{S}_z^{\; (1)}\otimes I\right ) \left ( \vert\pm z\rangle_1\otimes\vert\pm z\rangle_2\right )=\hat{S}_z^{\; (1)} \vert\pm z\rangle_1\otimes\vert\pm z\rangle_2

and

\small \left (I\otimes\hat{S}_z^{\; (2)}\right ) \left ( \vert\pm z\rangle_1\otimes\vert\pm z\rangle_2\right )=\vert\pm z\rangle_1\otimes\hat{S}_z^{\; (2)} \vert\pm z\rangle_2

respectively, where \small I is the 2×2 identity matrix.

The total \small z-component spin angular momentum operator \small \hat{S}_z^{\; (T)} that acts on the eigenstates in the Hilbert space \small H_1\otimes H_2 is therefore:

\small \hat{S}_z^{\; (T)}=\hat{S}_z^{\; (1)}\otimes \hat{S}_z^{\; (2)}\; \; \; \; \; \; \; \; 196

Substituting eq174 into the above equation,

\hat{S}_z^{\; (T)}=\frac{\hbar}{2}\begin{pmatrix} 1 &0 &0 &0 \\ 0 &1 &0 &0 \\ 0 &0 &-1 &0 \\ 0 &0 &0 &-1 \end{pmatrix}+\frac{\hbar}{2}\begin{pmatrix} 1 &0 &0 &0 \\ 0 &-1 &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &-1 \end{pmatrix}=\hbar\begin{pmatrix} 1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 0 &0 &0 &0 \\ 0 &0 &0 &-1 \end{pmatrix}\; \; \; \; \; \; \; \; 197

Similarly, from eq177

\hat{S}_x^{\; (T)}=\frac{\hbar}{2}\begin{pmatrix} 0 &0 &1 &0 \\ 0 &0 &0 &1 \\ 1 &0 &0 &0 \\ 0 &1 &0 &0 \end{pmatrix}+\frac{\hbar}{2}\begin{pmatrix} 0 &1 &0 &0 \\ 1 &0 &0 &0 \\ 0 &0 &0 &1 \\ 0 &0 &1 &0 \end{pmatrix}=\frac{\hbar}{2}\begin{pmatrix} 0 &1 &1 &0 \\ 1 &0 &0 &1 \\ 1 &0 &0 &1 \\ 0 &1 &1 &0 \end{pmatrix}\; \; \; \; \; \; \; \; 199

and from eq178

\hat{S}_y^{\; (T)}=\frac{\hbar}{2}\begin{pmatrix} 0 &0 &-i &0 \\ 0 &0 &0 &-i \\ i &0 &0 &0 \\ 0 &i &0 &0 \end{pmatrix}+\frac{\hbar}{2}\begin{pmatrix} 0 &-i &0 &0 \\ i &0 &0 &0 \\ 0 &0 &0 &-i \\ 0 &0 &i &0 \end{pmatrix}=\frac{\hbar}{2}\begin{pmatrix} 0 &-i &-i &0 \\ i&0 &0 &-i \\ i &0 &0 &-i \\ 0 &i &i &0 \end{pmatrix}\; \; \; \; \; \; \; \; 200

Next, we shall derive the matrix for \hat{{S}^{2}}^{(T)}. With regard to eq196, since s_{z,max}=S, \hat{S}_{z,max}^{\;\; \; \; \; \; \; \; (1)}\otimes I+I\otimes\hat{S}_{z,max}^{\;\; \; \; \; \; \; \; (2)} is equivalent to \hat{S}^{(T)}=\hat{S}^{(1)}\otimes I+I\otimes\hat{S}^{(2)}. Therefore,

\hat{{S}^{2}}^{(T)}=\left ( \hat{S}^{(1)}\otimes I+I\otimes\hat{S}^{(2)} \right )^{2}=\hat{{S}^{(1)}}^{2}\otimes I+2\hat{S}^{(1)}\otimes\hat{S}^{(2)}+I\otimes\hat{{S}^{(2)}}^{2}\; \; \; \; \; \; \; \; 201

Substitute eq173 and eq179 in eq201,

\hat{{S}^{2}}^{(T)}=\frac{3}{2}\hbar^{2}\begin{pmatrix} 1 &0 &0 &0 \\ 0 &1 &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1 \end{pmatrix}+\frac{\hbar^{2}}{2} \left ( \hat{\sigma}_x\otimes\hat{\sigma}_x+\hat{\sigma}_y\otimes\hat{\sigma}_y+\hat{\sigma}_z\otimes\hat{\sigma}_z \right ) \; \; \; \; \; \; \; \; 202

Substitute eq180 in the above equation and simplifying,

\hat{{S}^{2}}^{(T)}=\hbar^{2}\begin{pmatrix} 2 &0 &0 &0 \\ 0 &1 &1 &0 \\ 0 &1 &1 &0 \\ 0 &0 &0 &2 \end{pmatrix}\; \; \; \; \; \; \; \; 203

 

Question

What is the formula for \hat{S}_z^{\; T} that acts on the eigenstates in the Hilbert space H_1\otimes H_2\otimes H_3?

Answer

\hat{S}_z^{\; T}=\hat{S}_z^{\; 1}\otimes I\otimes I+I\otimes\hat{S}_z^{\; 2}\otimes I+I\otimes I\otimes\hat{S}_z^{\; 3}

 

Question

Show that \hat{{S}^{2}}^{(T)} commutes with \hat{S}_z^{\; (T)}.

Answer

Using eq196 and eq201 in simple notation, \left [\hat{{S}^{2}}^{(T)},\hat{S}_z^{\; (T)}\right ]=\left [\hat{S}_1^{\; 2}+2\hat{S}_1\cdot\hat{S}_2+\hat{S}_2^{\; 2},\hat{S}_{1z}+\hat{S}_{2z}\right ]. Expanding RHS of this equation and considering the following:

  1. Use eq179 to find expressions for \hat{S}_1^{\; 2}, \hat{S}_2^{\; 2} and \hat{S}_1\cdot\hat{S}_2.
  2. \hat{S}_{1i} and \hat{S}_{2i}, where i=x,y,z, act on different vector spaces and hence they commute with each other.
  3. [\hat{A}\hat{B},\hat{C}]=[\hat{A},\hat{C}]\hat{B}+\hat{A}[\hat{B},\hat{C}] and [\hat{A},\hat{B}]=-[\hat{B},\hat{A}].
  4.  Making use of eq165, eq166 and eq167.

We have

\left [ \hat{{S}^{2}}^{(T)},\hat{S}_z^{\; (T)} \right ]=0\; \; \; \; \; \; \; \; 204

 

 

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Sequential Stern-Gerlach experiments

A sequential Stern-Gerlach experiment involves passing a spin-\frac{1}{2} particle through multiple inhomogeneous magnetic fields, each of a certain orientation. In the first example, a beam of spin-\frac{1}{2} particles, e.g. silver atoms, undergoes a first measurement as it passes through an inhomogeneous magnetic field, which is parallel to the z-axis (see diagram below).

The beam is split into two, and the S_z=\frac{\hbar}{2} beam is allowed to travel through another inhomogeneous magnetic field, which is also z-directional (the S_z=-\frac{\hbar}{2} beam is blocked). The general state \chi of the valence silver electron prior to passing through the first magnetic field is given by eq171:

\chi=c_1\alpha+c_2\beta

where \alpha and \beta are basis vectors representing the electron spin eigenstates of \vert s=1/2,m_s=1/2\rangle and  \vert s=1/2,m_s=-1/2\rangle respectively; \vert c_1\vert^{2} is the probability of finding the electron with the state \vert s=1/2,m_s=1/2\rangle, and \vert c_2\vert^{2} is the probability of finding the electron with the state \vert s=1/2,m_s=-1/2\rangle, with \vert c_1\vert^{2}+\vert c_2\vert^{2}=1.

Since the direction of the first magnetic field is arbitrary assigned and the beam of silver atoms emerging from the source is not polarised, we have equal probability of silver atoms with valence electrons in each eigenstate emerging, i.e. \vert c_1\vert^{2}=\vert c_2\vert^{2}=0.5. Noting the orthonormality of basis vectors, the expectation value of S_z with reference to eq168 is:

\langle S_z\rangle=\langle\chi\vert\hat{S}_z\vert\chi\rangle=\langle\sqrt{0.5}\alpha+\sqrt{0.5}\beta\vert\sqrt{0.5} \frac{\hbar}{2}\alpha+\sqrt{0.5} \left ( -\frac{\hbar}{2}\right )\beta\rangle=0

The state \chi of the valence silver electron entering the second magnetic field is \chi=c_1\alpha+c_2\beta, where \vert c_1\vert^{2}=1 and \vert c_2\vert^{2}=0. Therefore, the expectation value of S_z is \langle S_z\rangle=\frac{\hbar}{2}.

For the second example, the S_z=\frac{\hbar}{2} beam emerging from the first magnetic field is passed through a second magnetic field that is rotated 90o, i.e. in the x-direction (see diagram below).

To determine the result of the second measurement, we must express the state of the valence electron in the S_z=\frac{\hbar}{2} beam in terms of \vert\alpha_x\rangle and \vert\beta_x\rangle, or in general where the second magnetic field is rotated parallel to an arbitrary direction \hat{\boldsymbol{\mathit{r}}} (\hat{\boldsymbol{\mathit{r}}} is the unit vector in spherical coordinates, i.e. \hat{\boldsymbol{\mathit{r}}}=sin\theta cos\phi\boldsymbol{\mathit{i}}+sin\theta sin\phi\boldsymbol{\mathit{j}}+cos\theta \boldsymbol{\mathit{k}}), in terms of \vert \alpha\rangle_{\hat{\boldsymbol{\mathit{r}}}} and \vert \beta\rangle_{\hat{\boldsymbol{\mathit{r}}}}.

 

Question

How do we derive the unit vector \hat{\boldsymbol{\mathit{r}}} in spherical coordinates?

Answer

Substitute eq77 in the definition of a unit vector,

\hat{\boldsymbol{\mathit{r}}}=\frac{\boldsymbol{\mathit{r}}}{\left |\boldsymbol{\mathit{r}}\right |}=\frac{x\boldsymbol{\mathit{i}}+y\boldsymbol{\mathit{j}}+z\boldsymbol{\mathit{k}}}{r}=sin\theta cos\phi\boldsymbol{\mathit{i}}+sin\theta sin\phi\boldsymbol{\mathit{j}}+cos\theta\boldsymbol{\mathit{k}}\; \; \; \; \; \; \; \; 182b

 

In other words, we need eq171 to be in the form:

\chi_\alpha=c_1\alpha_{\hat{\boldsymbol{\mathit{r}}}}+c_2\beta_{\hat{\boldsymbol{\mathit{r}}}}\; \; \; \; \; \; \; \; 183

This implies that we have to construct the operator \hat{S}_{\hat{\boldsymbol{\mathit{r}}}}, which acts on the component of spin angular momentum along \hat{\boldsymbol{\mathit{r}}}. To do so, we take the projection of \boldsymbol{\mathit{S}} onto \hat{\boldsymbol{\mathit{r}}}:

\boldsymbol{\mathit{S}}\cdot\hat{\boldsymbol{\mathit{r}}}=(\boldsymbol{\mathit{i}}S_x+\boldsymbol{\mathit{j}}S_y+\boldsymbol{\mathit{k}}S_z)\cdot(sin\theta cos\phi\boldsymbol{\mathit{i}}+\sin\theta sin\phi\boldsymbol{\mathit{j}}+cos\theta\boldsymbol{\mathit{k}})

\boldsymbol{\mathit{S}}\cdot\hat{\boldsymbol{\mathit{r}}}=S_xsin\theta cos\phi+S_y\sin\theta sin\phi+S_zcos\theta

Hence, the operator is

\hat{S}_{\hat{\boldsymbol{\mathit{r}}}}=\hat{S}_xsin\theta cos\phi+\hat{S}_y\sin\theta sin\phi+\hat{S}_zcos\theta

Substituting eq174, eq177 and eq178 in the above equation to further construct \hat{S}_{\hat{\boldsymbol{\mathit{r}}}} in matrix form,

\hat{S}_{\hat{\boldsymbol{\mathit{r}}}}=\frac{\hbar}{2}\begin{pmatrix} cos\theta &e^{-i\phi}sin\theta \\ e^{i\phi}sin\theta &-cos\theta \end{pmatrix}\; \; \; \; \; \; \; \; 184

The eigenvalue equation is:

\frac{\hbar}{2}\begin{pmatrix} cos\theta &e^{-i\phi}sin\theta \\ e^{i\phi}sin\theta &-cos\theta \end{pmatrix}\chi_\alpha=\begin{pmatrix} \lambda &0 \\ 0 &\lambda \end{pmatrix}\chi_\alpha\; \; \; \; \; \; \; \; 185

where \lambda is the eigenvalue; and the corresponding characteristic equation is:

\frac{\hbar}{2}\begin{vmatrix} cos\theta-\lambda &e^{-i\phi}sin\theta \\ e^{i\phi}sin\theta &-cos\theta-\lambda \end{vmatrix}=0

\lambda=\pm\frac{\hbar}{2}\; \; \; \; \; \; \; \; 186

Substitute eq183 and eq186 in eq185,

\frac{\hbar}{2}\begin{pmatrix} cos\theta &e^{-i\phi}sin\theta \\ e^{i\phi}sin\theta &-cos\theta \end{pmatrix}\begin{pmatrix} c_1\\c_2 \end{pmatrix}=\pm\frac{\hbar}{2}\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}\begin{pmatrix} c_1\\c_2 \end{pmatrix}

\begin{pmatrix} c_1cos\theta +c_2e^{-i\phi}sin\theta \\ c_1e^{i\phi}sin\theta -c_2cos\theta \end{pmatrix}=\pm\begin{pmatrix} c_1\\c_2 \end{pmatrix}\; \; \; \; \; \; \; \; 187

So,

c_1cos\theta +c_2e^{-i\phi}sin\theta =\pm c_1\; \; \; \; \; \Rightarrow \; \; \; \; \; c_2=c_1\frac{\pm1-cos\theta}{e^{-i\phi}sin\theta}

Since 1-cos\theta=2sin^{2}\frac{\theta}{2} and sin\theta=2sin\frac{\theta}{2}cos\frac{\theta}{2},  we have

c_2=c_1e^{i\phi}\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}

Substituting the above equation in \vert c_1\vert^{2}+\vert c_2\vert^{2}=1 and using \vert e^{i\phi}\vert ^{2}=1

\vert c_1\vert^{2}=cos^{2}\frac{\theta}{2}\; \; \; \; \; \; \; \; 188

 

Question

Show that \vert e^{i\phi}\vert^{2}=1.

Answer

If a,b\in \mathbb{R}, we have \vert a+ib\vert=\sqrt{a^{2}+b^{2}}.

\vert e^{i\phi}\vert^{2}=\vert cos\phi+isin\phi\vert^{2}=1

 

Substitute eq188 in \vert c_1\vert^{2} +\vert c_2\vert^{2}=1

\vert c_2\vert^{2}=sin^{2}\frac{\theta}{2}\; \; \; \; \; \; \; \; 189

Therefore, eq183 becomes:

\chi_\alpha=cos\frac{\theta}{2}\alpha_{\hat{\boldsymbol{\mathit{r}}}}+sin\frac{\theta}{2}\beta_{\hat{\boldsymbol{\mathit{r}}}}\; \; \; \; \; \; \; \; 190

Note that we could have equally used c_1e^{i\phi}sin\theta-c_2cos\theta=\pm c_2 in eq187 to arrive at eq188 and eq189. Using eq188, the probability of measuring the \alpha_{\hat{\boldsymbol{\mathit{r}}}=90^{\circ} } state is

\vert c_1\vert^{2}=cos^{2}\frac{90^{\circ}}{2}=0.5\; \; \; \; \; \; \; \; 191

Using eq189, the probability of measuring the \beta_{\hat{\boldsymbol{\mathit{r}}}=90^{\circ} } state is

\vert c_2\vert^{2}=sin^{2}\frac{90^{\circ}}{2}=0.5\; \; \; \; \; \; \; \; 192

Therefore, we observe that the S_z=\frac{\hbar}{2} beam is split equally into two beams by the second magnetic field. Using the same logic, the S_x=\frac{\hbar}{2} beam will again split into two beams by the third magnetic field:

\vert c_1\vert^{2}=cos^{2}\frac{270^{\circ}}{2}=0.5\; \; \; \; \; \; and\; \; \; \;\; \; \vert c_2\vert^{2}=sin^{2}\frac{270^{\circ}}{2}=0.5

In general, the probability of observing the \alpha_{\hat{\boldsymbol{\mathit{r}}}} state is depicted in the graph below.

 

Question

Do eq188 and eq189 apply to the output of the  beam passing through the second magnetic field in the first example?

Answer

Yes. In the first example, \vert c_1\vert^{2}=cos^{2}\frac{0^{\circ}}{2}=1 and \vert c_2\vert^{2}=sin^{2}\frac{0^{\circ}}{2}=0.

 

 

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Pauli matrices

Pauli matrices are matrix representations of the spin operators \small \hat{S}_x, \small \hat{S}_y and \small \hat{S}_z.

To derive the Pauli matrices, we let \small \alpha and \small \beta be basis vectors representing the electron spin eigenstates of \small \vert s=1/2,m_s=1/2\rangle and \small \vert s=1/2,m_s=-1/2\rangle respectively, with the following assignments:

\small \alpha=\begin{pmatrix} 1\\0 \end{pmatrix}\; \; \; \; \; \; \beta=\begin{pmatrix} 0\\1 \end{pmatrix}

The general state of an electron can then be written as a linear combination of the two spin states:

\small \chi=c_1\alpha+c_2\beta\; \; \; \; \; \; \; \; 171

where \small \vert c_1\vert^{2} is the probability of finding the electron with the state \small \vert s=1/2,m_s=1/2\rangle, and \small \vert c_2\vert^{2} is the probability of finding the electron with the state \small \vert s=1/2,m_s=-1/2\rangle, with

\small \vert c_1\vert^{2}+\vert c_2\vert^{2}=1\; \; \; \; \; \; \; \; 172

Having defined the two eigenstates, we can work out the corresponding matrix representation of \small \hat{S}^{2} using eq169 by letting \small \hat{S}^{2}=\begin{pmatrix} a &b \\ c &d \end{pmatrix}:

\small \begin{pmatrix} a &b \\ c &d \end{pmatrix}\begin{pmatrix} 1\\0 \end{pmatrix}=\frac{3}{4}\hbar^{2}\begin{pmatrix} 1\\0 \end{pmatrix}\; \; \; \; \; \Rightarrow \; \; \; \; \; \begin{pmatrix} a\\c \end{pmatrix}=\begin{pmatrix} \frac{3}{4}\hbar^{2}\\ 0 \end{pmatrix}

\small \begin{pmatrix} a &b \\ c &d \end{pmatrix}\begin{pmatrix} 0\\1 \end{pmatrix}=\frac{3}{4}\hbar^{2}\begin{pmatrix} 0\\1 \end{pmatrix}\; \; \; \; \; \Rightarrow \; \; \; \; \; \begin{pmatrix} b\\d \end{pmatrix}=\begin{pmatrix} 0\\ \frac{3}{4}\hbar^{2} \end{pmatrix}

So,

\small \hat{S}^{2}=\frac{3}{4}\hbar^{2}\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}\; \; \; \; \; \; \; \; 173

Similarly, using eq168 for \small \hat{S}_z, we have

\small \hat{S}_z=\frac{\hbar}{2}\begin{pmatrix} 1 &0 \\ 0 &-1 \end{pmatrix}\; \; \; \; \; \; \; \; 174

For \small \hat{S}_x and \small \hat{S}_y, we make use of eq170 where

\small \hat{S}_+\begin{pmatrix} 0\\1 \end{pmatrix}=\hbar\sqrt{\frac{3}{4}+\frac{1}{2}\left ( -\frac{1}{2}+1 \right )}\begin{pmatrix} 1\\0 \end{pmatrix}=\hbar\begin{pmatrix} 1\\0 \end{pmatrix}\; \; \; \; \; \; \Rightarrow \; \; \; \; \; \hat{S}_+=\hbar\begin{pmatrix} 0 &1 \\ 0& 0 \end{pmatrix}\; \; \; \; \; \; \; \; 175

\small \hat{S}_-\begin{pmatrix} 1\\0 \end{pmatrix}=\hbar\sqrt{\frac{3}{4}-\frac{1}{2}\left (\frac{1}{2}-1 \right )}\begin{pmatrix} 0\\1 \end{pmatrix}=\hbar\begin{pmatrix} 0\\1 \end{pmatrix}\; \; \; \; \; \; \Rightarrow \; \; \; \; \; \hat{S}_-=\hbar\begin{pmatrix} 0 &0 \\ 1& 0 \end{pmatrix}\; \; \; \; \; \; \; \; 176

Since \small \hat{S}_\pm=\hat{S}_x\pm i\hat{S}_y, we have \small \hat{S}_x=\hat{S}_+-i\hat{S}_y=\hat{S}_+-(\hat{S}_x-\hat{S}_-) or \small \hat{S}_x=\frac{\hat{S}_++\hat{S}_-}{2}. Similarly, \small \hat{S}_y=\frac{\hat{S}_+-\hat{S}_-}{2i}. Using eq175 and eq176,

\small \hat{S}_x=\frac{\hbar}{2}\begin{pmatrix} 0 &1 \\ 1 & 0 \end{pmatrix}\; \; \; \; \; \; \; \; 177

\small \hat{S}_y=\frac{\hbar}{2}\begin{pmatrix} 0 &-i \\ i & 0 \end{pmatrix}\; \; \; \; \; \; \; \; 178

In an earlier article, when we constructed quantum orbital angular momentum component operators \small \hat{L}_x, \small \hat{L}_y and \small \hat{L}_z, we replaced the position and linear momentum components of the classical angular momentum components \small L_x, \small L_y and \small L_z with their corresponding operators. We also suggested that we could have constructed an angular momentum operator using \small \boldsymbol{\mathit{L}}=\boldsymbol{\mathit{i}}L_x+\boldsymbol{\mathit{j}}L_y+\boldsymbol{\mathit{k}}L_z, such that \small \hat{\boldsymbol{\mathit{L}}}=\boldsymbol{\mathit{i}}\hat{L}_x+\boldsymbol{\mathit{j}}\hat{L}_y+\boldsymbol{\mathit{k}}\hat{L}_z, but for certain reasons chose to construct \small \hat{L}^{2} instead. For electron spin, it is useful to construct the spin operator using that suggestion:

\small \hat{\boldsymbol{\mathit{S}}}=\boldsymbol{\mathit{i}}\hat{S}_x+\boldsymbol{\mathit{j}}\hat{S}_y+\boldsymbol{\mathit{k}}\hat{S}_z=\frac{\hbar}{2}\hat{\boldsymbol{\mathit{\sigma}}}\; \; \; \; \; \; \; \; 179

where \small \hat{\boldsymbol{\mathit{\sigma}}}=\boldsymbol{\mathit{i}}\hat{\sigma}_x+\boldsymbol{\mathit{j}}\hat{\sigma}_y+\boldsymbol{\mathit{k}}\hat{\sigma}_z, with

\small \hat{\sigma}_x=\begin{pmatrix} 0 &1 \\ 1 &0 \end{pmatrix},\; \; \; \; \;\hat{\sigma}_y=\begin{pmatrix} 0 &-i \\ i &0 \end{pmatrix},\; \; \; \; \;\hat{\sigma}_z=\begin{pmatrix} 1 &0 \\ 0 &-1 \end{pmatrix}\; \; \; \; \; \; \; \; 180

\small \hat{\sigma}_x, \small \hat{\sigma}_y and \small \hat{\sigma}_z are called Pauli matrices. They represent observables and are Hermitian (the complex conjugate of each matrix returns the same matrix). Eq179 and eq180 are used to analyse the results of successive Stern-Gerlach experiments.

 

Question

Show that \small [\hat{L}^{2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}]=0 and \small [\hat{S}^{2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}]=0 but \small \left [ \hat{{L}^{2}}^{T},\hat{\boldsymbol{\mathit{L}}_1}\cdot\hat{\boldsymbol{\mathit{S}}_1}+\hat{\boldsymbol{\mathit{L}}_2}\cdot\hat{\boldsymbol{\mathit{S}}_2}\right ]\neq 0.

Answer

For \small \left [ \hat{L}^{2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}\right ], substitute in \small \hat{\boldsymbol{\mathit{L}}}=\boldsymbol{\mathit{i}}\hat{L}_x+\boldsymbol{\mathit{j}}\hat{L}_y+\boldsymbol{\mathit{k}}\hat{L}_z and \small \hat{\boldsymbol{\mathit{S}}}=\boldsymbol{\mathit{i}}\hat{S}_x+\boldsymbol{\mathit{j}}\hat{S}_y+\boldsymbol{\mathit{k}}\hat{S}_z and expand the expression.  From eq102, \small \hat{L}^{2} commutes with all components of \small \hat{\boldsymbol{\mathit{L}}}; and also all components of \small \hat{\boldsymbol{\mathit{S}}} because orbital angular momentum operators and spin angular momentum operators act on different vector spaces. Thus, \small \left [ \hat{L}^{2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}\right ]=0. Similarly, from eq165, eq166 and eq167, \small \hat{S}^{2} commutes with all components of \small \hat{\boldsymbol{\mathit{L}}} and \small \hat{\boldsymbol{\mathit{S}}} and so \small \left [ \hat{S}^{2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}\right ]=0.

Now,

\small \hat{{L}^{2}}^{T}=\hat{\boldsymbol{\mathit{L}}}^{T}\cdot\hat{\boldsymbol{\mathit{L}}}^{T}=\left (\hat{\boldsymbol{\mathit{L}}_1}+\hat{\boldsymbol{\mathit{L}}_2}\right )\cdot\left (\hat{\boldsymbol{\mathit{L}}_1}+\hat{\boldsymbol{\mathit{L}}_2}\right )=\hat{L}_{1}^{\: 2} +\hat{L}_{2}^{\: 2}+2\left (\hat{\boldsymbol{\mathit{L}}_1}\cdot\hat{\boldsymbol{\mathit{L}}_2}\right ) \; \; \; \; \; \; \; \; 181

Substituting eq181 in \small \left [{\hat{L^{2}}^{T}},\hat{\boldsymbol{\mathit{L}}_1}\cdot\hat{\boldsymbol{\mathit{S}}_1}+\hat{\boldsymbol{\mathit{L}}_2}\cdot\hat{\boldsymbol{\mathit{S}}_2}\right ]\neq 0 and expanding the expression, we find, after some algebra, that

\small \left [{\hat{L^{2}}^{T}},\hat{\boldsymbol{\mathit{L}}_1}\cdot\hat{\boldsymbol{\mathit{S}}_1}+\hat{\boldsymbol{\mathit{L}}_2}\cdot\hat{\boldsymbol{\mathit{S}}_2}\right ]\neq 0\; \; \; \; \; \; \; \; 182

Similarly,

\left [ \hat{{S}^2}^T,\hat{\boldsymbol{\mathit{L}}}_1\cdot\hat{\boldsymbol{\mathit{S}}}_1+\hat{\boldsymbol{\mathit{L}}}_2\cdot\hat{\boldsymbol{\mathit{S}}}_2\right ]\neq0\; \; \; \; \; \; \; \; 182a

 

 

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Spin angular momentum

The electron, as shown in the Stern-Gerlach experiment, has a form of angular momentum called spin angular momentum, in addition to its orbital angular momentum.

The spin angular momentum of an electron is referred to as an intrinsic angular momentum because it is similar in certain ways to classical spin (e.g. it is associated with a magnetic dipole moment) but is purely a quantum mechanical effect. It is erroneous to view the electron as spinning about its own axis.

The theory of quantum mechanics postulates that spin angular momentum operators \small \hat{S}^{2},\hat{S}_x,\hat{S}_y,\hat{S}_z are linear and Hermitian, and obey the same commutation relations as described by eq99, eq100 and eq101, i.e.:

\small \left [\hat{S}_x,\hat{S}_y\right ]=i\hbar\hat{S}_z\; \; \; \; \; \; \; \; 165

\small \left [\hat{S}_y,\hat{S}_z\right ]=i\hbar\hat{S}_x\; \; \; \; \; \; \; \; 166

\small \left [\hat{S}_z,\hat{S}_x\right ]=i\hbar\hat{S}_y\; \; \; \; \; \; \; \; 167

Therefore, the corresponding equations of eq132, eq133 and eq148 are:

\small \hat{S}_z\vert s,m_s\rangle=m_s\hbar\vert s,m_s\rangle\; \; \; \; \; \; \; \; 168

\small \hat{S}^{2}\vert s,m_s\rangle=s(s+1)\hbar^{2}\vert s,m_s\rangle\; \; \; \; \; \; \; \; 169

\small \hat{S}_\pm\vert s,m_s\rangle=\hbar\sqrt{s(s+1)-m_s(m_s\pm1)}\vert s,m_s\pm1\rangle\; \; \; \; \; \; \; \; 170

respectively, where \small s=0,\frac{1}{2},1,\frac{3}{2},\cdots is the spin angular momentum quantum number; \small m_s=-s,-s+1,\cdots,s-1,s is the spin magnetic quantum number; \small \hat{S}_\pm=\hat{S}_x\pm i\hat{S}_y.

For an electron, \small s has been determined experimentally not to have a range of values but to be specifically equal to \small \frac{1}{2}. We say that electrons have spin \small \frac{1}{2}. Similarly, both protons and neutrons have spin \small \frac{1}{2}.

 

 

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Gyromagnetic ratio of the electron

The gyromagnetic ratio of the electron \small \gamma_e can be evaluated by the following experiment:

In the above diagram, a sample of hydrogen atoms is placed in a uniform magnetic field (1 T) and irradiated with microwaves at different frequencies. When the electromagnetic source is turned off, no absorption is detected. However, as the electromagnetic source radiation is varied in the microwave range, an absorption is observed at \small v=2.8\times 10^{10}Hz, indicating an electronic transition between two different energy states (transition frequency for proton is in the \small 10^{6}Hz range).

Since the electron in a hydrogen atom is in the 1s orbital (\small l=0), the atom’s angular momentum is attributed to its electron spin angular momentum (the magnetic dipole moment of the nucleus is relatively weak). From eq67, the classical relation between the energy of a charged particle \small U in a magnetic field \small \boldsymbol{\mathit{B}} and the particle’s angular momentum \small \boldsymbol{\mathit{L}} is \small U=-\gamma\boldsymbol{\mathit{B}}\cdot\boldsymbol{\mathit{L}}, whose spin analogue is:

\small U=-\gamma_e\boldsymbol{\mathit{B}}\cdot\boldsymbol{\mathit{S}}=-\gamma_e[B_x(x,y,z)\boldsymbol{\mathit{i}}+B_y(x,y,z)\boldsymbol{\mathit{j}}+B_z(x,y,z)\boldsymbol{\mathit{k}}]\cdot(S_x\boldsymbol{\mathit{i}}+S_y\boldsymbol{\mathit{j}}+S_z\boldsymbol{\mathit{k}})

Analysing the effect of the uniform magnetic field (with magnitude \small B_0) on the energy states of hydrogen in the \small z-direction, the above equation becomes:

\small U=-\gamma_eB_0S_z\; \; \; \; \; \; \; \; 163

From eq75, each of the eigenvalues of \small \hat{L}^{2} is the square of the magnitude of the orbital angular momentum of an electron, which makes each of the eigenvalues of \small \hat{L}_z the \small z-component of the magnitude of the orbital angular momentum of an electron. Since \small \hat{S}_z is the analogue of \small \hat{L}_z, we postulate that the eigenvalues of \small \hat{S}_z is the \small z-component of the magnitude of the spin angular momentum of an electron, which from eq168, is \small m_s\hbar=\pm\frac{\hbar}{2}. Therefore, we have

\small hv=U_f-U_i=-\gamma_eB_0\left ( +\frac{\hbar}{2} \right )+\gamma_eB_0\left ( -\frac{\hbar}{2} \right )=-\gamma_eB_0\hbar

\small \gamma_e=-\frac{2\pi v}{B_0}=-1.759292\times 10^{11}Ckg^{-1}

 

Question

Why is the spin magnetic momentum quantum number \small m_s\hbar=-\frac{\hbar}{2} associated with the lower energy state \small U_i?

Answer

The classical gyromagnetic ratio of a charged particle is \small \gamma=\frac{q}{2m}. Since the electron has a negative charge, its gyromagnetic ratio \small \gamma_e is negative. Therefore, \small U_i< U_f if \small m_s\hbar=-\frac{\hbar}{2}.

 

If we replace \small q with \small -e and \small m with \small m_e in \small \gamma, we have \small \gamma=-8.7941\times 10^{10}Ckg^{-1}. So, \small \gamma_e is about twice the value of \small \gamma. Due to this difference, the classical notion of the electron spinning on its own axis (which is equivalent to a current loop) has no physical reality. The gyromagnetic ratio of the electron is formerly defined as:

\small \gamma_e=-g_e\frac{e}{2m_e}\; \; \; \; \; \; \; \; 164

where \small g_e is the g-value of the electron, which is measured in a recent experiment to be 2.00231930436256 with an uncertainty of 1.7×10-13.

This experiment also provides evidence that the spin angular momentum quantum number of an electron is \small \frac{1}{2}. Since the sole transition is between the two spin states of the electron, and that \small m_s=-s,-s+1,\cdots,s-1,s,

\small m_{s,final}-m_{s,initial}=(+s)-(-s)=1\; \; \; \; \Rightarrow \; \; \; \; s=\frac{1}{2}

 

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The Stern-Gerlach experiment

The Stern-Gerlach experiment, which was conducted in 1922 by Otto Stern and Walther Gerlach, showed that electrons have quantised spin angular momentum. In a typical Stern-Gerlach experiment, silver is vapourised in an oven, creating silver atoms, which after being collimated, pass through an inhomogeneous magnetic field and eventually deposit on a detector plate (see diagram below).

A silver atom is electrically neutral and has a single unpaired valence electron, whose orbital angular momentum is zero, as it resides in an s-orbital. The other 46 electrons occupy 4 closed shells and have zero total orbital angular momentum. These electronic properties of silver indicate that the beam of atoms will travel undeflected through the magnet. However, the beam is split symmetrically into two, with the atoms deposited on the detector plate in two horizontal lines. This propounds, according to classical physics (see eq66), that a silver atom possesses a non-zero magnetic dipole moment, which implies that the atom has some form of intrinsic angular momentum other than orbital angular momentum. Furthermore, eq66 and the symmetric splitting of the beam together suggest that there are two non-zero magnetic dipole moment involved, both of equal magnitude but in opposite directions. Therefore (with reference to eq65), a portion of the silver atoms must be in a different energy state versus the rest of the atoms.

 

Question

Why electrons occupying any closed shell of principal quantum number \small n (where \small n\in \mathbb{Z},n\geq 1 )  have zero total orbital angular momentum?

Answer

We know that the relation between a shell \small n and its sub-shells \small l is \small 0\leq l\leq n-1. We also know from eq132 that \small \hat{L}_z\vert l,m_l\rangle=m_l\hbar\vert l,m_l\rangle, where \small m_l=-l,-l+1,-l+2,\cdots,l. Furthermore, every sub-shell consists of one or more orbitals, each of which can hold 2 electrons. Therefore, for any closed shell, the sum of all \small m_l is zero and so, the total orbital angular momentum is zero.

 

To explain the results of the experiment, we need to derive a quantum mechanical expression that allows us to analyse the energy of a silver atom passing through the inhomogenous magnetic field. A reasonable starting point is eq68:

\small U=-\gamma(B_0+\alpha z)L_z

The quantum mechanical expression of the above classical equation, which is based on the current loop model, is obtained by replacing the potential energy term \small U and the \small z-component of angular momentum term \small L_z with the energy operator \small \hat{H} (Hamiltonian) and the \small z-component angular momentum operator \small \hat{L}_z respectively:

\small \hat{H}=-\gamma(B_0+\alpha z)\hat{L}_z

If we postulate that the intrinsic angular momentum operator \small \hat{S}_z is the analogue of \small \hat{L}_z, we can rewrite the above equation as:

\small \hat{H}_s=-\gamma_e(B_0+\alpha z)\hat{S}_z

We have also replaced the classical gyromagnetic ratio \small \gamma with a factor \small \gamma_e. The significance of this will be apparent at the end of the article. Assuming that the Hamiltonian \small \hat{H}_s acts on the intrinsic angular momentum eigenvector \small \vert s,m_s\rangle for a duration of \small 0\leq t\leq T when the silver atom passes through the magnet, it needs to be represented by a time-dependent operator:

\small \hat{H}_s(t)=-\gamma_e(B_0+\alpha z)\hat{S}_z\; \; \; \; \; \; \; \; 158

Since \small \hat{S}_z is the analogue of \small \hat{L}_z, we can also assume that the eigenvalue equations corresponding to eq132 and eq133 are:

\small \hat{S}_z\vert s,m_s\rangle=m_s\hbar\vert s,m_s\rangle\; \; \; \; \; \; \; \; \; 159

\small \hat{S}^{2}\vert s,m_s\rangle=s(s+1)\hbar^{2}\vert s,m_s\rangle\; \; \; \; \; \; \; \; \; 160

where \small m_s=-s,-s+1,-s+2,\cdots,s and \small s=0,\frac{1}{2},1,\frac{3}{2},\dots (unlike quantum numbers for orbital angular momentum, we did not restrict \small s and \small m_s to integers because we have no reason to do so).

As mentioned earlier, the silver atoms are split into two beams, implying two different energy states. In quantum mechanics, the state of atoms in each beam is described by a distinct set of quantum numbers. Hence, the state of atoms in one beam must be characterised by \small \vert s,m_s=+s\rangle and the other by \small \vert s,m_s=-s\rangle. However, we cannot tell which atom is which prior to \small t=0. It is therefore appropriate to express the time-dependent solution to eq158 prior to \small t=0 as a linear combination of the two states:

\small \chi(t)=c_1\alpha e^{-i\frac{E_+t}{\hbar}}+c_2\beta e^{-i\frac{E_-t}{\hbar}}\; \; \; \; \; \; \; \; 161

where \small c_1 and \small c_2 are constants; \small \alpha and \small \beta are the time-independent components of \small \alpha e^{-i\frac{E_+t}{\hbar}} and \small \beta e^{-i\frac{E_-t}{\hbar}} respectively.

Since \small m_s=\pm s, the eigenvalues of \small \hat{S}_z are \small \pm s\hbar, and hence from eq158,

\small E_\pm=\mp \gamma_e(B_0+\alpha z)s\hbar\; \; \; \; \; \; \; \; 162

In summary, the 2 energy states described in eq162 are due to a silver atom’s intrinsic angular momentum, which is characterised by the quantum numbers \small s and \small m_s=\pm s. Since, the magnetic dipole moment of a silver atom’s nucleus is very weak (see below for details), the intrinsic angular momentum is attributed to that of an electron (in this case, the single unpaired valence electron). George Uhlenbeck and Samuel Goudsmit, together with Wolfgang Pauli called it spin. The quantum number s is subsequently found in another experiment to be \small \frac{1}{2}, which makes \small \pm\frac{1}{2}.

 

Question

What is the definition of \small \gamma_e and how it is different from \small \gamma?

Answer

According to classical physics, we would expect \small \gamma to be equal to \small \gamma_e. However, the value of \small \gamma_e is determined in another experiment to be about twice the classical value. Due to this difference, the classical notion of the electron spinning on its own axis (which is equivalent to a current loop) has no physical reality.

 

Question

i) How does nuclear spin affect the experiment?

ii) Show that the change in energy of a nuclear magnetic dipole moment in the presence of a uniform magnetic field is E_{\pm}=-m_I\hbar\gamma B.

Answer

i) From eq61, the magnetic dipole moment in the \small z-direction that is due to an electron’s spin angular momentum is \small \left |\mu_{e,z} \right |=2\left | -\frac{e}{2m_e}S_z \right |=\frac{e\hbar}{2m_e}. The magnetic dipole moment of an atom’s nucleus is usually due to only one nucleon (magnetic dipole moment of an atom with even-even nucleons is negligible):

\small \left |\mu_{n,z} \right |=\left | \frac{e}{2m_p}S_z \right |=\frac{e\hbar}{4m_p}

\small \frac{\mu_{e,z}}{\mu_{n,z}}=\frac{2m_p}{m_e}\approx 3672

Therefore, the typical nuclear magnetic dipole moment is much weaker than an electron magnetic dipole moment. A magnetic field much stronger (Stern-Gerlach experiment \small B=0.1T) is needed to achieve reasonable resolution.

ii) The corresponding nuclear spin energy operator equation of eq68 in a uniform magnetic field B is \hat{U}=-\gamma B\hat{S}_z. With reference to eq168, the eigenvalue of \hat{U} is -m_I\hbar\gamma B or E_{\pm}=-m_I\hbar\gamma B, since m_I=\pm\frac{1}{2}.

 

For sequential Stern-Gerlach experiments, see this article.

 

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Biot-Savart law

The Biot-Savart law is an equation describing the magnetic field generated by a current segment.

Consider a point \small P in the vicinity of a wire carrying a current \small I (see diagram below).

The magnitude of \small I represents the amount of charges flowing through the wire (blue line). This implies that the amount of charges in an infinitesimal length of the wire \small d\vec{l} is proportional to \small Id\vec{l}. Since the magnetic field \small d\vec{B} at \small P is generated by the amount of moving charges in the wire, we have \small d\vec{B}\propto Id\vec{l}, or rather \small d\vec{B}\propto Id\vec{l}sin\theta because \small d\vec{B} is due to component of \small d\vec{l} that is perpendicular to \vec{R} (use right-hand rule to visualise).

The magnetic flux generated by \small I radiates spherically outward from every point along the infinitesimal length of the wire. For a particular \small I in a segment \small d\vec{l}, the amount of magnetic field lines radiated is a constant. As \small R increases, the surface area of the sphere becomes bigger. Therefore, the radiation passing through \small P must be proportional to the unit area of the sphere at \small R, i.e. \small d\vec{B}\propto \frac{1}{4\pi R^{2}}\hat{R}, where \small \hat{R} is the unit vector in the direction of \small \vec{R}. Finally, we would expect \small d\vec{B} to vary according to the material of the wire and the medium of the space between the wire and \small P. If we assume the wire to have negligible resistance and that the system is in a vacuum, we have

\small d\vec{B}=\frac{\mu_0Id\vec{l}sin\theta\hat{R}}{4\pi R^{2}}=\frac{\mu_0Id\vec{l}sin\theta\vec{R}}{4\pi R^{3}}=\frac{\mu_0Id\vec{l}\times\vec{R}}{4\pi R^{3}}\; \; \; \; \; \; \; \; 150

where \small \mu_0 is called the vacuum permeability or magnetic permeability of free space, and it represents the proportionality constant for the measurement of \small d\vec{B} at \small P in a vacuum.

The scalar form of eq150 is

\small d\vec{B}=\frac{\mu_0Idlsin\theta}{4\pi R^{2}}\; \; \; \; \; \; \; \; 151

The total field at \small P is

\small \vec{B}=\frac{\mu_0}{4\pi}\int \frac{Id\vec{l}\times\vec{R}}{R^{3}}\; \; \; \; \; \; \; \; 152

Eq150, eq151 and eq152 are different forms of the Biot-Savart law.

If the wire is a circle, the direction of \small d\vec{B} at \small P is determined by the right hand rule and is perpendicular to the plane formed by \small d\vec{l} and \small \vec{R}, where the angle \small \theta between \small d\vec{l} and \small \vec{R} is \small 90^{\circ} (see diagram below).

From eq151,

\small dB=\frac{\mu_0Idlsin90^{\circ}}{4\pi R^{2}}=\frac{\mu_0Idl}{4\pi R^{2}}\; \; \; \; \; \; \; \; 153

Let’s analyse the components of \small d\vec{B}. If we sum all the vectors of \small d\vec{B}_\perp at \small P from all current elements around the wire, they cancel out. The vectors of \small d\vec{B}_\parallel at \small P, however adds. Therefore, in scalar form, \small B=\int dB_\parallel, where \small dB_\parallel=dBcos\alpha, which when combined with eq153 gives

\small dB_\parallel=\frac{\mu_0Icos\alpha}{4\pi R^{2}}dl\; \; \; \; \; \; \; \; 154

Since \small R=\sqrt{r^{2}+z^{2}} and \small cos\alpha=\frac{r}{\sqrt{r^{2}+z^{2}}}, eq154 becomes \small dB_\parallel=\frac{\mu_0Ir}{4\pi(r^{2}+z^{2})^{3/2}}dl and

\small B=\int dB_\parallel=\frac{\mu_0Ir}{4\pi(r^{2}+z^{2})^{3/2}}\int dl

Substituting \small \int dl=2\pi r in the above equation

\small B=\frac{\mu_0Ir^{2}}{2(r^{2}+z^{2})^{3/2}}\; \; \; \; \; \; \; \; 155

For the case of \small B at point \small O, we have \small z, which is the length \small OP, equals to zero. So,

\small B=\frac{\mu_0I}{2r}\; \; \; \; \; \; \; \; 156

For the case of \small z\gg r, eq155 becomes \small B=\frac{\mu_0IA}{2\pi z^{3}}, where \small A=\pi r^{2}. Substitute eq60 and eq61 in \small B,

\small B=\frac{\mu_0\gamma}{2\pi z^{3}}L\; \; \; \; \; \; \; \; 157

Eq156 and eq157 are useful in constructing the orbit-orbit coupling term and the spin-orbit coupling term of the Hamiltonian.

 

 

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