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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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Larmor precession of an electron

Larmor precession is the change in orientation of the axis of the magnetic moment of a particle with respect to the axis of an external magnetic field.

Consider an electron at rest with intrinsic spin angular momentum of in a uniform magnetic field \small \boldsymbol{\mathit{B}} directed along the -axis. The electron’s spin gives rise to a magnetic dipole moment \small \boldsymbol{\mathit{\mu}}, which is antiparallel to . As a result, the magnetic field interacts with the electron’s magnetic moment and generates a torque \small \boldsymbol{\mathit{\tau}}=\boldsymbol{\mathit{\mu}}\times\boldsymbol{\mathit{B}} (see diagram below).

Eq61 becomes \small \boldsymbol{\mathit{\mu}}=\gamma_e\boldsymbol{\mathit{s}}, where \small \gamma_e is the gyromagnetic ratio of the electron. Taking the derivative on both sides of this equation with respect to \small t,

\small \frac{d\boldsymbol{\mathit{\mu}}}{dt}=\gamma_e\frac{d\boldsymbol{\mathit{s}}}{dt}

From eq64 and eq71, we have \small \frac{d\boldsymbol{\mathit{L}}}{dt}=\boldsymbol{\mathit{\mu}}\times \boldsymbol{\mathit{B}}, whose spin analogue is \small \frac{d\boldsymbol{\mathit{s}}}{dt}=\boldsymbol{\mathit{\mu}}\times \boldsymbol{\mathit{B}}. So,

\small \frac{d\boldsymbol{\mathit{\mu}}}{dt}=\gamma_e\, \boldsymbol{\mathit{\mu}}\times \boldsymbol{\mathit{B}}

 

Question

Explain in detail why precesses around .

Answer

Since , the torque arises from the interaction between the magnetic dipole moment and the external magnetic field causes the spin angular momentum to change in time. Because this torque is perpendicular to the magnetic field direction (taken as the -axis), it has no component along that axis. Consequently, the projection of the spin angular momentum onto the field direction remains constant. The only way for to change in time while maintaining a constant projection along the -axis is for it to precess about the direction of the magnetic field.

 

With reference to the above diagram, the change in arc length with respect to \small t is \small \frac{d\mu}{dt}=\mu sin\theta\frac{d\phi}{dt}. Hence,

\small \gamma_e\mu Bsin\theta=\mu sin\theta\frac{d\phi}{dt}

\small \int_{0}^{2\pi}d\phi=\gamma_eB\int_{0}^{T}dt

\small \omega_L=\vert\gamma_e B\vert=\frac{eB}{m_e}\; \; \; \; \; \; \; \; 149

where \small T is the period, \small m_e is the mass of an electron and \small \omega_L=\frac{2\pi}{T} is the Larmor frequency, which is defined as a positive value.

Note that the Larmor frequency is sometimes defined as \small v_L=\frac{1}{T}, which is then \small v_L=\frac{\vert\gamma_e B\vert}{2\pi}.

 

 

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The vector model of quantum angular momentum

The vector model of angular momentum is a diagrammatic representation of the implications of the commutation relation of \small \hat{L}^{2} with any one of the 3 component angular momentum operators.

In a previous article, we showed that it is possible to simultaneously specify eigenvalues of \small \hat{L}^{2} and \small \hat{L}_z because the 2 operators commute. Diagram I depicts the eigenvalues of \small \hat{L} (i.e. the square root of eigenvalues of \small \hat{L}^{2}) and \small \hat{L}_z.

where axes are in \small \hbar units; \small \boldsymbol{\mathit{L}} is the angular momentum vector with magnitude \small \sqrt{l(l+1)}, and has a \small z-component of magnitude \small m_l\in \mathbb{Z}.

Since any one of the component angular momenta does not commute with any of the other two, we cannot simultaneously specify more than one component of angular momentum (other than the trivial case of \small l_x=l_y=l_z=0). The angular momentum vector therefore lies on the cones in diagram II at any azimuthal angle.

 

Question

Can \small \boldsymbol{\mathit{L}} lie on the \small z-axis?

Answer

No. If \small \boldsymbol{\mathit{L}} lies on the \small z-axis, \small l_z is some non-zero integer while \small l_x=l_y=0. This means we have specified all 3 angular momentum components, which is impossible.

 

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Matrix elements of angular momentum ladder operators

The matrix elements of angular momentum ladder operators play a crucial role in quantum mechanics, facilitating transitions between different angular momentum states and providing insight into the underlying symmetry of quantum systems.

From eq132 of the previous article, we have

\small \hat{L}_z\vert l,m_l+1\rangle=(m_l+1)\hbar\vert l,m_l+1\rangle\; \; \; \; \; \; \; \; 134

From eq113 where k=1 and eq130, we have

\small \hat{L}_z\hat{L}_+\vert l,m_l\rangle=(m_l+1)\hbar\hat{L}_+\vert l,m_l\rangle\; \; \; \; \; \; \; \; 135

Comparing eq134 and eq135, both eigenstates \small \vert l,m_l+1\rangle and \small \hat{L}_+\vert l,m_l\rangle are associated with the same eigenvalue \small (m_l+1)\hbar. Therefore, one must be proportional to the other (e.g. the eigenstates \small e^{-ikx} and \small Ae^{-ikx} have the same eigenvalue), i.e.

\small \hat{L}_+\vert l,m_l\rangle=c_+\vert l,m_l+1\rangle\; \; \; \; \; \; \; \; 136

Similarly, for the lowering operator (using eq114), we have,

\small \hat{L}_-\vert l,m_l\rangle=c_-\vert l,m_l-1\rangle\; \; \; \; \; \; \; \; 137

where \small c_+ and \small c_- are scalars; or in summary

\small \hat{L}_\pm\vert l,m_l\rangle=c_\pm\vert l,m_l\pm 1\rangle\; \; \; \; \; \; \; \; 138

If the complete set of eigenstates is normalised, multiplying the above equation on the left by \small \langle l,m_l\pm1\vert gives the matrix elements of \small \hat{L}_\pm:

\small \langle l,m_l\pm 1\vert\hat{L}_\pm\vert l,m_l\rangle=c_\pm\; \; \; \; \; \; \; \; 139

From eq116, we have \small \hat{L}_-\hat{L}_+\vert l,m_l\rangle=(\hat{L}^{2}-\hat{L}_z^{\; 2}-\hbar\hat{L}_z)\vert l,m_l\rangle. Using eq132 and eq133,

\small \hat{L}_-\hat{L}_+\vert l,m_l\rangle=[l(l+1)-m_l(m_l+1)]\hbar^{2}\vert l,m_l\rangle\; \; \; \; \; \; \; \; 140

From eq136 and eq137

\small \hat{L}_-\hat{L}_+\vert l,m_l\rangle=c_+\hat{L}_-\vert l,m_l+1\rangle=c_+c_-\vert l,m_l\rangle\; \; \; \; \; \; \; \; 141

Comparing eq140 and eq141

\small c_+c_-=[l(l+1)-m_l(m_l+1)]\hbar^{2}\; \; \; \; \; \; \; \; 142

 

Question

Show that \small c_+c_-=\vert c_+\vert^{2}.

Answer

From eq139, where we let \small m_l=m_l+1

\small \langle l,m_l\vert\hat{L}_-\vert l,m_l+1\rangle=c_-

Substituting eq109 in the above equation

\small \langle l,m_l\vert\hat{L}_x\vert l,m_l+1\rangle-i\langle l,m_l\vert\hat{L}_y\vert l,m_l+1\rangle=c_-

Since \small \hat{L}_x and \small \hat{L}_y are Hermitian,

\small \left \{ \langle l,m_l+1\vert\hat{L}_x\vert l,m_l\rangle+i\langle l,m_l+1\vert\hat{L}_y\vert l,m_l\rangle\right \}^{*}=c_-

Substituting eq108 in the above equation

\small \langle l,m_l+1\vert\hat{L}_+\vert l,m_l\rangle^{*}=c_-

Substituting eq139 in the above equation

\small c_+^{\; *}=c_-\; \; \; \; \; \; \; \; 143

Therefore, \small c_+c_-=\vert c_+\vert^{2}.

 

Substituting \small c_+c_-=\vert c_+\vert^{2} in eq142

\small c_+=\hbar \sqrt{l(l+1)-m_l(m_l+1)}

Substituting the above equation in eq136,

\small \hat{L}_+\vert l,m_l\rangle=\hbar \sqrt{l(l+1)-m_l(m_l+1)}\; \vert l,m_l+1\rangle \; \; \; \; \; \; \; \; 144

To find \small c_-, we repeat the above steps for eq139 onwards and using eq115 to give

\small \hat{L}_+\hat{L}_-\vert l,m_l\rangle=\left ( \hat{L}^{2}-\hat{L}_z^{\; 2} +\hbar\hat{L}_z\right )\vert l,m_l\rangle=[l(l+1)-m_l(m_l-1)]\hbar^{2}\vert l,m_l\rangle\; \; \; \; \; \; \; \; 145

From eq137,

\small \hat{L}_+\hat{L}_-\vert l,m_l\rangle=c_-\hat{L}_+\vert l,m_l-1\rangle=c_-c_+\vert l,m_l\rangle\; \; \; \; \; \; \; \; 146

Comparing eq145 and eq146,

\small c_-c_+=[l(l+1)-m_l(m_l-1)]\hbar^{2}

Substituting eq143, where \small c_+=c_-^{\; *}, in the above equation

\small c_-=\hbar\sqrt{l(l+1)-m_l(m_l-1)}

Substituting the above equation in eq137,

\small \hat{L}_-\vert l,m_l\rangle=\hbar\sqrt{l(l+1)-m_l(m_l-1)}\vert l,m_l-1\rangle\; \; \; \; \; \; \; \; 147

Combining eq144 and eq147

\small \hat{L}_\pm\vert l,m_l\rangle=\hbar\sqrt{l(l+1)-m_l(m_l\pm1)}\vert l,m_l\pm1\rangle\; \; \; \; \; \; \; \; 148

 

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Eigenvalues of quantum orbital angular momentum operators

The eigenvalues of quantum orbital angular momentum operators are fundamental to understanding the quantisation of angular momentum in quantum mechanics, as they dictate the allowed energy levels and spatial distributions of particles in atomic and molecular systems.

As mentioned in an earlier article, if \small \left [ \hat{L}^{2},\hat{L}_z \right ]=0, a common complete set of eigenfunctions can be selected for the two operators. Let \small Y be the common set of normalised eigenfunctions with eigenvalues \small b and \small c for \small \hat{L}^{2} and \small \hat{L}_z respectively.

From eq75,

\small \hat{L}^{2}-\hat{L}_z^{\; 2} =\hat{L}_x^{\; 2}+\hat{L}_y^{\; 2}

Multiplying the above equation by \small Y on the right, \small Y^{*} on the left and integrating over all space, we have

\small \langle L^{2}\rangle-\langle L_z^{\; 2}\rangle=\langle L_x^{\; 2}\rangle+\langle L_y^{\; 2}\rangle

\small b-c^{2}=\langle L_x^{\; 2}\rangle+\langle L_y^{\; 2}\rangle

Even though \small Y is not an eigenfunction of \small \hat{L}_x^{\; 2}, we can still find the expectation value of \small \hat{L}_x^{\; 2}, which is

\small \langle L_x^{\; 2}\rangle=\int Y^{*}\hat{L}_x(\hat{L}_xY)d\tau=\int (\hat{L}_xY)(\hat{L}_xY)^{*}d\tau=\int \vert \hat{L}_xY\vert^{2}d\tau\geq 0

Note that we have used the fact that \small \hat{L}_x is Hermitian for the 2nd equality (see eq37). Similarly, \small \langle L_y^{\; 2}\rangle\geq 0. So,

\small b-c^{2}\geq 0\; \; \; \; \; or\; \; \; \; \; -\sqrt{b}\leq c\leq \sqrt{b}

Therefore, \small c has an upper bound and a lower bound. Since the eigenvalues of \small \hat{L}_z has an upper bound and a lower bound, from eq113 and eq114 we have

\small \hat{L}_zY_{max}=c_{max}Y_{max}\; \; \; \; \; \; \; \; 122

\small \hat{L}_zY_{min}=c_{min}Y_{min}\; \; \; \; \; \; \; \; 123

where \small Y_{max}=\hat{L}_+^{\; k_{max}}Y, \small c_{max}=c+k_{max}\hbar, \small Y_{min}=\hat{L}_-^{\; k_{min}}Y and \small c_{min}=c-k_{min}\hbar.

If we operate on eq122 with \small \hat{L}_+, we supposedly have

\small \hat{L}_z(\hat{L}_+Y_{max})=(c_{max}+\hbar)(\hat{L}_+Y_{max})

However, the above equation contradicts the upper bound eigenvalue of \small \hat{L}_z of \small c_{max}. This implies that we must truncate the ladder beyond eq122. Since \small c_{max}+\hbar\neq 0, we must have

\small \hat{L}_+Y_{max}=0\; \; \; \; \; \; \; \; 124

Using the same argument when operating on eq123 with \small \hat{L}_-, we have

\small \hat{L}_-Y_{min}=0 \; \; \; \; \; \; \; \; 125

If we further operate on eq124 with \small \hat{L}_-, we have

\small \hat{L}_-\hat{L}_+Y_{max}=0

Substitute eq116 in the above equation,

\small (\hat{L}^{2}-\hat{L}_z^{\; 2}-\hbar\hat{L}_z)Y_{max}=0

Using eq119 where \small k=k_{max}, and eq122,

\small (b-c_{max}^{\; \; \; \; \; \; 2}-\hbar c_{max})Y_{max}=0

Since \small Y_{max}\neq 0

\small b-c_{max}^{\; \; \; \; \; \; 2}-\hbar c_{max}=0 \; \; \; \; \; \; \; \; 126

Similarly, operating on eq125 with \small \hat{L}_+ and using eq115, eq119 and eq123, we have

\small b-c_{min}^{\; \; \; \; \; \; 2}+\hbar c_{min}=0 \; \; \; \; \; \; \; \; 127

Subtracting eq127 from eq126, we have \small (c_{max}+c_{min})(c_{min}-c_{max}-\hbar)=0. Since \small c_{max}> c_{min} and \small (c_{min}-c_{max}-\hbar)< 0, we have \small c_{max}+c_{min}=0 or

\small c_{max}=-c_{min}\; \; \; \; \; \; \; \; 128

As we know from eq122 and eq123, the value of \small c_{max}-c_{min} of a particular system is dependent on the number of consecutive operations on \small \hat{L}_zY by \small \hat{L}_+ or \small \hat{L}_-, with each operation raising or lowering the eigenvalue of \small \hat{L}_z by \small \hbar. Therefore,

\small c_{max}-c_{min}=0,\hbar,2\hbar,\cdots=2l\hbar\; \; \; \; \; \; \; \; 129

where \small l=0,\frac{1}{2},1,\frac{3}{2},\cdots

Substituting eq128 in eq129, we have \small c_{max}=l\hbar and \small c_{min}=-l\hbar. Therefore, the eigenvalues of \small \hat{L}_z are

\small c=-l\hbar,-l\hbar+\hbar,-l\hbar+2\hbar,\cdots,l\hbar=-l\hbar,(-l+1)\hbar,(-l+2)\hbar,\cdots,l\hbar

\small c=m_l\hbar\; \; \; \; \; \; \; \; 130

where \small m_l=-l,-l+1,-l+2,\cdots,l.

Substituting \small c_{min}=-l\hbar in eq127,

\small b=l(l+1)\hbar^{2}\; \; \; \; \; \; \; \; 131

where \small l=0,\frac{1}{2},1,\frac{3}{2},\cdots.

As mentioned in the previous article, the raising and lowering operators also apply to the spin angular momentum \small \boldsymbol{\mathit{S}} and the total angular momentum \small \boldsymbol{\mathit{J}}. We would therefore expect the eigenvalues of \small \hat{S}^{2} and \small \hat{S}_z to be \small s(s+1)\hbar^{2} and \small m_s\hbar respectively, and the eigenvalues of \small \hat{J}^{2} and \small \hat{J}_z to be \small j(j+1)\hbar^{2} and \small m_j\hbar respectively. However, the quantum numbers \small l and \small m_l for the orbital angular momentum \small \boldsymbol{\mathit{L}}, but not the quantum numbers for \small \boldsymbol{\mathit{S}} and \small \boldsymbol{\mathit{J}}, are restricted to integers. Therefore,

 

\small \boldsymbol{\mathit{L}} \small m_l\in \mathbb{Z} \small l\in \mathbb{Z}
\small \boldsymbol{\mathit{S}} \small m_s=-s,-s+1,-s+2,\cdots,s \small s=0,\frac{1}{2},1,\frac{3}{2},\cdots
\small \boldsymbol{\mathit{J}} \small m_j=-j,-j+1,-j+2,\cdots,j \small j=0,\frac{1}{2},1,\frac{3}{2},\cdots

 

Question

Why are the quantum numbers for \small \boldsymbol{\mathit{L}} restricted to integers?

Answer

The eigenvalue equation for \small \hat{L}_z (see eq95) is:

\small \frac{\hbar}{i}\frac{\partial}{\partial\phi}\psi=m_l\hbar\psi

where \small \psi=Ae^{im_l\phi}.

Since \small \psi must be single-valued,

\small Ae^{im_l\phi}=Ae^{im_l(\phi+2\pi)}

\small e^{im_l2\pi}=1

\small cosm_l2\pi + isinm_l2\pi=1

The solution to the above equation is \small m_l\in \mathbb{Z}. Furthermore, \small l is also an integer because \small m_l=-l,-l+1,-l+2,\cdots,l. In other words, there are values of for a given value of .

 

We would arrive at the same results (eq130 and eq131) if we have chosen \small \hat{L}_x or \small \hat{L}_y instead of \small \hat{L}_z. The significance of eq130 and eq131 is that we can simultaneously assign eigenvalues of \small \hat{L}^{2} and \small \hat{L}_z (or \small \hat{L}^{2} and \small \hat{L}_x or \small \hat{L}^{2} and \small \hat{L}_y) if the 2 operators commute. This, together with the fact that any pair of component angular momentum operators does not commute, implies that we cannot simultaneously specify eigenvalues of \small \hat{L}^{2} and more than one component angular momentum operators.

In conclusion, substituting eq130 in eq112 and eq131 in eq117

\small \hat{L}_zY=m_l\hbar Y

\small \hat{L}^{2}Y=l(l+1)\hbar^{2} Y\;\;\;\;\;\;\;\;131a

Since \small Y is a function of \small l and \small m_l, we can express the above eigenvalue equations as:

\small \hat{L}_z\vert l,m_l\rangle=m_l\hbar \vert l,m_l\rangle \; \; \; \; \; \; \; \; 132

\small \hat{L}^{2}\vert l,m_l\rangle=l(l+1)\hbar^{2} \vert l,m_l\rangle \; \; \; \; \; \; \; \; 133

 

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Quantum orbital angular momentum ladder operators

The ladder operators of quantum orbital angular momentum are defined as:

\small \hat{L}_+=\hat{L}_x+i\hat{L}_y\; \; \; \; \; \; \; \; 108

\small \hat{L}_-=\hat{L}_x-i\hat{L}_y\; \; \; \; \; \; \; \; 109

where \small \hat{L}_+ is the raising operator and \small \hat{L}_- is the lowering operator.

To demonstrate why the operators are named as such, we substitute eq108 in \small \left [\hat{L}_+,\hat{L}_z\right ]:

\small \left [\hat{L}_+,\hat{L}_z\right ]=\left [\hat{L}_x+i\hat{L}_y,\hat{L}_z\right ]=\left [\hat{L}_x,\hat{L}_z\right ]+i\left [\hat{L}_y,\hat{L}_z\right ]

Substituting eq100 and eq101 in the above equation, and noting that \small \left [ \hat{L}_a,\hat{L}_b \right ]=-\left [ \hat{L}_b,\hat{L}_a \right ], yields

\small \hat{L}_+\hat{L}_z-\hat{L}_z\hat{L}_+=-i\hbar\hat{L}_y-\hbar\hat{L}_x

\small \hat{L}_+\hat{L}_z=\hat{L}_z\hat{L}_+-\hbar\hat{L}_+\; \; \; \; \; \; \; \; 110

Similarly, we find

\small \hat{L}_-\hat{L}_z=\hat{L}_z\hat{L}_-+\hbar\hat{L}_-\; \; \; \; \; \; \; \; 111

If \small Y is an eigenfunction of \small \hat{L}_z with eigenvalue \small c,

\small \hat{L}_z Y=cY\; \; \; \; \; \; \; \; 112

Operating on eq112 with \small \hat{L}_+, we have \small \hat{L}_+\hat{L}_z Y=c\hat{L}_+Y. Substituting eq110 in the this equation and rearranging gives \small \hat{L}_z\hat{L}_+ Y=(c+\hbar)\hat{L}_+Y. So, \small \hat{L}_+Y is an eigenfunction of \small \hat{L}_z with eigenvalue \small (c+\hbar). Operating on eq112 with \small \hat{L}_+ transforms \small Y into another eigenfunction \small \hat{L}_+ Y with an eigenvalue higher than c by \small \hbar. If we operate on eq112 with \small \hat{L}_+ twice, we have

\small \hat{L}_+\hat{L}_z\hat{L}_+ Y=(c+\hbar)\hat{L}_+^{\; 2}Y

\small \left ( \hat{L}_z\hat{L}_+-\hbar\hat{L}_+\right )\hat{L}_+ Y=(c+\hbar)\hat{L}_+^{\; 2}Y

\small \hat{L}_z\hat{L}_+^{\; 2}Y=(c+2\hbar)\hat{L}_+^{\; 2}Y

By mathematical induction,

\small \hat{L}_z\hat{L}_+^{\; k}Y=(c+k\hbar)\hat{L}_+^{\; k}Y \; \; \; \; \; k=0,1,2,\cdots\; \; \; \; \; \; \; \; 113

Similarly, if we operate on eq112 \small k times with \small \hat{L}_-, we have

\small \hat{L}_z\hat{L}_-^{\; k}Y=(c-k\hbar)\hat{L}_-^{\; k}Y \; \; \; \; \; k=0,1,2,\cdots\; \; \; \; \; \; \; \; 114

In other words, the raising operator progressively raises the eigenvalue of \small Y by \small \hbar, while the lowering operator progressively lowers the eigenvalue of \small Y by \small \hbar, i.e. each operator generate a ladder of eigenvalues.

 

Question

Show that \small \hat{L}_+\hat{L}_-=\hat{L}^{2}-\hat{L}_z^{\; 2}+\hbar\hat{L}_z.

Answer

Substituting eq108 and eq109 in \small \hat{L}_+\hat{L}_-,

\small \hat{L}_+\hat{L}_-=\hat{L}_x^{\; 2}-i\hat{L}_x\hat{L}_y+i\hat{L}_y\hat{L}_x+\hat{L}_y^{\; 2}

Substituting eq75 in the above equation and then eq99 in the resultant equation, and noting that \small \left [ \hat{L}_a,\hat{L}_b \right ]=-\left [ \hat{L}_b,\hat{L}_a \right ], results in

\small \hat{L}_+\hat{L}_-=\hat{L}^{2}-\hat{L}_z^{\; 2}+\hbar\hat{L}_z\; \; \; \; \; \; \; \; 115

Similarly,

\small \hat{L}_-\hat{L}_+=\hat{L}^{2}-\hat{L}_z^{\; 2}-\hbar\hat{L}_z\; \; \; \; \; \; \; \; 116

 

If \small Y is simultaneously also an eigenfunction of \small \hat{L}^{2} with eigenvalue \small b,

\small \hat{L}^{2}Y=bY\; \; \; \; \; \; \; \; 117

 

Question

Show that \small \hat{L}^{2} commutes with \small \hat{L}_\pm ^{\; k}.

Answer

For \small k=1,  we have \small \left [ \hat{L}^{2},\hat{L}_ \pm \right ]=\left [ \hat{L}^{2},\hat{L}_ x\pm i\hat{L}_ y\right ]=\left [ \hat{L}^{2},\hat{L}_ x\right ]\pm i\left [ \hat{L}^{2},\hat{L}_ y\right ]. Since \small \left [ \hat{L}^{2},\hat{L}_ x \right ]=0 and \small \left [ \hat{L}^{2},\hat{L}_ y \right ]=0,

\small \left [ \hat{L}^{2},\hat{L}_ \pm \right ]=0

For \small k=2, we apply the identity \small \left [ \hat{A},\hat{B}\hat{C} \right ]=\left [ \hat{A},\hat{B}\right ]\hat{C}+\hat{B}\left [ \hat{A},\hat{C} \right ]:

\small \left [ \hat{L}^{2},\hat{L}_ \pm ^{\; 2}\right ]=\left [ \hat{L}^{2},\hat{L}_ \pm\right ]\hat{L}_ \pm+\hat{L}_ \pm\left [\hat{L}^{2},\hat{L}_ \pm\right]=0

By mathematical induction,

\small \left [ \hat{L}^{2},\hat{L}_\pm^{\; k} \right ]=0\; \; \; \; \; k=0,1,2,\cdots\; \; \; \; \; \; \; \; 118

 

Operating on eq117 with \small \hat{L}_\pm^{\; k} and using eq118

\small \hat{L}_\pm^{\; k} \hat{L}^{2}Y=b\hat{L}_\pm^{\; k}Y

\small \hat{L}^{2}\hat{L}_\pm^{\; k}Y=b\hat{L}_\pm^{\; k}Y\; \; \; \; \; k=0,1,2,\cdots\; \; \; \; \; \; \; \; 119

The raising and lowering operators also apply to the spin angular momentum \small \boldsymbol{\mathit{S}}, because the spin angular momentum component operators are postulated to obey the form of commutation relations as described by eq99, eq100 and eq101 (see eq165, eq166 and eq167). Similarly, the raising and lowering operators apply to the total angular momentum \small \boldsymbol{\mathit{J}} (see Q&A below).

 

Question

Show that \small \left [ \hat{J}_x,\hat{J}_y \right ]=i\hbar\hat{J}_z.

Answer

Let

\small \hat{J}_i=\hat{M}_{1i}+\hat{M}_{2i}\; \; \; \; \; \; \; \; 120

where \small i=x,y,z; \small \hat{M}_{1i} and \small \hat{M}_{2i} are component operators of \small \hat{M}^{(1)} and \small \hat{M}^{(2)} respectively.

\small \hat{M}^{(1)} and \small \hat{M}^{(2)} are operators of two sources of angular momentum, e.g. \small \hat{M}^{(1)} and \small \hat{M}^{(2)} are the orbital angular momentum operator and spin angular momentum operator respectively of a particle.

\small \left [ \hat{J}_x,\hat{J}_y \right ]=\left ( \hat{M}_{1x}+\hat{M}_{2x} \right )\left ( \hat{M}_{1y}+\hat{M}_{2y} \right )-\left ( \hat{M}_{1y}+\hat{M}_{2y} \right )\left ( \hat{M}_{1x}+\hat{M}_{2x} \right )

Expanding and rearranging the RHS of the above equation,

\small \left [ \hat{J}_x,\hat{J}_y \right ]=\left [ \hat{M}_{1x},\hat{M}_{1y} \right ]+\left [ \hat{M}_{2x},\hat{M}_{2y} \right ]+\left [ \hat{M}_{1x},\hat{M}_{2y} \right ]+\left [ \hat{M}_{2x},\hat{M}_{1y} \right ]

Since the 3rd term on RHS of the above equation involves operators acting on different vector spaces (e.g. spatial coordinates vs spin coordinates), they must commute. The same goes for the 4th term. So,

\small \left [ \hat{J}_x,\hat{J}_y \right ]=\left [ \hat{M}_{1x},\hat{M}_{1y} \right ]+\left [ \hat{M}_{2x},\hat{M}_{2y} \right ]

According to eq99 and eq165, \small \left [ \hat{M}_{1x},\hat{M}_{1y} \right ]=i\hbar\hat{M}_{1z} and \small \left [ \hat{M}_{2x},\hat{M}_{2y} \right ]=i\hbar\hat{M}_{2z}. So,

\small \left [ \hat{J}_{x},\hat{J}_{y} \right ]=i\hbar\hat{J}_{z}\; \; \; \; \; \; \; \; 121

Similarly, we have \small \left [ \hat{J}_{y},\hat{J}_{z} \right ]=i\hbar\hat{J}_{x} and \small \left [ \hat{J}_{z},\hat{J}_{x} \right ]=i\hbar\hat{J}_{y}.

 

 

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Commutation relations of quantum orbital angular momentum operators

The three orbital angular momentum component operators do not commute with one another.

To show that \left [ \hat{L}_x,\hat{L}_y \right ]\neq 0, we substitute eq72 and eq73 in \left [ \hat{L}_x,\hat{L}_y \right ], giving \left [ \hat{L}_x,\hat{L}_y \right ]=\hbar^{2}\left [ x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right ], which when combined with eq74, returns

\left [ \hat{L}_x,\hat{L}_y \right ]=i\hbar\hat{L}_z\; \; \; \; \; \; \; \; 99

Repeating the above procedure for and , we get

\left [ \hat{L}_y,\hat{L}_z \right ]=i\hbar\hat{L}_x\; \; \; \; \; \; \; \; 100

\left [ \hat{L}_z,\hat{L}_x \right ]=i\hbar\hat{L}_y\; \; \; \; \; \; \; \; 101

Hence, each of the three orbital angular momentum component operators do not commute with the other two. Next, to show that \hat{L}^{2} commutes with all 3 orbital angular momentum component operators, we begin with

\small \left [ \hat{L}^{2},\hat{L}_x \right ]=\left (\hat{L}_x^{\; 2}+\hat{L}_y^{\; 2}+\hat{L}_z^{\; 2}\right )\hat{L}_x-\hat{L}_x\left (\hat{L}_x^{\; 2}+\hat{L}_y^{\; 2}+\hat{L}_z^{\; 2}\right )=\left [ \hat{L}_y^{\; 2},\hat{L}_x \right ]+\left [ \hat{L}_z^{\; 2},\hat{L}_x \right ]

Using the identity \small \left [ \hat{L}_a^{\; 2},\hat{L}_b \right ]=\hat{L}_a\left [ \hat{L}_a,\hat{L}_b \right ]+\left [ \hat{L}_a,\hat{L}_b \right ]\hat{L}_a,

\small \left [ \hat{L}^{2},\hat{L}_x \right ]=\hat{L}_y\left [ \hat{L}_y,\hat{L}_x \right ]+\left [ \hat{L}_y,\hat{L}_x \right ]\hat{L}_y +\hat{L}_z\left [ \hat{L}_z,\hat{L}_x \right ]+\left [ \hat{L}_z,\hat{L}_x \right ]\hat{L}_z

Substituting eq99 and eq101 in the above equation, and noting that \small \left [ \hat{L}_a,\hat{L}_b \right ]=-\left [ \hat{L}_b,\hat{L}_a \right ], yields \small \left [ \hat{L}^{2},\hat{L}_x \right ]=0. Repeating the steps for \small \left [ \hat{L}^{2},\hat{L}_y \right ] and \small \left [ \hat{L}^{2},\hat{L}_z \right ] gives

\small \left [ \hat{L}^{2},\hat{L}_x \right ]=0\; \; \; \;\left [ \hat{L}^{2},\hat{L}_y \right ]=0\; \; \; \;\left [ \hat{L}^{2},\hat{L}_z \right ]=0\; \; \; \; \; \; \; \; 102

As mentioned in an earlier article, a common complete set of eigenfunctions can be selected for two operators only if they commute. Therefore, \small \hat{L}^{2} shares a common set of eigenfunctions with each of \small \hat{L}_x, \small \hat{L}_y and \small \hat{L}_z, but we cannot select a common set of eigenfunctions for any pair of angular momentum component operators.

 

Question

Show that each of the three orbital angular momentum component operators commute with \small p^{2}, \small \hat{p}^{2}, \small r, \small r^{2} and \small \frac{1}{r}, where \small p^{2}=p_x^{\; 2}+p_y^{\; 2}+p_z^{\; 2} and \small r^{2}=x^{\; 2}+y^{\; 2}+z^{\; 2}.

Answer

Substituting eq74 in \small \left [ \hat{L}_z,x \right ], \small \left [ \hat{L}_z,y \right ], \small \left [ \hat{L}_z,z \right ], \small \left [ \hat{L}_z,p_x \right ], \small \left [ \hat{L}_z,p_y \right ] and \small \left [ \hat{L}_z,p_z \right ] (noting that \small p_i=m\frac{i}{t}, where \small i=x,y,z), and carrying out the derivatives, yields

\small \left [ \hat{L}_z,x \right ]=i\hbar y\; \; \; \; \;\left [ \hat{L}_z,y \right ]=-i\hbar x\; \; \; \; \;\left [ \hat{L}_z,z \right ]=0

\small \left [ \hat{L}_z,p_x \right ]=i\hbar p_y\; \; \; \; \;\left [ \hat{L}_z,p_y \right ]=-i\hbar p_x\; \; \; \; \;\left [ \hat{L}_z,p_z \right ]=0

Using the identities \small \left [ \hat{A},\hat{B}+\hat{C}+\hat{D} \right ]=\left [ \hat{A},\hat{B} \right ]+\left [ \hat{A},\hat{C} \right ]+\left [ \hat{A},\hat{D} \right ] and \small \left [ \hat{A},\hat{B}\hat{C} \right ]=\left [ \hat{A}\hat{B} \right ]\hat{C}+\hat{B}\left[\hat{A},\hat{C} \right ],

\small \left [ \hat{L}_z,r^{2} \right ]=\left [ \hat{L}_z,x^{2}+y^{2}+z^{2}\right ]=0

\small \left [ \hat{L}_z,r \right ]=\left [ \hat{L}_z,\sqrt{x^{2}+y^{2}+z^{2}}\right ]=0

\small \left [ \hat{L}_z,\frac{1}{r} \right ]=\left [ \hat{L}_z,\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\right ]=0

\small \left [ \hat{L}_z,p^{2} \right ]=\left [ \hat{L}_z,p_x^{\; 2}+p_y^{\; 2}+p_z^{\; 2}\right ]=0\; \; \; \; \; \; \; \; 103

\small \left [ \hat{L}_z,\hat{p}^{2} \right ]=0 can be inferred from eq103. Repeating the same logic for \small \hat{L}_x and \small \hat{L}_y. we have

\small \left [ \hat{L}_i,r^{2} \right ]=0\; \; \; \;\left [ \hat{L}_i,r \right ]=0\; \; \; \;\left [ \hat{L}_i,\frac{1}{r} \right ]=0\; \; \; \;\left [ \hat{L}_i,p^{2} \right ]=0\; \; \; \;\left [ \hat{L}_i,\hat{p}^{2}\right ]=0\; \; \; \; \; \; \; \; 104

 

The commutation relations in the above Q&A are applicable to hydrogenic systems. For a system of 2-electrons, there are cross terms like:

\small \left [ \hat{L}_{1z},\frac{1}{r_2} \right ]=\left [ \hat{L}_{1z},\frac{1}{\sqrt{x_2^{\; 2}+y_2^{\; 2}+z_2^{\; 2}}} \right ]=0\; \; \; \; \; \; \; \; 105

which are useful in determining the commutation relations between \small \hat{L}^{2} and the multi-electron Hamiltonian, for example \small \left [ \hat{L}_{1z}+\hat{L}_{2z},\frac{1}{r_1}+\frac{1}{r_2} \right ]=0.

 

Question

Show that \small \left [ \hat{L}^{2},\hat{p}^{2} \right ]=0 and \small \left [ \hat{L}^{2},\frac{1}{r} \right ]=0.

Answer

Using eq75 and the identities \small \left [ \hat{A},\hat{B}+\hat{C}+\hat{D} \right ]=\left [ \hat{A},\hat{B} \right ]+\left [ \hat{A},\hat{C} \right ]+\left [ \hat{A},\hat{D} \right ] and \small \left [ \hat{A},\hat{B}\hat{C} \right ]=\left [ \hat{A}\hat{B} \right ]\hat{C}+\hat{B}\left[\hat{A},\hat{C} \right ],

\small \left [ \hat{L}^{2},\hat{p}^{2} \right ]=\left [ \hat{L}_x^{\; 2}+\hat{L}_y^{\; 2}+\hat{L}_z^{\; 2},\hat{p}^{2} \right ]=0\; \; \; \; \; \; \; \; 106

\small \left [ \hat{L}^{2},\frac{1}{r} \right ]=\left [ \hat{L}_x^{\; 2}+\hat{L}_y^{\; 2}+\hat{L}_z^{\; 2},\frac{1}{r}\right ]=0\; \; \; \; \; \; \; \; 107

 

 

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