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Many-electron systems

An atom with z>1 is an example of a many-electron system.

The electron configuration of a multi-electron atom, e.g. carbon: 1s22s22p2, specifies a general distribution of electrons in the atom’s orbitals. It is slightly vague in terms of energy levels, as different states may arise from a given electron configuration. In the case of carbon, the two 2p electrons could reside in any of the three 2p orbitals, with any spin orientation combinations that do not violate the Pauli Exclusion Principle. For example, the electrons in 2p can be arranged as [\uparrow\; \; \uparrow\; \times], [\uparrow\downarrow\; \times\; \: \times], [\uparrow\; \; \downarrow\; \times], etc.

To evaluate the relative non-relativistic energy levels of these different states for carbon, let’s assume the following:

  1. The core electrons of 1s22s2 do not affect the relative energy levels. This is because there is only one way to arrange these electrons, with the same arrangement common to all the different states. Consequently, the electronic configuration of carbon is effectively reduced to p2.
  2. The coupling of orbital angular momenta of the p2 electrons occurs separately from the coupling of spin angular momenta of the electrons, where \boldsymbol{\mathit{L}}=\sum_{i=1}^{n}\boldsymbol{l}_i and \boldsymbol{\mathit{S}}=\sum_{i=1}^{n}\boldsymbol{s}_i.
  3. The states \vert L,M_L,S,M_S\rangle are represented by total wavefunctions that are antisymmetric with respect to electron label exchange.

The non-relativistic multi-electron Hamiltonian \hat{H}_T of eq240 is constructed without considering orbital angular momentum coupling and spin angular momentum coupling. To satisfy our second assumption, we modify \hat{H}_T as follows:

\hat{H}_T^{'}=-\frac{\hbar^{2}}{2m_e}\sum_{i=1}^{n}\nabla_i^{\; 2}+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}C_{ij}\boldsymbol{\mathit{l}}_i\cdot\boldsymbol{\mathit{l}}_j-\sum_{i=1}^{n}\frac{Ze^2}{4\pi\varepsilon_0r_i}+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{e^{2}}{4\pi\varepsilon_0r_{ij}}

The extra term is a correction to the energy levels of states due to angular momentum coupling. The correction for spin-spin interactions, however, does not require an additional term, as it is accounted for in the last term (will be explained shortly).

 

Question

How is the new term derived?

Answer

Orbital angular momentum coupling is the interaction of the orbital magnetic moments of a pair of electrons. The interaction creates a potential in the form of eq67, where

U_{oo}=A\boldsymbol{\mathit{l}}_1\cdot\boldsymbol{\mathit{B}}

where A=\frac{e}{2m_e}, \boldsymbol{\mathit{l}}_1 is the angular momentum of one electron and \boldsymbol{\mathit{B}} is the magnetic field due to the orbital motion of the other electron.

Since \boldsymbol{\mathit{B}}\propto\boldsymbol{\mathit{l}}_2 (see article on Biot-Savart law), U_{oo}=C_{12}\boldsymbol{\mathit{l}}_1\cdot\boldsymbol{\mathit{l}}_2, where C_{12} is the constant of proportionality. For a system of more than two electrons, U_{oo}=\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}C_{ij}\boldsymbol{\mathit{l}}_i\cdot\boldsymbol{\mathit{l}}_j.

 

Rewriting \hat{H}_T^{'} in terms of eq50 and eq51,

\hat{H}_T^{'}=\hat{H}_{radial}+\sum_{i=1}^{n}\frac{\hat{L}_i^{\;2}}{2mr_i^{\;2}}+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}C_{ij}\boldsymbol{\mathit{l}}_i\cdot\boldsymbol{\mathit{l}}_j+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{e^{2}}{4\pi\varepsilon_0r_{ij}}

For the p2 configuration, which involves equivalent electrons, r_1=r_2=r. So,

\hat{H}_T^{'}=\hat{H}_{radial}+D\sum_{i=1}^{n}\hat{L}_i^{\;2}+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}C_{ij}\boldsymbol{\mathit{l}}_i\cdot\boldsymbol{\mathit{l}}_j+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{e^{2}}{4\pi\varepsilon_0r_{ij}}

where the constant D=\frac{1}{2mr^{2}}.

Comparing the second and third terms of \hat{H}_T^{'} with eq181, C_{ij}=\frac{1}{mr^{2}}. Therefore,

\hat{H}_T^{'}=\hat{H}_{radial}+D\hat{{L}^{2}}^{(T)}+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{e^{2}}{4\pi\varepsilon_0r_{ij}}\; \; \; \; \; \; \; \; 248

For a given electron configuration like p2, the values of r_i do not change. Therefore, the relative energies of states arising from an electron configuration are determined by the second and third terms of \hat{H}_T^{'}. This reduces \hat{H}_T^{'} to

\hat{H}_{relative}=\hat{H}_{angular}+\hat{H}_{ee\: repulsion}

We know from eq244 that \left [\hat{{L}^{2}}^{(T)}, \hat{H}_T \right]=0 and hence \left [\hat{{L}^{2}}^{(T)}, \hat{H}_{relative} \right]=0, which results in our ability to choose a common set of eigenstates for \hat{{L}^{2}}^{(T)} and \hat{H}_{relative}. Since the eigenvalue of \hat{{L}^{2}}^{(T)} is a function of the total orbital angular momentum quantum number L, E_{angular} is a function of L.

 

Question

Show that \left [\hat{{L}^{2}}^{(T)}, \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}C_{ij}\boldsymbol{\mathit{l}}_i\cdot\boldsymbol{\mathit{l}}_j\right ]=0 for a two-electron system.

Answer

Using eq181, eq99, eq100, eq101, the identities [\hat{A}+\hat{B}+\hat{C},\hat{D}]=[\hat{A},\hat{D}]+[\hat{B},\hat{D}]+[\hat{C},\hat{D}] and [\hat{A}\hat{B},\hat{C}]=[\hat{A},\hat{C}]\hat{B}+\hat{A}[\hat{B},\hat{C}], and that the component operators of \hat{L}_1 and \hat{L}_2 commute because they act on different vector spaces, we have \left [\hat{{L}^{2}}^{(T)}, \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}C_{ij}\boldsymbol{\mathit{l}}_i\cdot\boldsymbol{\mathit{l}}_j\right ]=0.

 

Lastly, two factors contribute to r_{ij}=\vert\boldsymbol{\mathit{r}}_i-\boldsymbol{\mathit{r}}_j\vert of the term \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{e^{2}}{4\pi\varepsilon_0r_{ij}}.  The first, called Coulomb interaction, can be explained by classical mechanics, where two electrons orbiting in the same direction meet less often than when they are orbiting in opposite directions. This implies that \vert\boldsymbol{\mathit{r}}_i-\boldsymbol{\mathit{r}}_j\vert_{ave}(same)>\vert\boldsymbol{\mathit{r}}_i-\boldsymbol{\mathit{r}}_j\vert_{ave}(opposite). Furthermore, the individual angular momentum of each electron orbiting in the same direction adds vectorially to give a higher total angular momentum (and hence a higher value of the quantum number L) than electrons orbiting in opposite directions. Hence, the Coulomb contribution to \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{e^{2}}{4\pi\varepsilon_0r_{ij}} is dependent on the quantum number L. The second factor affecting r_{ij}=\vert\boldsymbol{\mathit{r}}_i-\boldsymbol{\mathit{r}}_j\vert is the exchange interaction. As mentioned in an earlier article, the exchange interaction is due to the symmetry of the total wavefunction of the state \vert L,M_L,S,M_S\rangle. We can infer from eq247 that E_{ee\: repulsion} for total wavefunctions with anti-symmetric spin wavefunctions are greater than E_{ee\: repulsion} for total wavefunctions with symmetric spin wavefunctions. Hence, E_{ee\: repulsion} is dependent on the total spin angular momentum quantum number S.

This above analysis of energy levels of different states (with respect to \hat{H}_T) can be extended to any electron configuration of equivalent electrons. In short, for a given electron configuration with equivalent electrons, states with the same value of L and the same value of S have the same energy.

 

Question

How about a particular electron configuration of non-equivalent electrons, where r_i\neq r_j?

Answer

The concept of states with the same value of L and the same value of S having the same energy also applies, to reasonable estimation, to a given electron configuration of non-equivalent electrons because the eigenvalues of U_{oo} is relatively small compared to the eigenvalues of the other terms in \hat{H}_T^{'}. In fact, U_{oo} is usually omitted in the Hamiltonian, whether we are dealing with equivalent or non-equivalent electrons. The eigenvalue of the term 2(\hat{\boldsymbol{\mathit{L}}}_1\cdot\hat{\boldsymbol{\mathit{L}}}_2) in \hat{{L}^{2}}^{(T)}=\hat{L}_1^{\;2}+\hat{L}_2^{\;2}+2(\hat{\boldsymbol{\mathit{L}}}_1\cdot\hat{\boldsymbol{\mathit{L}}}_2) is associated with orbit-orbit coupling energy. If U_{oo} is relatively small compared to the eigenvalues of the other terms in \hat{H}_T^{'}, \hat{{L}^{2}}^{(T)}\approx \hat{L}_1^{\;2}+\hat{L}_2^{\;2} and \hat{H}_T^{'}\approx -\frac{\hbar^{2}}{2m_e}\sum_{i=1}^{n}\nabla_i^{\; 2}-\sum_{i=1}^{n}\frac{Ze^2}{4\pi\varepsilon_0r_i}+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{e^{2}}{4\pi\varepsilon_0r_{ij}}.

 

In summary, we have shown that different states that arise from a particular electron configuration have the same energy if they have the same value of L and the same value of S. However, we have not discussed the complete solution to \hat{H}_T , which involves deriving the exact formula of total wavefunctions of \hat{H}_T and the exact eigenvalues of \hat{H}_T. The solution to \hat{H}_T can be found using the Hartree-Fock method, which will be explained in another article.

 

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Exchange force (quantum mechanics)

Exchange force is the interaction between particles of a system, as a result of the symmetry of the wavefunction describing the system. This interaction results in a change in the expectation values of inter-particle distances, and hence, a change in energy eigenvalues.

Consider a system of two electrons. The exchange force between electrons is also known as spin correlation. Since the electrons are indistinguishable, the normalised spatial wavefunction can be expressed generally as:

\psi_{\pm}(r_1,r_2)=\frac{1}{\sqrt{2}}[\phi_a(r_1)\phi_b(r_2)\pm\phi_b(r_1)\phi_a(r_2)]

where \psi_+ and \psi_- are symmetric spatial wavefunction and anti-symmetric spatial wavefunction respectively of the system of a pair of identical electrons, \phi_a and \phi_b are orthonormal one-electron wavefunctions, and the notation of \phi_a(r_1)\phi_b(r_2) denotes electron 1 in the state \phi_a and electron 2 in the state \phi_b.

To evaluate the effect of wavefunction symmetry on the expectation values of inter-particle distances, we analyse, for convenience, the average value of the square of the inter-particle distance:

\langle(\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2)^{2}\rangle=\frac{1}{2}\int \psi_{\pm}^{\;*}(r_1^{\;2}+r_2^{\;2}-2\boldsymbol{\mathit{r}}_1\cdot\boldsymbol{\mathit{r}}_2)\psi_{\pm} dr_1dr_2

Substitute \psi_{\pm} in the above equation and expanding,

where \langle r_n^{\;2}\rangle_k=\int \phi_k^{\;*}(r_n) r_n^{\;2}\phi_k(r_n)dr_n and \langle \boldsymbol{\mathit{r}}_n \rangle_k=\int \phi_k^{\;*}(r_n)\boldsymbol{\mathit{r}}_n\phi_k(r_n)dr_n, with n=1,2 and k=a,b.

Applying the Hermitian property of \boldsymbol{\mathit{r}}_1 and \boldsymbol{\mathit{r}}_2 for the last two terms of the above equation,

where \langle\boldsymbol{\mathit{r}}_n\rangle_{ab}=\int \phi_a^{\;*}(r_n)\boldsymbol{\mathit{r}}_n\phi_b(r_n)dr_n.

Since the two particles are indistinguishable, \langle r_1^{\;2}\rangle_k=\langle r_2^{\;2}\rangle_k=\langle r^{2}\rangle_k. Similarly, \langle \boldsymbol{\mathit{r}}_1\rangle_k=\langle \boldsymbol{\mathit{r}}_2\rangle_k=\langle \boldsymbol{\mathit{r}}\rangle_k and \langle \boldsymbol{\mathit{r}}_1\rangle_{ab}=\langle \boldsymbol{\mathit{r}}_2\rangle_{ab}=\langle \boldsymbol{\mathit{r}}\rangle_{ab}. So,

\langle(\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2)^{2}\rangle=\langle r^{2}\rangle_a+\langle r^{2}\rangle_b-2\langle\boldsymbol{\mathit{r}}\rangle_a\langle\boldsymbol{\mathit{r}}\rangle_b\mp2\vert\langle\boldsymbol{\mathit{r}}\rangle_{ab}\vert^{2}\; \; \; \; \; \; \; \; 246

where we have used the logic that \frac{1}{2}\langle r^{2}\rangle_k\pm\frac{1}{2}\langle r^{2}\rangle_k=\langle r^{2}\rangle_k because the difference of the terms is zero.

The consequence of eq246 is that \langle(\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2)^{2}\rangle_{\psi_+}<\langle(\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2)^{2}\rangle_{\psi_-}, which implies that

\frac{1}{\; \; \vert(\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2)\vert_{\psi_+}}>\frac{1}{\; \; \vert(\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2)\vert_{\psi_-}}\; \; \; \; \; \; \; \; 247

The eigenvalues of the Hamiltonian of eq239 are therefore expected to be different for a singlet state and a triplet state because the total wavefunction of the system, according to Pauli’s exclusion principle, must be anti-symmetric with respect to particle exchange.

 

 

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Non-relativistic multi-electron Hamiltonian

The three dimensional Hamiltonian operator \hat{H}_T for an n-electron atom (excluding spin-orbit and other interactions) is:

\hat{H}_T=\frac{1}{2m_e}\sum_{i=1}^{n}\hat{p}_i^{\;2}-\sum_{i=1}^{n}\frac{Ze^{2}}{4\pi\varepsilon_0r_i}+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{e^{2}}{4\pi\varepsilon_0r_{ij}}\; \; \; \; \; \; \; \; 239

where \hat{p}_i^{\;2}=\hat{p}_{ix}^{\;2}+\hat{p}_{iy}^{\;2}+\hat{p}_{iz}^{\;2}=\left ( \frac{\hbar}{i}\frac{\partial}{\partial x} \right )^{2}+\left ( \frac{\hbar}{i}\frac{\partial}{\partial y} \right )^{2}+\left ( \frac{\hbar}{i}\frac{\partial}{\partial z} \right )^{2}, r_i=\sqrt{x_i^{\;2}+y_i^{\;2}+z_i^{\;2}} and r_{ij}=\vert\boldsymbol{\mathit{r}}_i-\boldsymbol{\mathit{r}}_j\vert=\sqrt{(x_i-x_j)^{2}+(y_i-y_j)^{2}+(z_i-z_j)^{2}}

or

\hat{H}_T=-\frac{\hbar^{2}}{2m_e}\sum_{i=1}^{n}\nabla_i^{\;2}-\sum_{i=1}^{n}\frac{Ze^{2}}{4\pi\varepsilon_0r_i}+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{e^{2}}{4\pi\varepsilon_0r_{ij}}\; \; \; \; \; \; \; \; 240

where \nabla_i^{\;2}=\frac{\partial^{2}}{\partial x_i^{\;2}}+\frac{\partial^{2}}{\partial y_i^{\;2}}+\frac{\partial^{2}}{\partial z_i^{\;2}}.

It is the total energy operator of the atom. The first term consists of the kinetic energy operators of the atom’s electrons and acts on a function of r, \theta and \phi. The second term, which is the electrostatic potential energy for the attractions between the electrons and the protons, acts on a function of just r because we have assumed an infinitely heavy nucleus that is reduced to a point at the origin. The third term, which acts on a function of r, \theta and \phi, is the electrostatic potential energy of electron repulsions. To illustrate the double summation for the last term, we consider a four-electron system with electrons e1, e2, e3 and e4. The possible interacting potentials, in the form of ej–ek, are:

which can be represented by the double summation {\color{Red} \sum_{i=1}^{n-1}}{\color{Blue} \sum_{j=i+1}^{n}}. Note that this double summation can also be represented as \frac{1}{2}{\color{Blue} \sum_{j\neq i}}{\color{red} \sum_{i=1}^{n}}, which is illustrated as:

or {\color{Blue} \sum_{j> i}}{\color{red} \sum_{i=1}^{n}}, which is

To show that eq239 commutes with the components of the total orbital angular momentum operator \hat{\boldsymbol{\mathit{L}}}^{(T)}, we consider a two electron system. As we know, \hat{\boldsymbol{\mathit{L}}}^{(T)}=\hat{\boldsymbol{\mathit{L}}}_1+\hat{\boldsymbol{\mathit{L}}}_2, where \hat{\boldsymbol{\mathit{L}}}_1 and \hat{\boldsymbol{\mathit{L}}}_2 are the individual orbital angular momentum of the system. Using eq76, we have

\hat{\boldsymbol{\mathit{L}}}^{(T)}=\hat{\boldsymbol{\mathit{i}}}\hat{L}_{1x}+\hat{\boldsymbol{\mathit{j}}}\hat{L}_{1y}+\hat{\boldsymbol{\mathit{k}}}\hat{L}_{1z}+\hat{\boldsymbol{\mathit{i}}}\hat{L}_{2x}+\hat{\boldsymbol{\mathit{j}}}\hat{L}_{2y}+\hat{\boldsymbol{\mathit{k}}}\hat{L}_{2z}

\hat{\boldsymbol{\mathit{L}}}^{(T)}=\hat{\boldsymbol{\mathit{i}}}(\hat{L}_{1x}+\hat{L}_{2x}) +\hat{\boldsymbol{\mathit{j}}}(\hat{L}_{1y}+\hat{L}_{2y}) +\hat{\boldsymbol{\mathit{k}}}(\hat{L}_{1z}+\hat{L}_{2z})

where (\hat{L}_{1x}+\hat{L}_{2x}), (\hat{L}_{1y}+\hat{L}_{2y}) and (\hat{L}_{1z}+\hat{L}_{2z}) are components of \hat{\boldsymbol{\mathit{L}}}^{(T)}.

For the uncoupled case, where the orbital angular momenta of the two electrons do not interact, eq239 becomes \hat{H}_{T,uc}=\frac{1}{2m_e}\sum_{i=1}^{2}\hat{p}_i^{\;2}-\sum_{i=1}^{2}\frac{Ze^{2}}{4\pi\varepsilon_0r_i}. From eq103, eq104 and eq105, we have [\hat{L}_{mk},\frac{1}{r}]=0 and [\hat{L}_{mk},\hat{p}^{2}]=0, where m=1,2 and k=x,y,z. This implies that the components of \hat{\boldsymbol{\mathit{L}}}_1 and \hat{\boldsymbol{\mathit{L}}}_2 commutes with \hat{H}_{T,uc}, and that \left [ \hat{L}_k^{\;(T)}=\hat{L}_{1k}+\hat{L}_{2k},\frac{1}{r} \right ]=0 and \left [ \hat{L}_k^{\;(T)},\hat{p}^{2} \right ]=0, and hence \left [ \hat{L}_k^{\;(T)},\sum_{i=1}^{2}\frac{1}{r}_i \right ]=0 and \left [ \hat{L}_k^{\;(T)},\sum_{i=1}^{2}\hat{p}_i^{\;2} \right ]=0. Therefore, when we analyse the commutation relations of the components of \hat{\boldsymbol{\mathit{L}}}_1 and \hat{\boldsymbol{\mathit{L}}}_2 with \hat{H}_T, we are left with the commutation relations of the components of \hat{\boldsymbol{\mathit{L}}}_1 and \hat{\boldsymbol{\mathit{L}}}_2 with \frac{1}{\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert}.

 

Question

Show that the components of \hat{\boldsymbol{\mathit{L}}}^{(T)}, but not those of \hat{\boldsymbol{\mathit{L}}}_1 and \hat{\boldsymbol{\mathit{L}}}_2, commute with \frac{1}{\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert}.

Answer

Using eq74,

\left [ \hat{L}_{1z},\frac{1}{\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert} \right ]=\frac{\hbar}{i\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert^{3/2} }\left [ -x_1(y_1-y_2)+y_1(x_1-x_2) \right ]\neq0\; \; \; \; \; \; \; \; 241

\left [ \hat{L}_{2z},\frac{1}{\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert} \right ]=\frac{\hbar}{i\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert^{3/2} }\left [ x_2(y_1-y_2)-y_2(x_1-x_2) \right ]\neq0\; \; \; \; \; \; \; \; 242

Similarly,  the x,y-components of the uncoupled operators \hat{\boldsymbol{\mathit{L}}}_1 and \hat{\boldsymbol{\mathit{L}}}_2 do not commute with with \frac{1}{\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert}. Therefore, all components of \hat{\boldsymbol{\mathit{L}}}_1 and \hat{\boldsymbol{\mathit{L}}}_2 do not commute with the Hamiltonian \hat{H}_T.

Using the identity [\hat{A}+\hat{B},\hat{C}]=[\hat{A},\hat{C}]+[\hat{B},\hat{C}] and substituting eq241 and eq242 into \left [ \hat{L}_{1z}+\hat{L}_{2z},\frac{1}{\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert} \right ], we have \left [ \hat{L}_{1z}+\hat{L}_{2z},\frac{1}{\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert} \right ]=0. Similarly, we find that \left [ \hat{L}_{1x}+\hat{L}_{2x},\frac{1}{\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert} \right ]=0 and \left [ \hat{L}_{1y}+\hat{L}_{2y},\frac{1}{\vert\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2\vert} \right ]=0. So,

\left [ \hat{L}_x^{\;(T)},\hat{H}_T \right ]=\left [ \hat{L}_y^{\;(T)},\hat{H}_T \right ]=\left [ \hat{L}_z^{\;(T)},\hat{H}_T \right ]=0 \; \; \; \; \; \; \; \; 243

 

Therefore, the components of the coupled total orbital angular momentum operator commutes with the Hamiltonian \hat{H}_T. If so, we can evaluate \left [ \hat{{L}^{2}}^{(T)},\hat{H}_T \right ] using the identities [\hat{A}+\hat{B}+\hat{C},\hat{D}]=[\hat{A},\hat{D}]+[\hat{B},\hat{D}]+[\hat{C},\hat{D}] and [\hat{A}\hat{B},\hat{C}]=[\hat{A},\hat{C}]\hat{B}+\hat{A}[\hat{B},\hat{C}] and eq243, which gives us

\left [ \hat{{L}^{2}}^{(T)},\hat{H}_T \right ]=0\; \; \; \; \; \; \; \; 244

Finally, \hat{H}_T also commutes with \hat{{S}^{2}}^{(T)} and \hat{S}_z^{\;(T)} because \hat{H}_T and spin angular momentum operators act on different vector spaces.

 

Question

Show that \hat{\boldsymbol{\mathit{L}}}^{(T)}\cdot\hat{\boldsymbol{\mathit{S}}}^{(T)} commutes with \hat{H}_T, where \hat{\boldsymbol{\mathit{L}}}^{(T)}=\sum_{i=1}^{n}\hat{\boldsymbol{\mathit{L}}}_i and .\hat{\boldsymbol{\mathit{S}}}^{(T)}=\sum_{i=1}^{n}\hat{\boldsymbol{\mathit{S}}}_i.

Answer

\left [ \hat{\boldsymbol{\mathit{L}}}^{T}\cdot\hat{\boldsymbol{\mathit{S}}}^{T},\hat{H}_T\right ]=\left [\hat{L_x^{\;(T)}}\hat{S_x^{\;(T)}}+\hat{L_y^{\;(T)}}\hat{S_y^{\;(T)}}+\hat{L_z^{\;(T)}}\hat{S_z^{\;(T)}},\hat{H}_T \right ]

Using the identities [\hat{A}+\hat{B}+\hat{C},\hat{D}]=[\hat{A},\hat{D}]+[\hat{B},\hat{D}]+[\hat{C},\hat{D}] and [\hat{A}\hat{B},\hat{C}]=[\hat{A},\hat{C}]\hat{B}+\hat{A}[\hat{B},\hat{C}], and noting that every component of \hat{\boldsymbol{\mathit{L}}} commutes with every component of \hat{\boldsymbol{\mathit{S}}} because they act on different vector spaces, and that the components of \hat{\boldsymbol{\mathit{S}}} commutes with \hat{H}_T because they act on different vector spaces, we have

\left [ \hat{\boldsymbol{\mathit{L}}}^{T}\cdot\hat{\boldsymbol{\mathit{S}}}^{T},\hat{H}_T\right ]=0\; \; \; \; \; \; \; \; 245

 

 

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Pauli exclusion principle

The Pauli exclusion principle states that no two identical fermions (particles with spin s=\frac{1}{2},\frac{3}{2},\frac{5}{2},\cdots) can occupy the same state.

The principle is a consequence of the postulate that the wavefunction of a system of fermions is antisymmetric with respect to label exchange, i.e. the wavefunction changes sign when the labels are exchanged.

The identical property of particles is fulfilled by the linear combinations:

\psi_{\pm}(r_1,r_2)=c\left [ \psi_{r_1}(1)\psi_{r_2}(2)\pm\psi_{r_1}(2)\psi_{r_2}(1)\right ]

where the labels 1 and 2 denote the first particle and second particle respectively, \psi_{r_j}(i) denotes particle i in the state \psi_{r_j} and there is equal probability \vert c\vert^{2} of the wavefunction \psi_{\pm} collapsing to \psi_{r_1}(1)\psi_{r_2}(2) and \psi_{r_1}(2)\psi_{r_2}(1) upon measurement.

 

Question

Show that \psi_+ is symmetric with respect to label exchange, while \psi_- is antisymmetric with respect to label exchange.

Answer

Exchanging the labels of \psi_+=c\left [ \psi_{r_1}(1)\psi_{r_2}(2)+\psi_{r_1}(2)\psi_{r_2}(1)\right ] and \psi_-=c\left [ \psi_{r_1}(1)\psi_{r_2}(2)-\psi_{r_1}(2)\psi_{r_2}(1)\right ],

c\left [ \psi_{r_1}(2)\psi_{r_2}(1)+\psi_{r_1}(1)\psi_{r_2}(2)\right ]=\psi_+

c\left [ \psi_{r_1}(2)\psi_{r_2}(1)-\psi_{r_1}(1)\psi_{r_2}(2)\right ]=-\psi_-

 

Since the wavefunction of a system of fermions is antisymmetric with respect to label exchange, it must be expressed by

\psi_-(r_1,r_2)=c\left [ \psi_{r_1}(1)\psi_{r_2}(2)-\psi_{r_1}(2)\psi_{r_2}(1)\right ]

When \psi_{r_1}=\psi_{r_2}, the wavefunction \psi_-=0, which contradicts the proposition that an eigenfunction is a non-zero function. Therefore, no two identical fermions can occupy the same state.

If the system is an atom, the total wavefunction (spatial + spin) describing two identical electrons (s=\frac{1}{2}) is

\Psi(n,l,m_l,m_s)=\psi(n,l,m_l)\sigma(m_s)

\psi(n,l,m_l)=\psi_{\pm}=\phi_{n_1,l_1,m_{l1}}(1)\phi_{n_2,l_2,m_{l2}}(2)\pm\phi_{n_1,l_1,m_{l1}}(2)\phi_{n_2,l_2,m_{l2}}(1) is the composite spatial wavefunction of the system, where \phi is a one-electron spatial wavefunction (also known as hydrogenic wavefunction). \sigma(m_s) is the composite spin wavefunction of the system and is explicitly given by eq222 through eq225. Since \Psi must be antisymmetric with respect to label exchange

\Psi=\left\{\begin{matrix} \psi_+\sigma_d\\\psi_-\sigma_{a\: or\: b\: or\: c} \end{matrix}\right.

When n_1=n_2, l_1=l_2 and m_{l1}=m_{l2} only \psi_+\sigma_d is non-zero. Since \sigma_d refers to the singlet state (i.e. a state with anti-parallel spins, where m_{s1}\neq m_{s2}), no two identical electrons in an atom can have the same set of quantum numbers. This is consistent with the general statement that no two identical fermions can occupy the same state because a distinct quantum state of an atom is characterised by a particular set of quantum numbers, e.g. n=1, l=1, m_l=0 and m_s=\frac{1}{2}.

 

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Exchange operator (quantum mechanics)

The exchange operator \hat{P}_{ik} acts on a function resulting in the swapping of labels of any two identical particles, i.e.

\hat{P}_{ik}\psi(q_1,\cdots,q_i,\cdots,q_k,\cdots,q_n)=\psi(q_1,\cdots,q_k,\cdots,q_i,\cdots,q_n)

where the label q_n refers to the n-th particle.

If we apply \hat{P}_{ik} twice on the function,

\hat{P}_{ik}^{\; \; 2}f(\cdots,q_i,\cdots,q_k,\cdots)=\hat{P}_{ik}f(\cdots,q_k,\cdots,q_i,\cdots)=f(\cdots,q_i,\cdots,q_k,\cdots)

Therefore, \hat{P}_{ik}^{\; \; 2}=\hat{I}, where \hat{I} is the identity operator.

 

Question

Show that the eigenvalues of \hat{P}_{ik} are \lambda=\pm1.

Answer

If is an eigenfunction of , then \hat{P}_{ik}^{\; \; 2}\psi=\lambda\hat{P}_{ik}\psi=\lambda^{2}\psi. Since \hat{P}_{ik}^{\; \; 2}=\hat{I}, we have \psi=\lambda^{2}\psi. As an eigenfunction must be non-zero,  \lambda=\pm1.

 

Experiment data reveals that the wavefunction of a system of two identical fermions (particles with spin s=\frac{1}{2},\frac{3}{2},\frac{5}{2},\cdots)  is antisymmetric with respect to label exchange (i.e. the eigenvalue is -1 when the exchange operator acts on the wavefunction), while the wavefunction of a system of two identical bosons (s=0,1,2,\cdots) is symmetric with respect to label exchange (eigenvalue of +1). The antisymmetric property of fermion wavefunctions and the symmetric property of boson wavefunctions can be regarded as postulates of quantum mechanics.

For identical fermions, \hat{P}_{ik}\psi(\cdots,q_i,\cdots,q_k,\cdots)=-\psi(\cdots,q_k,\cdots,q_i,\cdots). Since the way we label identical particles cannot affect the state of the system, the commutation relation between \hat{P}_{ik} and the Hamiltonian is

\left [\hat{P}_{ik},\hat{H}\right ]\psi=\hat{P}_{ik}\hat{H}\psi-\hat{H}\hat{P}_{ik}\psi=E\hat{P}_{ik}\psi+\hat{H}\psi=-E\psi+E\psi=0

This implies that, for a system of identical fermions, we can select a common complete set of eigenfunctions for \hat{P}_{ik} and \hat{H}, with the eigenfunctions being antisymmetric under label exchange.

 

Question

Show that \hat{P}_{ik} commutes with \hat{{S}^{2}}^{(T)}, \hat{S}_z^{\; (T)}, \hat{S}_1^{\; 2}, \hat{S}_2^{\; 2}, \hat{\boldsymbol{\mathit{S}}}_1\cdot\hat{\boldsymbol{\mathit{S}}}_2, \hat{{L}^{2}}^{(T)}, \hat{L}_z^{\; (T)}, \hat{L}_{1z}, \hat{S}_{1z}, \hat{\boldsymbol{\mathit{S}}}^{(T)}\cdot\hat{\boldsymbol{\mathit{L}}}^{(T)}, \hat{J}^{2} and \hat{J}_z.

Answer

\left [ \hat{P}_{12},\hat{S}_1^{\; 2} \right ]\psi=s_1(s_1+1)\hbar^{2}\hat{P}_{12}\psi+\hat{S}_1^{\;2}\psi=-s_1(s_1+1)\hbar^{2}\psi+s_1(s_1+1)\hbar^{2}\psi=0

Similarly, \left [ \hat{P}_{12},\hat{S}_2^{\; 2} \right ]=0. For \hat{{S}^{2}}^{(T)} and \hat{S}_z^{\;(T)}, we have \hat{{S}^{2}}^{(T)}\psi=S(S+1)\hbar^{2}\psi and \hat{S}_z^{\;(T)}\psi=m_S\hbar\psi respectively. So \left [\hat{P}_{12} ,\hat{{S}^{2}}^{(T)}\right ]=0 and \left [\hat{P}_{12} ,\hat{S}_z^{\;(T)}\right ]=0. Since \hat{{S}^{2}}^{(T)}=\hat{S}_1^{\;2}+2\hat{\boldsymbol{\mathit{S}}}_1\cdot\hat{\boldsymbol{\mathit{S}}}_2+\hat{S}_2^{\;2}, we have \left [\hat{P}_{12},\hat{\boldsymbol{\mathit{S}}}_1\cdot\hat{\boldsymbol{\mathit{S}}}_2\right ]=0. Similarly, \hat{P}_{12} commutes with \hat{{L}^{2}}^{(T)}, \hat{L}_z^{\;(T)}, \hat{L}_{1z} and \hat{S}_{1z}.

Next, \left [\hat{P}_{12},\hat{\boldsymbol{\mathit{S}}}^{(T)}\cdot\hat{\boldsymbol{\mathit{L}}}^{(T)}\right ]=\left [\hat{P}_{12},\left ( \hat{\boldsymbol{\mathit{S}}}_1+\hat{\boldsymbol{\mathit{S}}}_2\right )\cdot\left ( \hat{\boldsymbol{\mathit{L}}}_1+\hat{\boldsymbol{\mathit{L}}}_2\right )\right ]. Expanding this equation using eq76 and eq179, and noting that \left [\hat{P}_{12},\hat{L}_{mk}\right ]=0 and \left [\hat{P}_{12},\hat{S}_{mk}\right ]=0, where m=1,2 and k=x,y,z, we have \left [\hat{P}_{12},\hat{\boldsymbol{\mathit{S}}}^{(T)}\cdot\hat{\boldsymbol{\mathit{L}}}^{(T)}\right ]=0. It follows that \left [\hat{P}_{12},\hat{J}^{2}\right ]=\left [\hat{P}_{12},\hat{{L}^{2}}^{(T)}\right ]+\left [\hat{P}_{12},\hat{{S}^{2}}^{(T)}\right ]+\left [\hat{P}_{12},2\hat{\boldsymbol{\mathit{S}}}^{(T)}\cdot\hat{\boldsymbol{\mathit{L}}}^{(T)}\right ]=0 and that \left [\hat{P}_{12},\hat{J}_z\right ]=\left [\hat{P}_{12},\hat{L}_z^{\; (T)}\right ]+\left [\hat{P}_{12},\hat{S}_z^{\; (T)}\right ]=0.

 

 

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BB84 protocol

BB84 is a quantum key distribution protocol that is based on photon polarisation states. It was invented by Charles Bennett and Gilles Brassard in 1984. The protocol can be broken down into the following steps:

1) The sender transmits a raw key in the form of a string of photons via an optical fibre to the receiver. Each photon is randomly chosen by the sender to be polarised along one of four directions: i) z-direction or \updownarrow  (\theta=0^{\circ}), ii) x-direction or \leftrightarrow (\theta=90^{\circ}), iii) ⤢ (\theta=45^{\circ}), iv) ⤡ (\theta=-45^{\circ}). This can be accomplished by passing photons generated by a laser L through an attenuator A (a material that reduces the intensity of the laser beam through absorption or reflection to produce single photons), followed by a birefringent crystal B_1 to separate photons into two orthogonal polarisations, one of which is passed through a Pockels cell P_1 for the appropriate rotations (see diagram below).

2) The polarised photons are assigned the following qubit values: \updownarrow=0, \leftrightarrow=1, ⤢ = 0, ⤡ = 1, which can be grouped into the following bases:

Basis 0 1
+ \updownarrow \leftrightarrow
\times ⤢ ⤡

3) The receiver, using Pockels cell B_2, randomly selects one of two rotations (90^{\circ} representing the basis +, or 45^{\circ} representing the basis \times) to analyse each photon sent by the sender. The outputs of the detectors D_1 and D_2 are set to 1 and 0 respectively.

Assuming a raw key of twelve bits are sent, we may have the following case:

Sender Polarisation \small \updownarrow \small \leftrightarrow \small \leftrightarrow ⤡ ⤢ \small \leftrightarrow ⤢ \small \updownarrow ⤡ ⤡ \small \updownarrow \small \leftrightarrow
Qubit value 0 1 1 1 0 1 0 0 1 1 0 1
Receiver Basis \small + \small + \small \times \small + \small + \small \times \small \times \small \times \small \times \small + \small \times \small +
Qubit value 0 1 1 0 0 0 0 0 1 1 1 1
Retained qubits 0 1 – – – – 0 – 1 – – 1

1) The sender and receiver share their bases with each other publicly (e.g. over the internet).

2) Only the qubits corresponding to polarisations with the same basis are retained. To complete the protocol, all other qubits can be converted to 0. The final key in the above example is 010000001001, which is used to encrypt a message (assuming of the same qubit-length) using modular addition.

Let’s suppose a third party intercepts the sender’s qubits and measures them using the same method as the receiver. The interceptor then has to guess the sender’s polarisation for each measured qubit and resend them to the receiver (see table below). Since the interceptor does not know the bases selected by the sender, the bits he measured are only useful if he is lucky enough to choose the same bases for the qubits that the sender and receiver eventually retained.

Sender Polarisation \small \updownarrow \small \leftrightarrow \small \leftrightarrow ⤡ ⤢ \small \leftrightarrow ⤢ \small \updownarrow ⤡ ⤡ \small \updownarrow \small \leftrightarrow
Qubit value 0 1 1 1 0 1 0 0 1 1 0 1
Interceptor Basis \small \times \small + \small \times \small \times \small + \small + \small + \small \times \small + \small + \small \times \small \times
Qubit value 0 1 0 1 1 1 0 1 0 0 0 1
Resends ⤢ \small \leftrightarrow ⤢ ⤡ \small \leftrightarrow \small \leftrightarrow \small \updownarrow ⤡ \small \updownarrow \small \updownarrow ⤢ ⤡
Receiver Basis \small + \small + \small \times \small + \small + \small \times \small \times \small \times \small \times \small + \small \times \small +
Qubit value 1 1 0 0 1 0 0 1 0 0 0 1
Retained qubits 1 1 – – – – 0 – 0 – – 1
Discrepancies Y N – – – – N – Y – – N

To detect the presence of an interceptor, the sender and receiver share pre-determined segments of qubits with each other. The probability that the sender and receiver select the same bases but the interceptor chooses different bases is \small \frac{1}{4}, of which half the time the interceptor resend polarisations of the ‘wrong’ basis to the receiver. Therefore, the probability of the sender and the receiver detecting discrepancies in the retained qubits when they share segments of their qubit values is \small \frac{1}{8} (assuming the shared qubits are representative of a randomly selected segment), which is very significant if the qubit-length of the raw key is relatively long. For example, if the raw key has 72 qubits, of which 40 common qubits are eventually shared between the sender and the receiver, 5 retained qubits will be erroneous. When discrepancies are detected, the sender and receiver will discard the key and start over.

 

 

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Photon polarisation

Photon polarisation is the quantum mechanical treatment of light polarisation. Photoelectric effect experiments have shown that the direction of electron emission depends on the polarisation of the incident light. In quantum mechanics, classical polarisation directions are referred to as quantum states, with the states \vert z\rangle and \vert x\rangle representing linear polarisation parallel to the z-axis and the x-axis respectively (see diagram below).

If we consider a unit circle, the state of a photon \vert\psi\rangle polarised at \theta with respect to the z-axis can therefore be written as a linear combination of \vert z\rangle and \vert x\rangle:

\vert\psi\rangle=cos\theta\vert z\rangle+sin\theta\vert x\rangle\; \; \; \; \; \; \; \; 237

where \vert z\rangle and \vert x\rangle become basis state vectors.

Since these basis vectors are unit vectors \vert z\rangle=\begin{pmatrix} 1\\0 \end{pmatrix} and \vert x\rangle=\begin{pmatrix} 0\\1 \end{pmatrix}, we can express the state of the photon as \vert \psi\rangle=\begin{pmatrix} cos\theta\\sin\theta \end{pmatrix}.

 

Question

Can a single photon be unpolarised?

Answer

Each photon in a string of photons (e.g. from the sun) has a definite polarisation, even though one photon’s oscillation direction may be randomly oriented relatively to another photon’s oscillation direction.

 

As per any normalised quantum state of \vert\psi\rangle=\sum_{i=1}^{2}c_i\vert\phi_i\rangle, the square of each coefficient \vert c_i\vert^{2} is interpreted as the probability that a measurement of a system will yield an eigenvalue associated with the corresponding eigenstate \phi_i, i.e. cos^{2}\theta +sin^{2}\theta=1. If we measure the polarisation of a string of photons, whose generic states are given by eq237, using a calcite crystal (see diagram above; the optic axis lies in the plane of the viewer’s screen), photons with oscillations perpendicular to the plane of the screen emerge as part of the o-ray with the state \vert\psi\rangle=\vert x\rangle, while photons with oscillations parallel to the plane emerge as part of the e-ray with the state \vert\psi\rangle=\vert z\rangle. A photon oscillating at \theta=\pm45^{\circ} (\vert\psi\rangle=\frac{1}{\sqrt{2}}[\vert z\rangle\pm\vert x\rangle]) has 50% probability of emerging as part of the o-ray and 50% probability emerging as part of the e-ray.

 

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Pockels cell

A Pockels cell is an electro-optical device that rotates the polarisation of light. It is based on the Pockels Effect, where the birefringence of the crystal in the cell changes linearly with an applied voltage (see diagram below).

An example of a crystal exhibiting the Pockels Effect is ammonium dihydrogen phosphate NH4H2PO4, which has a tetragonal crystal structure. It is aligned with its optic axis parallel to the propagation direction of the light. In the absence of an external electric field, the incoming beam suffers no birefringence. When an external electric field is applied across the crystal, the electron distribution of the crystal is distorted to the extent that the relative values of n_o and n_e changes, with the degree of change being proportional to the applied voltage.

To explain the change in polarisation direction of light, let’s regard the vertically polarised light as a superposition of two circular light components that are rotating in phase with the same angular frequency \omega but in opposite directions (see above diagram). When the external electric field is turned on, one component travels faster than the other in the Pockels cell. For a cell of length l, the difference in time \Delta t travelled by each component is:

\Delta t=\frac{l}{c_o}-\frac{l}{c_e}\; \; \; \; \; \; \; \; 236

where c_o and c_e are the speeds of the two components in the cell.

Substitute the definition of refractive index (n_x=\frac{c}{c_x}, where c is the speed of light in vacuum) and eq235 in eq236

\Delta t=\left ( \frac{n_o}{c}-\frac{n_e}{c} \right )l=\Delta n\frac{l}{c}

Substitute the above equation in the definition of phase difference (\Delta \phi=\omega\Delta t=2\pi\frac{c}{\lambda}\Delta t) between the two components,

\Delta \phi=\Delta n\frac{2\pi l}{\lambda}

As the birefringence of the crystal \Delta n changes proportionally with voltage V, the phase difference \Delta \phi between the two circular components changes with an applied voltage. Since the angle of rotation of polarisation is \Delta\theta=\frac{\Delta \phi}{2} (you can visualise the rotation angle using the diagram above),

\Delta\theta=\kappa V\frac{\pi l}{\lambda}

where \kappa is a proportionality constant, which is specific to the material used.

 

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Birefringence

Birefringence is the optical property of a material in which an incident ray of light is split into two perpendicularly polarised rays by the material. One material exhibiting strong birefringence is calcite (CaCO3), which has a hexagonal unit cell as shown in the diagram below.

When a single crystal of calcite is oriented in a particular direction, we see layers of atoms arranged symmetrically. Each layer consists of carbonate molecules in a plane and calcium atoms directly above the carbon atoms (see above diagram). The layers are stacked in an ABBAABBA… fashion, where the oxygen atoms in B are rotated 60o relative the oxygen atoms in A. Due to the symmetry of the atoms, a ray of unpolarised light passing perpendicular to the layers (oscillation parallel to the layers) emerges as a single ray of unpolarised light, i.e. it suffers no birefringence. The direction in which the ray that suffers no birefringence travels is called the optic axis of the crystal.

Birefringence occurs when a ray of unpolarised light passes through a calcite crystal, which is oriented relative to the ray as shown in the diagram above. For simplicity, only two orthogonal oscillations of the incident ray are depicted, with the up-down arrows representing polarisation along the plane of the viewer’s screen and the blue dots representing polarisation into the plane of the screen. The surface of the calcite crystal facing the reader and the optic axis of the crystal are parallel to the plane of the screen. As mentioned earlier, oscillations that are perpendicular to the optic axis (i.e. blue dots) suffer no birefringence, with its path unaltered. The ray of such oscillations that emerges from the crystal is called an ordinary ray or o-ray.

The polarisation represented by the double arrows are not travelling perpendicular to the optic axis and encounters electron densities that are not symmetrically distributed. The induced oscillations in different parts of those electron densities reradiate light waves that combine to form a refracted wave. The refracted light travels through the crystal with a different speed v and emerges as a separate ray, which is called an extraordinary ray or e-ray. This implies that the crystal has two different refractive indices n_o and n_e. The birefringence \Delta n of a material is defined as:

\Delta n=n_e-n_o\; \; \; \; \; \; \; \; 235

 

Question

Why is the e-ray deflected ‘upwards’ in the crystal?

Answer

The incident ray, with oscillation parallel to the plane of the viewer’s screen, can be further resolved into a component that is perpendicular to the optic axis (oscillation parallel to the optic axis) and one that is parallel to the optic axis (oscillation perpendicular to the optic axis, see diagram below).

We can imagine that the oscillation of the latter component encounters a denser distribution of electrons as it travels along the optic axis in the crystal (refer to the structure of calcite in the earlier diagram) as compared to the other component. Therefore, v_{\parallel}> v_{\perp} and the resultant ray is deflected “upwards”.

 

 

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Light polarisation

Light polarisation is the classical description of oscillation directions of electromagnetic waves. Light from the sun or incandescent lamps is emitted by a large number of atoms, each producing oscillations in random directions at any given time. Since the oscillations at any point fluctuates rapidly and randomly, we call this unpolarised light (see diagram below).

 

Question

Do the arrows in the above diagram represent electric field vectors or magnetic field vectors?

Answer

It is more convenient, and in many cases more practical, to define polarisation of light in terms of the oscillation of electric field vectors (a magnetic field does not exert a force on charges at rest).

 

Polarised light can be produced by passing unpolarised light through a polariser. One of the most widely used polarisers is the H-sheet, which is a sheet of polyvinyl alcohol that is heated and stretched to the extent that the polymeric chains are aligned in a particular direction. Long chains of iodine, which are parallel to the alcohol chains, are then formed when the sheet is dipped in iodine solution. As the valence electrons in the iodine chains are only mobile along the chains, light that is polarised parallel to the chains is absorbed, while light that is polarised perpendicular to the chains is transmitted (see diagram below).

 

Question

What happens when light, which is polarised at an angle \theta to the iodine chains where 0^{\circ}< \theta< 90^{\circ}, passes through the H-sheet?

Answer

The polarised light before passing through the H-sheet can be resolved into a component that is parallel to the iodine chains and one that is perpendicular to the iodine chains. Therefore, the transmitted light is polarised perpendicular to the chains.

 

 

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