Advanced Chemistry - Mono Mole

July 14, 2022November 17, 2024

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,

$\begin{align}\langle(\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2)^{2}\rangle =&\frac{1}{2}\left ( \langle r_1^{\;2}\rangle_a\pm\langle r_2^{\;2}\rangle_a+\langle r_2^{\;2}\rangle_b\pm\langle r_1^{\;2}\rangle_b\right )-\langle\boldsymbol{\mathit{r}}_1\rangle_a\langle\boldsymbol{\mathit{r}}_2\rangle_b \mp\langle\boldsymbol{\mathit{r}}_1\rangle_b\langle\boldsymbol{\mathit{r}}_2\rangle_a\\&\mp\int \phi_a^{\;*}(r_1)\boldsymbol{\mathit{r}}_1\phi_b(r_1)dr_1\int \phi_b^{\;*}(r_2)\boldsymbol{\mathit{r}}_2\phi_a(r_2)dr_2\\&\mp\int \phi_b^{\;*}(r_1)\boldsymbol{\mathit{r}}_1\phi_a(r_1)dr_1\int \phi_a^{\;*}(r_2)\boldsymbol{\mathit{r}}_2\phi_b(r_2)dr_2\end{align}$

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,

$\begin{align}\langle(\boldsymbol{\mathit{r}}_1-\boldsymbol{\mathit{r}}_2)^{2}\rangle=&\frac{1}{2}(\langle r_1^{\;2}\rangle_a\pm\langle r_2^{\;2}\rangle_a+\langle r_2^{\;2}\rangle_b\pm\langle r_1^{\;2}\rangle_b)-\langle\boldsymbol{\mathit{r}}_1\rangle_a\langle\boldsymbol{\mathit{r}}_2\rangle_b\\&\mp\langle\boldsymbol{\mathit{r}}_1\rangle_b\langle\boldsymbol{\mathit{r}}_2\rangle_a\mp\langle\boldsymbol{\mathit{r}}_1\rangle_{ab}\langle\boldsymbol{\mathit{r}}_2\rangle_{ab}^{\;*}\mp\langle\boldsymbol{\mathit{r}}_1\rangle_{ab}^{\;*}\langle\boldsymbol{\mathit{r}}_2\rangle_{ab}\end{align}$

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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July 14, 2022April 5, 2026

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 $r_{ij}=\vert\boldsymbol{\mathit{r}}_i-\boldsymbol{\mathit{r}}_j\vert=\sqrt{(x_i-x_j)^{2}+(y_i-y_j)^{2}+(z_i-z_j)^{2}}$

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

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

Question

How is the potential energy term derived?

Answer

Let’s assume an electron with charge $-e$ produces a spherically symmetric electric field $\boldsymbol{\mathit{E}}$ , whose magnitude is proportional to the number of field lines crossing an element of area $\boldsymbol{\mathit{A}}$ perpendicular to the field. This is, in turn, proportional to the amount of charge enclosed by the surface:

$\oint\boldsymbol{\mathit{E}}\cdot d\boldsymbol{\mathit{A}}=E\times 4\pi r^2=-he=-\frac{e}{\varepsilon_0}$

where the proportionality constant $h$ is experimentally determined to be $\frac{1}{\varepsilon_0}$ , and $\varepsilon_0$ is the permittivity of free space (8.8542 x 10^-12Fm^-1).

If we place another electron in this field, the potential energy $V$ between the two electrons is the work done to move the second electron from infinity to a distance from the first electron:

$V=-\int^r_{\infty}Fdr=\int^r_{\infty}eEdr=-\int^r_{\infty}\frac{e^2}{4\pi\varepsilon_0r^2}dr=\frac{e^2}{4\pi\varepsilon_0r}$

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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