Identical Particles

Collide two particles, 1 and 2. If we have a detector \(\mathcal{D}\) which detects the particles after the collision. Did it detect 1, 2, or both? Exchange-degeneracy.

We have particle 1 at \(x=a\) and particle 2 at \(x=b\). If they are distinguishable then exchanging them gives a different setup: \(|a_{m_1}b_{m_2}\rangle \neq |a_{m_2}b_{m_1}\rangle\). If they are indistinguishable then exchanging them gives the same setup: \(|a_{m_1}b_{m_2}\rangle = |a_{m_2}b_{m_1}\rangle\).

Given \(N\) particles and \(N\) boxes then we have \(N!\) possible combinations to fill the boxes.

Given \(N\) particles and \(k\) boxes then we have \(\frac{N!}{(N-k)!} = k!{N\choose k}\) possible permutations to fill the boxes.

How many $k-$combinations can we get? With combinations the boxes are unlabled hence ab and ba are the same. Then we have \({N\choose k}\).

Note: \((x+y)^N = \sum_{k=0}^N {N\choose k}x^ky^{N-k}\).

\(E_{tot} = \frac{\hbar^2\pi^2}{mL^2}\). For each particle, \(\frac{\hbar^2\pi^2}{2mL^2}(n_1^2+n_2^2)\). Then, we have \(|11\rangle\) which is allowed for distinguishable and bosonic particles.

\(E_{tot} = \frac{5\hbar^2\pi^2}{2mL^2}\). For each particle, \(\frac{\hbar^2\pi^2}{2mL^2}(n_1^2+n_2^2)\). Then, we have \(|12\rangle,|21\rangle\) which is allowed for distinguishable. Boson: \(1/\sqrt{2}(|12\rangle+|21\rangle)\). Fermion: \(1/\sqrt{2}(|12\rangle-|21\rangle)\).

For more degrees of freedom, say spin, then a symmetric function would be both the coordinate and spin being symmetric or antisymmetric.

We will not consider relativisitic corrections, and have spin-dependent terms and a stationary nucleus.

\(H = \sum_i^Z \frac{p_i^2}{2m} - \sum_i^Z \frac{Ze^2}{R_i} + \sum_{i

If we didn’t have the last term, \(H = \sum_i^Z H_i\). Hence, \(\Psi = \Psi_1\cdots\Psi_Z\).

How many terms does the last term have? \(\frac{Z(Z-1)}{2}={Z\choose 2}\) terms.

The second term is roughly, \(\frac{Ze^2}{R}Z\) and the third is roughly \(\frac{e^2}{R}\frac{Z(Z-1)}{2}\).

The second term is 4 times larger for helium, not a great perturbation. As we increase \(Z\), we approach 2, which is definitely a great perturbation.

\(H_0\Psi = E\Psi\), \(H_0 = H_1 + H_2\), \(H = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \frac{e^2}{|\vec{R}_1-R_2|} - \frac{2e^2}{R_1} - \frac{2e^2}{R_2} = H_0 + V_p\), \(E = E_1^{n_1\ell_1} + E_2^{n_2\ell_2}\). Then, \(H_0 = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + V_c(R_1) + V_c(R_2)\).

Degeneracy:

1. \(n_1\neq n_2\vee\ell_1\neq\ell_2\): \(4(2\ell_1+1)(2\ell_2+1)\)
2. \(n_1=n_2\wedge\ell_1=\ell_2\): \((2\ell+1)(4\ell+1)\)

If we had a \(1s^2\) state then we would have \(|\Psi\rangle = \frac{1}{\sqrt{2}}\Psi_{n_1\ell_1m_1}\left(|+-\rangle - |-+\rangle\right)|100;100\rangle = \Psi_{1s}(\vec{r}_1)\Psi_{1s}(\vec{r}_2)\frac{1}{\sqrt{2}}(|+-\rangle - |-+\rangle)\). Spatial antisymmetric and spin symmetric OR Spatial symmetric and spin antisymmetric. The spin antisymmetric is only possible for the same spatial quantum numbers.

\(V_p = -\frac{2e^2}{R_1} - \frac{2e^2}{R_2} + \frac{e^2}{|\vec{R}_1-\vec{R}_2|} - V_c(R_1) - V_c(R_2)\).

Consider the 1s + 2s configuration. For our 4-fold degenerate space, we must find the 4 wavefunctions. \(|^1S\rangle = |\Psi_{spatial,S}\rangle\otimes|\Psi_{spin,A}\rangle = \Psi_{S}\otimes|s=0,m_S=0\rangle = \Psi_{S}\otimes|0,0\rangle = \Psi_{S}\otimes\frac{1}{\sqrt{2}}(|+-\rangle - |-+\rangle)\). \(|^2S\rangle = |\Psi_{spatial,A}\rangle\otimes|\Psi_{spin,S}\rangle = \frac{1}{\sqrt{2}}(|100;200\rangle - |200;100\rangle)\otimes\begin{cases}|10\rangle\\|11\rangle\\|1-1\rangle\end{cases}\).

So, \(\langle\Psi|V|\Psi\rangle = \langle\Psi_{spatial}|V|\Psi_{spatial}\rangle\). Note that the spin is unperturbed by our potential. From the spins, we see that our perturbation is already diagonal. I will call our two spatial wavefunctions by the spectroscopic notation. \(\langle ^1S|V|^1S\rangle,\langle ^3S|V|^3S\rangle\).

Then, \(\langle V\rangle = \frac{1}{2}\left[\int d^3\vec{r}_1d^3\vec{r}_2|\Psi_{1s}(\vec{r}_1)|^2V|\Psi_{2s}(\vec{r}_2)|^2 + \int d^3\vec{r}_1d^3\vec{r}_2|\Psi_{1s}(\vec{r}_2)|^2V|\Psi_{2s}(\vec{r}_1)|^2 \pm \int d^3\vec{r}_1d^3\vec{r}_2\Psi_{1s}^*(\vec{r}_1)\Psi_{2s}^*(\vec{r}_2)V\Psi_{1s}(\vec{r}_2)\Psi_{2s}(\vec{r}_1) \pm \int d^3\vec{r}_1d^3\vec{r}_2\Psi_{1s}^*(\vec{r}_2)\Psi_{2s}^*(\vec{r}_1)V\Psi_{1s}(\vec{r}_1)\Psi_{2s}(\vec{r}_2)\right] = \int d^3\vec{r}_1d^3\vec{r}_2|\Psi_{1s}(\vec{r}_1)|^2V|\Psi_{2s}(\vec{r}_2)|^2 \pm \int d^3\vec{r}_1d^3\vec{r}_2\Psi_{1s}^*(\vec{r}_1)\Psi_{2s}^*(\vec{r}_2)V\Psi_{1s}(\vec{r}_2)\Psi_{2s}(\vec{r}_1)\). Note that \(V\) has five terms so we get a trivial 10 integrals to evaluate.

Consider the first integral.

\begin{align*} \int d^6\vec{r}|\Psi_{1s}(\vec{r}_1)|^2V|\Psi_{2s}(\vec{r}_2)|^2 &= -2e^2\left(\int d^3\vec{r}_1\frac{|\Psi_{1s}(\vec{r}_1)|^2}{r_1} + \int d^3\vec{r}_2\frac{|\Psi_{1s}(\vec{r}_2)|^2}{r_2}\right) \\ &+ e^2 \int d^6\vec{r} \frac{|\Psi_{1s}(\vec{r}_1)|^2|\Psi_{2s}(\vec{r}_2)|^2}{|\vec{r}_1-\vec{r}_2|} \\ &- \int d^3\vec{r}_1|\Psi_{1s}(\vec{r}_1)|^2V_c(r_1) - \int d^3\vec{r}_2|\Psi_{2s}(\vec{r}_2)|^2V_c(r_2) \\ &= -2e^2I' + e^2K' - I_1 - I_2 = I + K - I_1 - I_2. \end{align*}

The cross terms,

\begin{align*} \int d^3\vec{r}_1d^3\vec{r}_2\Psi_{1s}^*(\vec{r}_1)\Psi_{2s}^*(\vec{r}_2)V\Psi_{1s}(\vec{r}_2)\Psi_{2s}(\vec{r}_1) &= -2e^2\left(\int d^3\vec{r}_1\frac{\Psi_{1s}(\vec{r}_1)\Psi_{2s}(\vec{r}_1)}{r_1}\int d^3\vec{r}_2\Psi_{2s}(\vec{r}_2)\Psi_{1s}(\vec{r}_2) + \int d^3\vec{r}_2\frac{\Psi_{1s}(\vec{r}_2)\Psi_{2s}(\vec{r}_2)}{r_2}\int d^3\vec{r}_1\Psi_{2s}(\vec{r}_1)\Psi_{1s}(\vec{r}_1)\right) (=0) \\ &+ e^2 \int d^6\vec{r} \frac{\Psi_{1s}(\vec{r}_1)\Psi_{2s}(\vec{r}_2)\Psi_{1s}(\vec{r}_2)\Psi_{2s}(\vec{r}_1)}{|\vec{r}_1-\vec{r}_2|} \\ &- \int d^3\vec{r}_1\Psi_{1s}(\vec{r}_1)\Psi_{2s}(\vec{r}_1)V_c(r_1)\int d^3\vec{r}_2\Psi_{1s}(\vec{r}_2)\Psi_{2s}(\vec{r}_2) - \int d^3\vec{r}_1\Psi_{1s}(\vec{r}_1)\Psi_{2s}(\vec{r}_1)V_c(r_1)\int d^3\vec{r}_1\Psi_{1s}(\vec{r}_1)\Psi_{2s}(\vec{r}_1) \\ &= -2e^20 + e^2J' - 0 - 0 = J. \end{align*}

Note that \(I_1, I_2\) are average energy for the central field for each electron. \(J\) is the exchange integral and is purely Quantum Mechanical.

Therefore, our full energies are \(E_+ = I-I_1-I_2+K+J\) and \(E_- = I-I_1-I_2+K-J\). Thus, we get a 4-fold degeneracy split into 1-fold and 3-fold degeneracy split by the exchange interaction.

So, if \(V_c(r_i) = -\frac{2e^2}{r_i}\) then \(V = \frac{e^2}{|\vec{r}_1-\vec{r}_2}\) and \(I-I_1-I_2 = 0\), Energies become \(K\pm J\). But note that this only contributes to an energy shift, not the split.

The singlet state has the higher energy, \(E_+\) since it is the symmetric spin wavefunction. The triplet state has the higher energy, \(E_-\) since it is the antisymmetric spin wavefunction.

In the context of helium, the singlet state is called parahelium and the triplet is orthohelium.

Consider 2 hydrogen nuclei (a,b) with an electron each (1,2).

So, \(\vec{r}_{a1}\) is the vector between nucleus a and electron 1. Our total set of vectors are: \(\vec{r}_{a1},\vec{r}_{a2},\vec{r}_{ab},\vec{r}_{b1},\vec{r}_{b2},\vec{r}_{12}\).

The Hamiltonian is then, \(-\frac{\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2) + e^2\frac{1}{r_{12}} - e^2\left(\frac{1}{\vec{r}_{a1}} + \frac{1}{\vec{r}_{a2}} + \frac{1}{\vec{r}_{b1}} + \frac{1}{\vec{r}_{b2}}\right)\) Let \(r_{ab}\) be a parameter of the system.

If \(r_{ab}\to\infty\) then \(H_{a1}=-\frac{\hbar^2}{2m_e}\nabla_1^2 - \frac{e^2}{r_{a1}},H_{b2} = -\frac{\hbar^2}{2m_e}\nabla_2^2 - \frac{e^2}{r_{b2}}\). Thus, \(H_0 = H_{a1} + H_{b2}\). The perturbation is then, \(V = \frac{e^2}{r_{12}} - e^2\left(\frac{1}{r_{a2}} + \frac{1}{r_{b1}}\right)\).

So, \(H_0\Psi_1 = E_0\Psi_1\), \(\Psi_1 = \Psi_a(\vec{r}_{a1})\Psi_b(\vec{r}_{b2})\). Then, \(H_0\Psi_1 = E_0\Psi_1 = 2E_I\Psi_1\).

The exchanged hamiltonian is then \(H_0' = H_{a2} + H_{b1}\). \(H_{a2} = -\frac{\hbar^2}{2m}\nabla_{2}^2 -\frac{e^2}{r_{a2}}, H_{b1} = \cdots\). Then, \(\Psi_2 = \Psi_a(\vec{r}_{a2})\Psi_b(\vec{r}_{b1})\) with the same energy, \(H_0'\Psi_2 = 2E_I\Psi_2\).

Then the full Hamiltonian is \(H = H_0 + V_{a1,b2} = H_0' + V_{a2,b1}\).

Our full wavefunction is then \(c_1\Psi_1 + c_2\Psi_2 + \phi\). Note that we will show \(|c_2| = |c_1|\).

\(H_1\Psi = E\Psi = (2E_I+\Delta E)\Psi\).

\begin{align} c_1H\Psi_1 + C_2H\Psi_2 + H\phi &= (2E_1+\Delta E)(c_1\Psi_1 + c_2\Psi_2+\phi) \\ c_1V_{a1,b2}\Psi_1 + c_2V_{a2,b1}\Psi_2 + H\phi &= 2E_1\phi + (\Delta E)(c_1\Psi_1 + c_2\Psi_2 + \phi) \nonumber \\ c_1V_{a1,b2}\Psi_1 &= \Delta Ec_1\Psi_1 \\ c_2V_{a2,b1}\Psi_2 &= \Delta Ec_2\Psi_2 \\ H\phi &= (2E_1 + \Delta E)\phi \end{align}

Reducing by neglecting second-order perturbation,

\begin{align} c_1V_{a1,b2}\Psi_1 &= \Delta Ec_1\Psi_1 \\ c_2V_{a2,b1}\Psi_2 &= \Delta Ec_2\Psi_2 \\ H_0\phi = H_0'\phi &= 2E_1 \phi \end{align}

Then,

\begin{align} (H_0-2E_I)\phi &= (\Delta E-V_{a1,b2})c_1\Psi_1 + (\Delta E-V_{a2,b1})c_2\Psi_2. \end{align}

Note the RHS is the inhomogeneity. If RHS is zero then, \(H_0\phi - 2E_1\phi\Rightarrow \phi = \Psi_1\) if \(H_0\) and \(\phi=\Psi_2\) if \(H_0'\).

For an inhomogenous equation, there is a solution only if the RHS is orthogonal to the solution of the homogenous equation. Then,

\begin{align} \int [(\Delta E-V_{a1,b2})c_1\Psi_1 + (\Delta E-V_{a2,b1})c_2\Psi_2]^*\Psi_1 d^3\vec{r}_1d^3\vec{r}_2 &= 0 \\ \int [(\Delta E-V_{a1,b2})c_1\Psi_1 + (\Delta E-V_{a2,b1})c_2\Psi_2]^*\Psi_2 d^3\vec{r}_1d^3\vec{r}_2 &= 0 \end{align}

Let,

\begin{align} K &= \int V_{a1,b2}|\Psi_1|^2d^3\vec{r}_1d^3\vec{r}_2 \\ K &= \int V_{a2,b1}|\Psi_2|^2d^3\vec{r}_1d^3\vec{r}_2 \\ \Delta E &= \int \Delta E|\Psi_{1,2}|^2 d^3\vec{r}_1d^3\vec{r}_2 \\ 0>J &= \int V_{a2,b1}\Psi_2^*\Psi_1d^3\vec{r}_1d^3\vec{r}_2 \\ 0>J &= \int V_{a1,b2}\Psi_1^*\Psi_2d^3\vec{r}_1d^3\vec{r}_2 \end{align}

Gives us,

\begin{align} \Delta E c_1 - c_1K + \Delta E c_2\int \Psi_2^*\Psi_1 d^3\vec{r}_1d^3\vec{r}_2 - c_2J &= 0 \\ \Delta E c_1 - c_1K + \Delta E c_2|s|^2 - c_2J &= 0 \\ \Delta E c_2 - c_2K + \Delta E c_1\int \Psi_1^*\Psi_2 d^3\vec{r}_1d^3\vec{r}_2 - c_1J &= 0 \Delta E c_2 - c_2K + \Delta E c_1|s|^2 - c_1J &= 0 \end{align}

So,

\begin{align} \Psi_1^*\Psi_2 d^3\vec{r}_1d^3\vec{r}_2 &= \int \Psi_a^*(\vec{r}_{a1})\Psi_b(\vec{r}_{b1})d^3\vec{r}_1\int\Psi_b^*(\vec{r}_{b2})\Psi_a(\vec{r}_{a2})d^3\vec{r}_2 \\ &= SS^* \\ &= |S|^2. \end{align}

Gives us,

\begin{align} [\Delta E - K]c_1 + [\Delta E |S|^2 - J]c_2 &= 0 \\ [\Delta E - K]c_2 + [\Delta E |S|^2 - J]c_1 &= 0. \begin{vmatrix} \Delta E-K & \Delta E|S|^2 - J \\ \Delta E|S|^2 - J & \Delta E - K \end{vmatrix} &= 0 \\ (\Delta E)_1 &= \frac{K+J}{1+|S|^2} \\ (\Delta E)_2 &= \frac{K-J}{1+|S|^2}. \end{align}

The second is the symmetric solution \(c_1=c_2\).

\(\Psi_{S,Spatial} = \frac{1}{\sqrt{2(1 + |S|^2)}}(\Psi_1 + \Psi_2)\). Then, \(E_S = 2E_1 + \frac{e^2}{r_{ab}} + \frac{K-J}{1+|S|^2} = 2E_1 + \frac{e^2}{r_{ab}} + K + J - |S|^2\frac{K+J}{1+|S|^2}\), which also includes the proton-proton energy. The second form is written so that you can treat the last term as a small correction.

\(\Psi_{A,Spatial} = \frac{1}{\sqrt{2(1-|S|^2)}}(\Psi_1-\Psi_2)\). \(E_A = 2E_1 + \frac{e^2}{r_{ab}} + \frac{K+J}{1+|S|^2} = 2E_1 + \frac{e^2}{r_{ab}} + K - J - |S|^2\frac{K-J}{1-|S|^2}\). Thus, the exchange energy is the differentiating factor and since \(J<0\) the antisymmetric wavefunction is higher energy.

Bond length: \(r_{ab,0}\approx 1.4 a_0\), where the verticle reaches the x-axis. The bottom of the well is like a harmonic oscillator with \(\frac{\hbar\omega_0}{2}\) given by the curvature of the bottom of the well: \(\omega_0 = \sqrt{\frac{1}{m}\frac{d^3E_S}{dr_{ab}^2}}\). The height is the Diassociation energy \(D=4.38\) eV.

	Theory	Experiment
\(r_{ab}\)	0.735 A	0.753 A
\(\omega_0\)	4280 1/cm	4390 1/cm
\(D\)	4.37 eV	4.38 eV