Time evolution of a pair of electrons

The Hamiltonian

Suppose we have a pair of electrons, fixed in close enough proximity so that their magnetic and electric fields interact.

Where there is no external magnetic field the two electron's own fields interact, in such a way that each electron provides the magnetic field for the other.

Under these conditions, we assume that the Hamiltonian is proportional to the product of the spin on the first electron, $\vec{\sigma}$, with the spin of the second, $\vec{\tau}$,

Classically, we would write $$H = E \left(\vec{\sigma} \cdot \vec{\tau}\right) = E \left(\boldsymbol{\sigma_x}\boldsymbol{\tau_x} + \boldsymbol{\sigma_y}\boldsymbol{\tau_y} + \boldsymbol{\sigma_z}\boldsymbol{\tau_z}\right)$$ where we have written $E$ as the constant of proportionality. The quantum analogue of this looks the same but uses the spin operators rather than spatial vectors.

The eigenspace of the Hamiltonian

We would like to find the eigenvalues and eigenvectors associated the Hamiltonian, $$H = E \left(\boldsymbol{\sigma_x}\boldsymbol{\tau_x} + \boldsymbol{\sigma_y}\boldsymbol{\tau_y} + \boldsymbol{\sigma_z}\boldsymbol{\tau_z}\right)$$

So, suppose that $$\ket{\Psi} = \alpha\ket{uu} + \beta\ket{ud} + \gamma\ket{du} + \delta\ket{dd}$$ is an eigenvector of $H$ with associated eigenvalue $\lambda$ so that $$H \ket{\Psi} = \lambda \ket{\Psi}$$

Note that \begin{align*}\boldsymbol{\sigma_x}\boldsymbol{\tau_x} \ket{\Psi} &= \boldsymbol{\sigma_x}\boldsymbol{\tau_x} \left(\alpha\ket{uu} + \beta\ket{ud} + \gamma\ket{du} + \delta\ket{dd}\right) = \alpha\ket{dd} + \beta\ket{du} + \gamma\ket{ud} + \delta\ket{uu}\\\boldsymbol{\sigma_y}\boldsymbol{\tau_y} \ket{\Psi} &= \boldsymbol{\sigma_y}\boldsymbol{\tau_y} \left(\alpha\ket{uu} + \beta\ket{ud} + \gamma\ket{du} + \delta\ket{dd}\right) = -\alpha\ket{dd} + \beta\ket{du} + \gamma\ket{ud} - \delta\ket{uu}\\\boldsymbol{\sigma_z}\boldsymbol{\tau_z} \ket{\Psi} &= \boldsymbol{\sigma_z}\boldsymbol{\tau_z} \left(\alpha\ket{uu} + \beta\ket{ud} + \gamma\ket{du} + \delta\ket{dd}\right) = \alpha\ket{uu} - \beta\ket{ud} - \gamma\ket{du} + \delta\ket{dd}\end{align*}

Summing these together and multiplying by $E$, we can write the left-hand side as $$H \ket{\Psi} = E\left(\alpha\ket{uu} + \left(2\gamma - \beta\right)\ket{ud} + \left(2\beta - \gamma\right)\ket{du} + \delta\ket{dd}\right)$$ and in matrix notation the equation becomes, $$E \, \begin{bmatrix} \alpha \\ \left(2\gamma - \beta\right) \\ \left(2\beta - \gamma\right) \\ \delta \end{bmatrix} = \lambda \begin{bmatrix} \alpha \\ \beta \\ \gamma \\ \delta \end{bmatrix}$$

So we have a set of four identities \begin{align*}\left(E - \lambda\right)\alpha &= 0\\\left(E + \lambda\right)\beta &= 2 E \gamma\\\left(E + \lambda\right)\gamma &= 2 E \beta\\\left(E - \lambda\right)\delta &= 0\end{align*}

There are two distinct cases, where either $\lambda = E$ or $\lambda \neq E$.

If $\lambda = E$, then, the four identities are reduced to three, each of which represents an eigenvector (check by plugging back into the equation), \begin{align*}\alpha(0) = 0 &\Rightarrow \alpha \in \mathbb{C} &\Rightarrow& \ket{H \equiv E} = \ket{uu}\\\left.\begin{matrix} 2 E \beta = 2 E \gamma \\ 2 E \gamma = 2 E \beta \end{matrix}\right\} &\Rightarrow \beta = \gamma&\Rightarrow& \ket{H \equiv E} = \frac{\ket{ud} + \ket{du}}{\sqrt{2}}\\\delta (0) = 0 &\Rightarrow \delta \in \mathbb{C} &\Rightarrow& \ket{H \equiv E} = \ket{dd}\end{align*}

So, the eigenspace of $H$ associated with $\lambda = E$ is $$\left\{H \equiv E\right\} = \mathbf{LS}\left\{\ket{uu},\frac{\ket{ud} + \ket{du}}{\sqrt{2}},\ket{dd}\right\}$$

If $\lambda \neq E$, then, the four identities become \begin{align*}\left(E - \lambda\right)\alpha = 0 &\Rightarrow \alpha = 0\\\left.\begin{matrix}\left(E + \lambda\right)\beta = 2 E \gamma\\\left(E + \lambda\right)\gamma = 2 E \beta\end{matrix}\right\} &\Rightarrow \beta^2 = \gamma^2 \Rightarrow \gamma = - \beta\\\left(E - \lambda\right)\delta = 0 &\Rightarrow \delta = 0\end{align*}

The eigenvalue can then be calculated since we must have $E + \lambda = -2 E$ or $$\lambda = -3 E$$

From this we get the single eigenvector, $$\ket{H \equiv -3E} = \frac{\ket{ud} - \ket{du}}{\sqrt{2}}$$

So the eigenspace of $H$ for $\lambda \neq E$, $$\left\{H \equiv -3E\right\} = \mathbf{LS}\left\{\frac{\ket{ud} - \ket{du}}{\sqrt{2}}\right\}$$

Singlet and triplet states

If we write - as we have earlier in the course - \begin{align*}\ket{T} &= \frac{\ket{ud} + \ket{du}}{\sqrt{2}}\\\ket{S} &= \frac{\ket{ud} - \ket{du}}{\sqrt{2}}\end{align*} and also \begin{align*}E_T &= E\\E_S &= -3E\end{align*} then the eigenspaces become \begin{align*}\left\{H \equiv E_T\right\} &= \mathbf{LS}\left\{\ket{uu},\ket{T},\ket{dd}\right\}\\\left\{H \equiv E_S\right\} &= \mathbf{LS}\left\{\ket{S}\right\}\end{align*}

We say an element of $\left\{H \equiv E_T\right\}$ is a triplet state and an element of $\left\{H \equiv E_S\right\}$ is called a singlet state.

The Hamiltonian can be written in terms of its eigenvalues, $$H = \begin{bmatrix}E_T & 0 & 0 & 0 \\0 & E_T & 0 & 0\\0 & 0 & E_T & 0\\0 & 0 & 0 & E_S\end{bmatrix}$$

The eigenvectors $\left\{\ket{uu},\ket{T},\ket{dd},\ket{S}\right\}$ form a basis for the whole space of states, so that any state can be expressed as a linear combination.

Suppose the state $\ket{\Psi}$ is defined on the standard basis as $$\ket{\Psi} = \alpha\ket{uu} + \beta\ket{ud} + \gamma\ket{du} + \delta\ket{dd}$$

We can write two of the four standard basis vectors in terms of the eigenvectors of $H$, \begin{align*}\ket{ud} &= \frac{\ket{T} + \ket{S}}{\sqrt{2}}\\\ket{du} &= \frac{\ket{T} - \ket{S}}{\sqrt{2}}\end{align*} and then these can be substituted, \begin{align*}\ket{\Psi} &= \alpha\ket{uu} + \beta\ket{ud} + \gamma\ket{du} + \delta\ket{dd}\\&= \alpha\ket{uu} + \beta\frac{\ket{T} + \ket{S}}{\sqrt{2}} + \gamma\frac{\ket{T} - \ket{S}}{\sqrt{2}} + \delta\ket{dd}\\&= \alpha\ket{uu} + \frac{\beta + \gamma}{\sqrt{2}}\ket{T} + \delta\ket{dd} + \frac{\beta - \gamma}{\sqrt{2}}\ket{S}\\&= \alpha\ket{uu} + \beta'\ket{T} + \delta\ket{dd} + \gamma'\ket{S}\end{align*} where $\beta' = \frac{\beta + \gamma}{\sqrt{2}}$ and $\gamma' = \frac{\beta - \gamma}{\sqrt{2}}$.

Time evolution

Now that we can express any spin-state, $$\ket{\Psi} = \alpha\ket{uu} + \beta\ket{ud} + \gamma\ket{du} + \delta\ket{dd}$$ in the basis of the Hamiltonian's eigenvectors, $$\ket{\Psi} = \alpha\ket{uu} + \beta'\ket{T} + \delta\ket{dd} + \gamma'\ket{S}$$ where $\beta' = \frac{\beta + \gamma}{\sqrt{2}}$ and $\gamma' = \frac{\beta - \gamma}{\sqrt{2}}$ we can consider the time of a pair of electrons, using Schrodinger's equation, $$\frac{\partial}{\partial t}\ket{\Psi} = -\mathrm{i}\frac{H}{\hbar} \ket{\Psi}$$ which has solutions \begin{align*}\ket{\Psi(t)} &= e^{-\mathrm{i}\frac{H}{\hbar}t}\ket{\Psi}\\&= e^{-\mathrm{i}\frac{H}{\hbar}t}\left(\alpha\ket{uu} + \beta'\ket{T} + \delta\ket{dd} + \gamma'\ket{S}\right)\\&= e^{-\mathrm{i}\frac{E_T}{\hbar}t}\left(\alpha\ket{uu} + \beta'\ket{T} + \delta\ket{dd}\right) + e^{-\mathrm{i}\frac{E_S}{\hbar}t}\left(\gamma'\ket{S}\right)\\&= e^{-\mathrm{i}\frac{E_T}{\hbar}t}\left(\alpha\ket{uu} + \beta'\left(\frac{\ket{ud} + \ket{du}}{\sqrt{2}}\right) + \delta\ket{dd}\right) + e^{-\mathrm{i}\frac{E_S}{\hbar}t}\left(\gamma'\left(\frac{\ket{ud} - \ket{du}}{\sqrt{2}}\right)\right)\end{align*} which can be re-arranged as $$\ket{\Psi(t)} = \alpha(t)\ket{uu} + \beta(t) \ket{ud} + \gamma(t) \ket{du} + \delta(t) \ket{dd}$$ where \begin{align*}\alpha(t) &= \alpha e^{-\mathrm{i}\frac{E_T}{\hbar}t}\\\beta(t) &= \frac{1}{2}\left(\left(\beta + \gamma\right) e^{-\mathrm{i}\frac{E_T}{\hbar}t} + \left(\beta - \gamma\right) e^{-\mathrm{i}\frac{E_S}{\hbar}t}\right)\\\gamma(t) &= \frac{1}{2}\left(\left(\beta + \gamma\right) e^{-\mathrm{i}\frac{E_T}{\hbar}t} - \left(\beta - \gamma\right) e^{-\mathrm{i}\frac{E_S}{\hbar}t}\right)\\\delta(t) &= \delta e^{-\mathrm{i}\frac{E_T}{\hbar}t}\end{align*}

Example

Suppose we know that the pair of electrons start out at in the state $$\ket{\Psi} = \ket{ud}$$

This means that $\beta = 1, \alpha = \gamma = \delta = 0$, and we can plug these coefficients into the above formulae to get $$\ket{\Psi(t)} = \frac{1}{2}\left(e^{-\mathrm{i}\frac{E_T}{\hbar}t} + e^{-\mathrm{i}\frac{E_S}{\hbar}t}\right)\ket{ud} + \frac{1}{2}\left(e^{-\mathrm{i}\frac{E_T}{\hbar}t} - e^{-\mathrm{i}\frac{E_S}{\hbar}t}\right)\ket{du}$$

To analyse this result, we can ask what the probability is that the electrons will be in the initial state $\ket{ud}$ at a given time, \begin{align*}|\braket{ud}{\Psi(t)}|^2 &= \braket{ud}{\Psi(t)}^*\braket{ud}{\Psi(t)}\\&= \frac{1}{4}\left(e^{\mathrm{i}\frac{E_T}{\hbar}t} + e^{\mathrm{i}\frac{E_S}{\hbar}t}\right)\left(e^{-\mathrm{i}\frac{E_T}{\hbar}t} + e^{-\mathrm{i}\frac{E_S}{\hbar}t}\right)\\&= \frac{1}{4}\left(2 + e^{\mathrm{i}\frac{E_T - E_S}{\hbar}t} + e^{-\mathrm{i}\frac{E_T - E_S}{\hbar}t}\right)\\&= \frac{1}{2}\left(1 + \frac{e^{\mathrm{i}\frac{4E_T}{\hbar}t} + e^{-\mathrm{i}\frac{4E_T}{\hbar}t}}{2}\right) & \text{ since } E_S = - 3E_T\\&= \frac{1}{2}\left(1 + \cos \frac{4E_T}{\hbar}t \right)\\&= \left\{\begin{matrix}1 &\text{ when } 4E_T t_n = 2n\pi &\text{ or } t_n = \frac{n\pi}{2E_T}\\0 &\text{ when } 4E_T t_n = \left(2n + 1\right)\pi &\text{ or } t_n = \frac{\left(n + \tfrac{1}{2}\right)\pi}{2E_T}\end{matrix}\right.\end{align*}

So, at times $t_n = \frac{n\pi}{2E_T}$, the electrons will be in the state $\ket{ud}$, and when $t_n = \frac{\left(n + \tfrac{1}{2}\right)\pi}{2E_T}$ the electrons will be the state $\ket{du}$.

Figure 9.2 - Probabilities of $\ket{ud}$ and $\ket{du}$

At other times, the electrons may be in either state with some probability.

We can show that the probability that the electrons will be in the state $\ket{du}$ (opposite to the initial state) at a given time is $$|\braket{du}{\Psi(t)}|^2 = \frac{1}{2}\left(1 - \cos \frac{4E_T}{\hbar}t \right)$$

We show both probabilities in figure 9.2, courtesy of Wolfram|Alpha, we have put $$x = x(t) = \frac{4E_T}{\hbar}t$$