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Motivation

Our midterm posed a (low mark “quick question”) that I didn’t complete (or at least not properly). This shouldn’t have been a difficult question, but I spend way too much time on it, costing me time that I needed for other questions.

It turns out that there isn’t anything fancy required for this question, just perseverance and careful work.

Guts

The question asked for the time evolution of a two particle state

$$\begin{equation}\label{eqn:twoSpinHamiltonian:20} \psi = \inv{\sqrt{2}} \lr{ \ket{\uparrow \downarrow} – \ket{\downarrow \uparrow} } \end{equation}$$

under the action of the Hamiltonian

$$\begin{equation}\label{eqn:twoSpinHamiltonian:40} H = – B S_{z,1} + 2 B S_{x,2} = \frac{\Hbar B}{2}\lr{ -\sigma_{z,1} + 2 \sigma_{x,2} } . \end{equation}$$

We have to know the action of the Hamiltonian on all the states

$$\begin{equation}\label{eqn:twoSpinHamiltonian:60} \begin{aligned} H \ket{\uparrow \uparrow} &= \frac{B \Hbar}{2} \lr{ -\ket{\uparrow \uparrow} + 2 \ket{\uparrow \downarrow} } \\ H \ket{\uparrow \downarrow} &= \frac{B \Hbar}{2} \lr{ -\ket{\uparrow \downarrow} + 2 \ket{\uparrow \uparrow} } \\ H \ket{\downarrow \uparrow} &= \frac{B \Hbar}{2} \lr{ \ket{\downarrow \uparrow} + 2 \ket{\downarrow \downarrow} } \\ H \ket{\downarrow \downarrow} &= \frac{B \Hbar}{2} \lr{ \ket{\downarrow \downarrow} + 2 \ket{\downarrow \uparrow} } \\ \end{aligned} \end{equation}$$

With respect to the basis $\setlr{ \ket{\uparrow \uparrow}, \ket{\uparrow \downarrow}, \ket{\downarrow \uparrow}, \ket{\downarrow \downarrow} }$, the matrix of the Hamiltonian is

$$\begin{equation}\label{eqn:twoSpinHamiltonian:80} H = \frac{ \Hbar B }{2} \begin{bmatrix} -1 & 2 & 0 & 0 \\ 2 & -1 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 2 & 1 \\ \end{bmatrix} \end{equation}$$

Utilizing the block diagonal form (and ignoring the $\Hbar B/2$ factor for now), the characteristic equation is

$$\begin{equation}\label{eqn:twoSpinHamiltonian:100} 0 = \begin{vmatrix} -1 -\lambda & 2 \\ 2 & -1 – \lambda \end{vmatrix} \begin{vmatrix} 1 -\lambda & 2 \\ 2 & 1 – \lambda \end{vmatrix} = \lr{(1 + \lambda)^2 – 4} \lr{(1 – \lambda)^2 – 4}. \end{equation}$$

This has solutions

$$\begin{equation}\label{eqn:twoSpinHamiltonian:120} 1 \pm \lambda = \pm 2, \end{equation}$$

or, with the $\Hbar B/2$ factors put back in

$$\begin{equation}\label{eqn:twoSpinHamiltonian:140} \lambda = \pm \Hbar B/2 , \pm 3 \Hbar B/2. \end{equation}$$

I was thinking that we needed to compute the time evolution operator

$$\begin{equation}\label{eqn:twoSpinHamiltonian:160} U = e^{-i H t/\Hbar}, \end{equation}$$

but we actually only need the eigenvectors, and the inverse relations. We can find the eigenvectors by inspection in each case from

$$\begin{equation}\label{eqn:twoSpinHamiltonian:180} \begin{aligned} H – (1) \frac{ \Hbar B }{2} &= \frac{ \Hbar B }{2} \begin{bmatrix} -2 & 2 & 0 & 0 \\ 2 & -2 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 2 & 0 \\ \end{bmatrix} \\ H – (-1) \frac{ \Hbar B }{2} &= \frac{ \Hbar B }{2} \begin{bmatrix} 0 & 2 & 0 & 0 \\ 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 2 & 2 \\ \end{bmatrix} \\ H – (3) \frac{ \Hbar B }{2} &= \frac{ \Hbar B }{2} \begin{bmatrix} -4 & 2 & 0 & 0 \\ 2 & -4 & 0 & 0 \\ 0 & 0 &-2 & 2 \\ 0 & 0 & 2 &-2 \\ \end{bmatrix} \\ H – (-3) \frac{ \Hbar B }{2} &= \frac{ \Hbar B }{2} \begin{bmatrix} 2 & 2 & 0 & 0 \\ 2 & 2 & 0 & 0 \\ 0 & 0 & 4 & 2 \\ 0 & 0 & 2 & 1 \\ \end{bmatrix}. \end{aligned} \end{equation}$$

The eigenkets are

$$\begin{equation}\label{eqn:twoSpinHamiltonian:280} \begin{aligned} \ket{1} &= \inv{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix} \\ \ket{-1} &= \inv{\sqrt{2}} \begin{bmatrix} 0 \\ 0 \\ 1 \\ -1 \\ \end{bmatrix} \\ \ket{3} &= \inv{\sqrt{2}} \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \\ \end{bmatrix} \\ \ket{-3} &= \inv{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \\ \end{bmatrix}, \end{aligned} \end{equation}$$

or

$$\begin{equation}\label{eqn:twoSpinHamiltonian:300} \begin{aligned} \sqrt{2} \ket{1} &= \ket{\uparrow \uparrow} + \ket{\uparrow \downarrow} \\ \sqrt{2} \ket{-1} &= \ket{\downarrow \uparrow} – \ket{\downarrow \downarrow} \\ \sqrt{2} \ket{3} &= \ket{\downarrow \uparrow} + \ket{\downarrow \downarrow} \\ \sqrt{2} \ket{-3} &= \ket{\uparrow \uparrow} – \ket{\uparrow \downarrow}. \end{aligned} \end{equation}$$

We can invert these

$$\begin{equation}\label{eqn:twoSpinHamiltonian:220} \begin{aligned} \ket{\uparrow \uparrow} &= \inv{\sqrt{2}} \lr{ \ket{1} + \ket{-3} } \\ \ket{\uparrow \downarrow} &= \inv{\sqrt{2}} \lr{ \ket{1} – \ket{-3} } \\ \ket{\downarrow \uparrow} &= \inv{\sqrt{2}} \lr{ \ket{3} + \ket{-1} } \\ \ket{\downarrow \downarrow} &= \inv{\sqrt{2}} \lr{ \ket{3} – \ket{-1} } \\ \end{aligned} \end{equation}$$

The original state of interest can now be expressed in terms of the eigenkets

$$\begin{equation}\label{eqn:twoSpinHamiltonian:240} \psi = \inv{2} \lr{ \ket{1} – \ket{-3} – \ket{3} – \ket{-1} } \end{equation}$$

The time evolution of this ket is

$$\begin{equation}\label{eqn:twoSpinHamiltonian:260} \begin{aligned} \psi(t) &= \inv{2} \lr{ e^{-i B t/2} \ket{1} – e^{3 i B t/2} \ket{-3} – e^{-3 i B t/2} \ket{3} – e^{i B t/2} \ket{-1} } \\ &= \inv{2 \sqrt{2}} \Biglr{ e^{-i B t/2} \lr{ \ket{\uparrow \uparrow} + \ket{\uparrow \downarrow} } – e^{3 i B t/2} \lr{ \ket{\uparrow \uparrow} – \ket{\uparrow \downarrow} } – e^{-3 i B t/2} \lr{ \ket{\downarrow \uparrow} + \ket{\downarrow \downarrow} } – e^{i B t/2} \lr{ \ket{\downarrow \uparrow} – \ket{\downarrow \downarrow} } } \\ &= \inv{2 \sqrt{2}} \Biglr{ \lr{ e^{-i B t/2} – e^{3 i B t/2} } \ket{\uparrow \uparrow} + \lr{ e^{-i B t/2} + e^{3 i B t/2} } \ket{\uparrow \downarrow} – \lr{ e^{-3 i B t/2} + e^{i B t/2} } \ket{\downarrow \uparrow} + \lr{ e^{i B t/2} – e^{-3 i B t/2} } \ket{\downarrow \downarrow} } \\ &= \inv{2 \sqrt{2}} \Biglr{ e^{i B t/2} \lr{ e^{-2 i B t/2} – e^{2 i B t/2} } \ket{\uparrow \uparrow} + e^{i B t/2} \lr{ e^{-2 i B t/2} + e^{2 i B t/2} } \ket{\uparrow \downarrow} – e^{- i B t/2} \lr{ e^{-2 i B t/2} + e^{2 i B t/2} } \ket{\downarrow \uparrow} + e^{- i B t/2} \lr{ e^{2 i B t/2} – e^{-2 i B t/2} } \ket{\downarrow \downarrow} } \\ &= \inv{\sqrt{2}} \lr{ i \sin( B t ) \lr{ e^{- i B t/2} \ket{\downarrow \downarrow} – e^{i B t/2} \ket{\uparrow \uparrow} } + \cos( B t ) \lr{ e^{i B t/2} \ket{\uparrow \downarrow} – e^{- i B t/2} \ket{\downarrow \uparrow} } } \end{aligned} \end{equation}$$

Note that this returns to the original state when $t = \frac{2 \pi n}{B}, n \in \mathbb{Z}$. I think I’ve got it right this time (although I got a slightly different answer on paper before typing it up.)

This doesn’t exactly seem like a quick answer question, at least to me. Is there some easier way to do it?