Interpreting the Time Evolution Operator on a hands-on example [closed]

Question

Given a time independent Hamiltonian:

$$H = \omega\,\left(\begin{array}{cc}0 \quad 1\\1\quad0\end{array}\right)$$

whats's the time evolution of a state $|\psi(t_0)\rangle = \left(\begin{array}{cc}\alpha_+\\\alpha_-\end{array}\right)$ and how can it be interpreted?

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My work so far: time evolution can be described with a time evolution operator:

$$|\psi(t)\rangle = U(t,t_0)\,|\psi\rangle$$

where $$U(t,t_0) = e^{\textstyle-i/\hbar\,(t-t_0)\,H}$$

since $H$ is a matrix the matrix exponential has to be calculated:

$$U = \left(\begin{array}{cc}\cos\left(\dfrac{\omega}{\hbar}\,(t-t_0)\right)\quad-i\,\sin\left(\dfrac{\omega}{\hbar}\,(t-t_0)\right)\\[12pt] -i\,\sin\left(\dfrac{\omega}{\hbar}\,(t-t_0)\right)\quad\cos\left(\dfrac{\omega}{\hbar}\,(t-t_0)\right)\end{array}\right)$$

Thus the time evolution simply can be determined by expanding the matrix product:

$$|\psi(t)\rangle = U(t,t_0)\,\left(\begin{array}{cc}\alpha_+\\\alpha_-\end{array}\right)$$

$\texttt{Right at all?}$ If so, what is actually portrayed? The Time Operator $U$ reminds on rotation:

$$R = \left(\begin{array}{cc}\cos(\varphi)\quad -\sin(\varphi) \\[12pt] \sin(\varphi)\quad \cos(\varphi)\end{array}\right)$$

How does this translate to complex numbers? A complex rotation matrix like:

$$C = \left(\begin{array}{cc}\cos(\varphi)\quad -i\,\sin(\varphi) \\[12pt] -i\,\sin(\varphi)\quad \cos(\varphi)\end{array}\right)$$ would already make sense because it has property of rotation matrices: $\text{det}(C) = 1$

But I'm afraid I'm missing the true core here.

Answer 1

It may help to look at a simple example. If your state vector is $\pmatrix{1\\0}$ at $t=0$, then you will have $$|\psi(t)\rangle = \pmatrix{\cos(\omega t)\\-i\sin(\omega t)}$$ In particular,
$$|\psi(0)\rangle = \pmatrix{1\\0} = |z\uparrow\rangle \quad |\psi(\pi/4\omega)\rangle = \frac{1}{\sqrt{2}}\pmatrix{1\\-i} = |y\downarrow\rangle$$ $$|\psi(\pi/2\omega)\rangle = \pmatrix{0\\-i} = \color{red}{-i}|z\downarrow\rangle \quad |\psi(3\pi/4\omega)\rangle = \frac{1}{\sqrt{2}}\pmatrix{-1\\-i} =\color{red}{-} |y\uparrow\rangle$$ $$|\psi(\pi/\omega)\rangle = \pmatrix{-1\\0} = \color{red}{-}|z\uparrow\rangle$$ where I've highlighted unimportant phase factors in red. A particle with spin initially spin-up along the $\hat z$ axis will evolve first to be spin-down along the $\hat y$-axis, then spin-down along the $\hat z$-axis, and finally spin-up along the $\hat y$-axis before returning to its original state (up to a phase factor).

In other words, the spin of the particle is precessing about the $\hat x$-axis - which makes excellent sense, because the Hamiltonian $H = \omega \sigma_x$ is proportional to $\sigma_x$, the infinitesimal generator of rotations about the $x$-axis.

The Time Operator $U$ reminds [me of a] rotation

That's a good insight. Indeed $U(t)$ is a rotation operator. However, the familiar rotation matrix $$R = \pmatrix{\cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)} = e^{\theta L}, \qquad L = \pmatrix{0 & -1 \\ 1 & 0 }$$

is obtained by exponentiating $L$ rather than $i\sigma_x$.

In more technical language, when we seek to implement rotations on a vector space what we need to do is find a representation of the rotation group on that vector space - that is, we need to identify rotations with matrices which act on the vector space that we're interested in. It turns out that if we want to implement rotations on $\mathbb R^2$ - which should be understood as 2D vectors in a plane - then the appropriate infinitesimal generator of rotations is $L$ (there is only one kind of rotation in a 2D plane) and the rotation matrix is given by $R=e^{\theta L}$.

On the other hand, if we want to implement rotations on $\mathbb C^2$ - which should be understood as a $2$-component spinor - then there are three infinitesimal generators of rotations given by $i\sigma_x,i\sigma_y,$ and $i\sigma_z$, and the corresponding rotation matrices are given by exponentiating these generators instead.

Answer 2

It is indeed good your intuition that the time evolution is a rotation but to get the general sense of what is happening you have to think in terms of Hilbert spaces and basis on such spaces.

What is happening is that the system is oscillating between the two basis state. Let's suppose that the values for $\alpha_+=1$ and $\alpha_-=0$. If you consider times such as: $$ t = \frac{\omega}{\hbar} t_0 + \frac{\hbar}{\omega} \frac{\pi}{2} n $$ for $n$ a even integer you will certainly find the system to be $|\psi(t)> = (1, 0 )^\top$ while if you consider: $$ t = \frac{\omega}{\hbar} t_0 + \frac{\hbar}{\omega} \frac{\pi}{2} n $$ for $n$ an odd integer you will find the system in state $|\psi(t)> = (0,1)^\top$. If we sketch then our Hilbert space as a simple 2D plane where the first axis corresponds to the first state and the same for the second axis we see that the geometric vector that represents the coordinates of the state is rotating.

Note that because of the form of the time evolution operator the determinant will always be one since time evolution preserves the norm.