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I was preparing for my quantum mechanics exam, and I came to think about this question regarding the spectral representation of time evolution operator. Let's say we are given the Hamiltonian:

$\hat{H} = S_{-} S_{+} $.

What will be the time evolution operator in the spectral representation? Here is my shot at it. What worries me is that, shouldn't this Hamiltonian have two eigenstates? I am only getting one. Anyways, here is my approach:

The raising and lowering spin operators can be written, in terms of the z-basis, as:

$S_{+} = \hbar |+ z \rangle \langle -z|$, $S_{-} = \hbar |- z \rangle \langle +z | $.

Then, the Hamiltonian becomes:

$\hat{H} = \hbar^2 |- z \rangle \langle +z |+ z \rangle \langle -z| = \hbar^2 |-z\rangle \langle -z|. $

Now, the only eigenvector of this equation is $|-z\rangle$, with the corresponding eigenvalue, $\hbar^2$. So, we can write the spectral representation of the time evolution operator, given by:

$e^{i\hat{H}t/\hbar} = e^{i\hbar t} |-z\rangle \langle -z|$.

Is this it? Or am I missing something here? Any help would be greatly appreciated!

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    $\begingroup$ $|+z \rangle$ could also be eigenvector with eigenvalue zero if your basis is orthonormal. Notice that the Hamiltonian in matrix representation is 2 x 2 hence you need two eigenvectors (since eigenvectors form a basis for the Hilbert space since Hamiltonian is self-adjoint...) $\endgroup$ – Mathphys meister Oct 15 '19 at 5:52
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$\newcommand{\bra}[1] {\left< #1 \right|} \newcommand{\ket}[1] {\left| #1 \right>} \newcommand{\bracket}[2] {\left< #1 \vert #2 \right>} $ Your calculation of the Hamiltonian $$\hat{H} = \hbar^2 \ket{-z}\bra{-z}$$ is correct so far.

One eigenvector is $\ket{-z}$, with eigenvalue $\hbar^2$.
But you missed (as Dani already wrote in his comment): another eigenvector is $\ket{+z}$, with eigenvalue $0$.

So you could also write (in a rather pedantic way): $$\hat{H} = 0\ket{+z}\bra{+z} + \hbar^2 \ket{-z}\bra{-z}$$

Then it is more obvious, that the time evolution operator becomes: $$e^{i\hat{H}t/\hbar} = 1 \ket{+z}\bra{+z} + e^{i\hbar t}\ket{-z}\bra{-z}.$$


Yet another equivalent approach is by using the matrix representation (with $\ket{+z}$ and $\ket{-z}$ as base vectors) and calculate the matrix exponential of a diagonal matrix.

We have $$\hat{H}=\begin{pmatrix}0 & 0 \\ 0 & \hbar^2\end{pmatrix}.$$ Then we get $$e^{i\hat{H}t/\hbar} = \exp\left[i\begin{pmatrix}0 & 0 \\ 0 & \hbar^2\end{pmatrix}t/\hbar\right] = \exp \begin{pmatrix}0 & 0 \\ 0 & i\hbar t\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & e^{i\hbar t}\end{pmatrix} $$

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  • $\begingroup$ Thank you. This clears a lot. $\endgroup$ – Sohair Abdullah Oct 15 '19 at 17:16
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My quantum mechanics is a little rusty but I will attempt an answer. The problem is that the Hamiltonian is not Hermitean. A Hermitean operator in this case would have two eigenvectors. You need to add the hermitean conjugate of the S_{-}S_{+} term to the hamiltonian. From what I remember this is common practice in condensed matter physics.

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  • $\begingroup$ The Hamiltonian is Hermitian, because $(S_-S_+)^\dagger = S_+^\dagger S_-^\dagger = S_-S_+$. $\endgroup$ – Thomas Fritsch Oct 15 '19 at 8:57
  • $\begingroup$ Oops...I agree... $\endgroup$ – Harshant Singh Oct 15 '19 at 10:05

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