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!

  • 1
    $\begingroup$ $|+z \rangle$ 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...) $\endgroup$ Oct 15, 2019 at 5:52

2 Answers 2


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

  • $\begingroup$ Thank you. This clears a lot. $\endgroup$ Oct 15, 2019 at 17:16

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.

  • $\begingroup$ The Hamiltonian is Hermitian, because $(S_-S_+)^\dagger = S_+^\dagger S_-^\dagger = S_-S_+$. $\endgroup$ Oct 15, 2019 at 8:57
  • $\begingroup$ Oops...I agree... $\endgroup$ Oct 15, 2019 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.