# Representation of time evolution operator and spectral decomposition

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!

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


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

• Thank you. This clears a lot. 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.

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