# What is an operator times one of its eigenstates?

I am trying to get a hold of caluclating with matrix elements. I have a Hamiltionian $\hat{H}$ in a two-dimensional Hilbert space, having eigenstates $\psi_1$ and $\psi_2$. My professor wrote down these equations:

$$\hat{H} \psi_1 = H_{11} \psi_1 + H_{12}\psi_2 \\ \hat{H} \psi_2 = H_{21} \psi_1 + H_{22}\psi_2$$

He said these are alternative way to write the matrix elements of $\hat{H}$. However, I fail to see why is this. I tries to look it up in a linear algebra textbook, maybe this is a special property of matrix multiplications.

I have tried the following:

$$H_{11} = \psi_1^* \hat{H} \psi_1 \\ H_{11} \psi_1 = \psi_1^* \hat{H} \psi_1 \psi_1$$

I did the same thing with $H_{12}$, added the equations together, but still can not see anything that resembles the original system.

Where do those two equations come from?

• You copied down the professor's equation wrong, or he/she wrote them wrong. The first two equations clearly contradict each other. – tparker May 24 '17 at 21:12
• Sorry, yes, edited it. – Ezze May 24 '17 at 21:18
• By definition, if $\{\psi_i\}$ are an orthonormal basis, then $H_{ij} \equiv \langle \psi_i | H | \psi_j \rangle$. Of course, your can immediately see that the matrix elements friend in the basis, so we should probably write $H_{ij}^{\psi}$. – DanielSank May 25 '17 at 1:36

## 3 Answers

The idea is that the two states $\{\psi_1,\psi_2 \}$ form a basis for the Hilbert space, this means that any other vector can be written as a linear combination of these two states, in particular the result of operating $H$ on, say, $\psi_{1}$

$$H\psi_1 = H_{11}\psi_1 + H_{12}\psi_2 \\ H\psi_2 = H_{21}\psi_1 + H_{22}\psi_2$$

Note that up to this point, the coefficients $H_{ij}$ are just complex number in the expansion. Now, if $\{\psi_1,\psi_2\}$ form an orthonormal basis we can write

$$\langle\psi_1 | H \psi_1\rangle = \langle\psi_1 |H_{11}\psi_1 + H_{12}\psi_2 \rangle = H_{11}\underbrace{\langle \psi_1|\psi_1\rangle}_{=1} + H_{12}\underbrace{\langle \psi_1|\psi_2\rangle}_{=0} = H_{11}$$

You can test the rest, but in general

$$H_{ij} = \langle \psi_i|H|\psi_j\rangle$$

• Thank s. Seeing your equations, I guess that the two basis vectors don't even have to be the eigenstates of the $H$ operator, they just have to be orthonormal? – Ezze May 24 '17 at 21:14
• @Ezze Correct! If they were eigenstates of $H$ then the matrix would be diagonal, since $\langle \psi_i | H |\psi_j\rangle = \langle \psi_i | E_j |\psi_j\rangle = E_j\delta_{ij}$ – caverac May 24 '17 at 21:40
• Oh so this is where I misunderstood my prof! He used the same symbol for energy eigenstates and a different set of states. He actually wrote these equations with respect to an arbitrary base, not in terms of energy eigenstates. This makes sense now. Thank you! – Ezze May 24 '17 at 21:44

I highly doubt that there being "alternative ways to write the matrix elements" mentioned by the professor has any conceptual significance. After choosing a basis of the two-dimensional Hilbert space, operators can be expressed as matrices,

$$H=\begin{pmatrix}H_{11} & H_{12}\\ H_{21} & H_{22}\end{pmatrix}$$ where $H_{11}, H_{22}$ are real and $H_{12}^*=H_{21}$.

The equations you cite simply mean that we have chosen the basis to be $\{\psi_1, \psi_2\}$. This is an orthonormal basis. In particular for a self-adjoint (as observables must be) matrix, there is always an orthonormal basis of eigenvectors. So the two eigenvectors in this basis are represented as $\psi_1=\begin{pmatrix}1\\0\end{pmatrix}$ and $\psi_2=\begin{pmatrix}0\\1\end{pmatrix}$. Note that given an orthonormal Basis of an n-dim. Hilbert space $\{\psi_1, \dots, \psi_n\}$ we may express the Matrix elements of any operator as scalar products: $H_{ij}:=\langle \psi_i, H\psi_j\rangle=\langle\psi_i|H|\psi_j\rangle$

Because $\psi_1$ and $\psi_2$ are eigenstates of the Hamiltonian, we know:

$$\hat{H}\psi_1 = E_1 \psi_1, \\ \hat{H}\psi_2 = E_2 \psi_2.$$

Let's consider the matrix element: $H_{ij} = \langle \psi_i|\hat{H}|\psi_j\rangle$.

If $\psi_n$ is an eigenstate of the Hamiltonian, then: $H_{ij} =\langle \psi_i|\hat{H}\psi_j\rangle = \langle \psi_i|E_j\psi_j\rangle$, which as $E_n$ is just a scalar, becomes: $$H_{ij} =E_n\langle \psi_i|\psi_j\rangle = \delta_{ij}E_n,$$

where I've used that the eigenstates of a Hermitian operator are orthogonal.

That means in your first equations, the off-diagonal $H_{12}$, and $H_{21}$ terms are both zero. So they are valid, but a bit of a weird thing to write out.

You can get far by thinking in terms of vectors. Here is what your equation would say: $$\left( \begin{array}{cc} E_1 & 0 \\ 0 & E_2 \end{array} \right) % \left( \begin{array}{cc} 1 \\ 0 \end{array} \right) = E_1\times \left( \begin{array}{cc} 1 \\ 0 \end{array} \right) + 0\times \left( \begin{array}{cc} 0 \\ 1 \end{array} \right).$$ So while it is valid, it seems a bit odd to include the zero.