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Proof of two nifty properties of Hermitian operators

December 18, 2020

Given a Hermitian operator $\hat{H}$:

1. The Eigenvalues are Real

Let’s say we apply our Hermitian operator $\hat{H}$ to a state $\vert\Psi\rangle$. We’re going to get some scalar of an eigenvalue of $\hat{H}$, call this $h$. Also consider what happens when we take the $\dagger$ operator, the hermitian conjugate, of the eigenvalue equation. We get these two equations, courtesy of the fact that since $\hat{H}$ is Hermitian, its conjugate transpose is equal to itself, $\hat{H}^\dagger=\hat{H}$:

\[\hat{H}\vert\psi\rangle = h\vert\psi\rangle \qquad\text{and}\qquad \langle\psi\vert\hat{H} = \langle\psi\vert h^*\]

Now let’s take the inner product of the left equation with $\langle\psi\vert$, and the inner product of the right equation with $\vert\psi\rangle$. This is a pretty cool trick.

\[\langle\psi\vert\hat{H}\vert\psi\rangle = h\langle\psi\vert\psi\rangle \qquad\text{and}\qquad \langle\psi\vert\hat{H}\vert\psi\rangle = \langle\psi\vert\psi\rangle h^*\]

The left hand side, $\langle\psi\vert\hat{H}\vert\psi\rangle$, is the same, so we can combine these statements.

\[\begin{align} h\langle\psi\vert\psi\rangle &= \langle\psi\vert\psi\rangle h^*\\ h\langle\psi\vert\psi\rangle - \langle\psi\vert\psi\rangle h^* &= 0\\ (h-h^*)\langle\psi\vert\psi\rangle &= 0 \end{align}\]

So we see that either $(h-h^*)=0$ or $\langle\psi\vert\psi\rangle = 0$. It wouldn’t make sense that our state $\vert\psi\rangle$ has a norm (length) of 0, and we can just define our state such that this is not the case. So the only thing left is that $h-h^*=0$, where $h$ is an eigenvalue of $\hat{H}$ and $h^*$ is its complex conjugate.

If $h=a+bi$, then $h^*=a-bi$ by definition of complex conjugate. So because $h-h^*=0$, $h$ must have an imaginary component of $b=0$, and thus $h$ must be real. ✔️

2. The Eigenstates are Orthogonal

Let’s look at what happens when we have two eigenvalue equations for two different states, $\psi_a$ and $\psi_b$. One equation has a bra and one has a ket.

\[\hat{H}\vert\psi_a\rangle=h_a \vert\psi_a\rangle \qquad\text{and}\qquad \langle\psi_b\vert\hat{H}=\langle\psi_b\vert h_b\]

Now take the inner product of the left equation with $\langle\psi_b\vert$ and the inner product of the right equation with $\vert\psi_a\rangle$.

\[\langle\psi_b\vert\hat{H}\vert\psi_a\rangle=h_a\langle\psi_b\vert\psi_a\rangle \qquad\text{and}\qquad \langle\psi_b\vert\hat{H}\vert\psi_a\rangle=\langle\psi_b\vert\psi_a\rangle h_b\]

So once again we can combine these statements.

\[\begin{align} h_a\langle\psi_b\vert\psi_a\rangle &=\langle\psi_b\vert\psi_a\rangle h_b\\ (h_b-h_a)\langle\psi_b\vert\psi_a\rangle&=0 \end{align}\]

In order for this to be true either $(h_b-h_a)=0$ or $\langle\psi_b\vert\psi_a\rangle=0$. This means the eigenvalues must either by degenerate (identical), or the inner product between the eigenstates must be 0. If the eigenvalues are not degenerate ($h_a \neq h_b$), then of course our eigenstates are orthogonal. But what if our eigenvalues are degenerate?

If the eigenvalues are equal, $h_a=h_b=h$, then any state in the basis of $\vert\psi_a\rangle$ and $\vert\psi_b\rangle$, which we will denote as $(a\vert\psi_a\rangle + b\vert\psi_b\rangle)$, will also have an eigenvalue of $h$.

\[\begin{align} \hat{H}(a\vert\psi_a\rangle+b\vert\psi_b\rangle)&=a\hat{H}\vert\psi_a\rangle+b\hat{H}\vert\psi_b\rangle\\ &=ah_a\vert\psi_a\rangle+bh_b\vert\psi_b\rangle\\ &= ah\vert\psi_a\rangle+bh\vert\psi_b\rangle\\ &=h(a\vert\psi_a\rangle+b\vert\psi_b\rangle) \end{align}\]

What this means is that if our eigenvalues are identical, then there are an infinite number of eigenstates that correspond to that eigenvalue. Not all these eigenstates are orthogonal to each other, but we can simply choose a pair that are orthogonal. ✔️