# QM: Linear operator, conditions on kets to make operator Hermitian

I've just started my first quantum mechanics course, and I'm not quite sure how to attack this homework task:

We suppose the Hamiltonian $$\hat{H}$$ is linear and acts as follows:

$$\hat{H}\vert{\psi}\rangle$$ = $$g\vert{\phi}\rangle$$, $$\hat{H}\vert{\phi}\rangle$$ = $$g^{*}\vert{\psi}\rangle$$, $$\hat{H}\vert{\psi_n}\rangle = 0$$

where $$g$$ is an arbitrary complex number, $$\vert{\psi}\rangle$$ and $$\vert{\phi}\rangle$$ is a pair of linearly independent states, both normalized to unity, but not necessarily orthogonal. $$\vert{\psi_n}\rangle$$ where $$n \in [1,N]$$ are all states orthogonal to both $$\vert{\psi}\rangle$$ and $$\vert{\phi}\rangle$$. What are the conditions that $$\vert{\psi}\rangle$$ and $$\vert{\phi}\rangle$$ must satisfy for $$\hat{H}$$ to be hermitian?

I know that for a operator to be Hermitian, it can only admit real eigenvalues and that eigenvectors corresponding to distinct eigenvalues are orthogonal. As I have not been able to wrap my head around this, I've googled some, and also found that it must be diagonalizable? Im not sure if this is something I can take use of... I really appreciate all answers and help!! :)

• … indeed \vert \phi\rangle… Sep 2 '21 at 19:06

Note : I am writing this answer late at night, so I apologize in advance if the explanation is unclear or unnecessarily complicated. I will probably edit it in the future to make it clearer.

$$|\phi\rangle$$and $$|\psi\rangle$$ span a subspace of dimension $$\leq 2$$. We can therefore find a ket $$|\tilde \psi\rangle$$and angles $$\theta,\varphi$$ such that $$(|\phi\rangle,|\tilde\psi\rangle)$$ is orthonormal and : $$|\psi\rangle = e^{i\varphi}\cos(\theta)|\phi\rangle +\sin(\theta)|\tilde \psi\rangle$$

Then, we have : $$H|\phi\rangle = g^*e^{i\varphi}\cos(\theta)|\phi\rangle +g^*\sin(\theta)|\tilde \psi\rangle$$ and : \begin{align} H|\tilde\psi\rangle &= \frac{1}{\sin(\theta)}H(|\psi\rangle-e^{i\varphi}\cos(\theta)|\phi\rangle) \\ &= \frac{g}{\sin\theta}|\phi\rangle-\frac{g^*e^{i\varphi}\cos\theta}{\sin\theta}|\psi\rangle\\ &= \frac{g}{\sin\theta}|\phi\rangle-\frac{g^*e^{i\varphi}\cos\theta}{\sin\theta}(e^{i\varphi}\cos(\theta)|\phi\rangle +\sin(\theta)|\tilde \psi\rangle) \\ &=\frac{1}{\sin(\theta)}(g - g^*e^{2i\varphi}\cos^2(\theta))|\phi\rangle - g^*e^{i\varphi}\cos(\theta)|\tilde\psi\rangle \end{align}

If $$H$$ is to be Hermitian, we need $$\langle \phi|H|\phi\rangle = g^*\langle\phi|\psi\rangle= g^*e^{i\varphi}\cos(\theta)$$ to be real. This means that $$g = e^{2i\varphi}g^*$$ and we have : $$\langle \phi|H|\tilde\psi\rangle =\frac{1}{\sin(\theta)}(g - g^*e^{2i\varphi}\cos^2(\theta)) = \frac{g}{\sin(\theta)}(1-\cos^2(\theta)) =g \sin(\theta) = \langle \tilde\psi|H|\phi\rangle^*$$ as we expect for a Hermitian operator. We also see that $$\langle \tilde\psi|H|\tilde\psi\rangle$$ is real, so $$H$$ is Hermitian.

To recap, $$H$$ is hermitian if, and only if, $$g\langle\psi|\phi\rangle \in\mathbb R$$.

Edit : Here is a simpler and (hopefully) clearer answer.

First, if $$H$$ is hermitian, then $$\langle \psi|H|\psi\rangle = g\langle\psi|\phi\rangle \in \mathbb R$$.

Now, let us assume that this condition holds.Then $$(g\langle \psi|\phi\rangle)^*= g\langle \psi|\phi\rangle$$ and $$(g^*\langle \phi|\psi\rangle)^*= g^*\langle \phi|\psi\rangle$$ Take two arbitrary kets : \begin{align}|1\rangle &= a_0|\psi\rangle + a_1 |\phi\rangle + \sum_{n\geq 2} a_n|\psi_n\rangle\\ |2\rangle &= b_0|\psi\rangle +b_1 |\phi\rangle + \sum_{n\geq 2} b_n|\psi_n\rangle \end{align} and compute : \begin{align} \langle 1 |H|2 \rangle &= \Big( a_0^* \langle \psi| + a_1^*\langle \phi|\Big)\Big(b_0 g|\phi\rangle + b_1 g^*|\psi\rangle\Big) \\ &= a_0^*b_1g^* \|\psi\|^2 + a_1^*b_0g\|\phi\|^2 + a_0^*b_0 g \langle \psi|\phi\rangle + a_1^*b_1 g^*\langle \phi|\psi\rangle\\ &= \Big[a_0b_1^*g \|\psi\|^2 + a_1b_0^*g^*\|\phi\|^2 + a_0b_0^* g \langle \psi|\phi\rangle + a_1b_1^* g^*\langle \phi|\psi\rangle \Big]^* \\ &= \Big[\Big( b_0^* \langle \psi| + b_1^*\langle \phi|\Big)\Big(a_0 g|\phi\rangle + a_1 g^*|\psi\rangle\Big) \Big]^*\\ &=(\langle 2 |H|1\rangle)^* &= \langle 1 |H^\dagger\rangle \end{align}

• I was going to suggest you write up your comments as an answer since you got me out of trouble... Thanks for that. I upvoted. Sep 2 '21 at 21:48
• Thank you so much @SolubleFish! :) Sep 3 '21 at 7:07