2
$\begingroup$

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!! :)

$\endgroup$
1
  • 1
    $\begingroup$ … indeed \vert \phi\rangle… $\endgroup$ Sep 2 at 19:06
2
$\begingroup$

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}

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Sep 2 at 21:48
  • $\begingroup$ Thank you so much @SolubleFish! :) $\endgroup$ Sep 3 at 7:07

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.