For orthonormal functions $\psi$ and $\phi$, by definition; $$\langle \psi|\phi\rangle=0$$ However, when a wavefunction, $f$, is expressed as a linear combination of these orthonormal wavefunctions: $$f=c_1\psi+c_2\phi$$ operating on this function with an arbitrary operator, $\hat A$ results in: $$\langle f|\hat A|f\rangle=\langle c_1\psi+c_2\phi|c_1\hat A\psi+c_2\hat A\phi\rangle=c_1^2\langle\psi|\hat A|\psi\rangle+c_1c_2\langle\psi|\hat A|\phi\rangle+c_2c_1\langle\phi|\hat A|\psi\rangle+c_2^2\langle\phi|\hat A|\phi\rangle$$

My question is, do the crossed terms equal zero? (i.e does $\langle\psi|\hat A|\phi\rangle$ and $\langle\phi|\hat A|\psi\rangle$ equal zero?)

I initially thought that, yes, they do equal zero because, suppose that $\psi$ and $\phi$ are eigenfunctions of the operator $\hat A$ then it follows that: $$\langle\phi|\hat A|\psi\rangle=\langle\phi|a|\psi\rangle=a\langle\phi|\psi\rangle=a(0)=0$$ However, in a problem sheet I have just done, the question relies on these terms not being zero. Is my logic wrong or does this only equal zero in the case where the functions are eigenfunctions of the operator?

  • 1
    $\begingroup$ What's the exact problem? I think your logic is right, if $\hat{A}|\psi\rangle \neq a|\psi\rangle$ then the term is not equal to $0$. $\endgroup$ Oct 30, 2015 at 19:23
  • $\begingroup$ The problem I was set isn't massively relevant I just want to know if $\langle\psi|\hat A|\phi\rangle=0$ or not. Are you saying that it is non zero if the function is not an eigenfunction? $\endgroup$
    – RobChem
    Oct 30, 2015 at 19:25
  • $\begingroup$ It could be zero, but there's no reason it should be if the operator is arbitrary. $\endgroup$ Oct 30, 2015 at 19:28
  • $\begingroup$ If the $|\phi|\rangle$ and $|\psi\rangle$ are eigenfunctions of $\hat A $ and are orthogonal, then you are right: $\langle\phi|\hat A|\psi\rangle=\langle\psi|\hat A|\phi\rangle = 0$. $\endgroup$
    – Philipp
    Oct 30, 2015 at 19:31
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    $\begingroup$ Let's assume that $\phi$ and $\psi$ are eigenstates of some hermitian operator $\hat{B}$. This operator will generate a complete basis set of eigenfunctions, which includes $\phi$, $\psi$, and infinity more wavefunctions, which we'll call $|f_1\rangle$, $|f_2\rangle$ and so on. Now, if we denote $A|\psi\rangle = |g\rangle$, then you can expand $|g\rangle$ in the $\hat{B}$ basis: $|g\rangle = a|\phi\rangle + b|\psi\rangle + c_1|f_1\rangle + c_2|f_2\rangle + \cdots$ Only if the value of $a$ in this expansion is $0$, will $\langle\phi|\hat{A}|\psi\rangle$ be equal to $0$. $\endgroup$ Oct 30, 2015 at 19:42

1 Answer 1


If you have two orthonormal states $|\psi\rangle$ and $|\phi\rangle$, $$ \langle \psi|\phi\rangle = 0, $$ there is no guarantee that the term $$ \langle\psi|\hat{A}|\phi\rangle $$ vanishes. All depends on the action of $\hat{A}$ on your state.

As you already noticed, if $|\phi\rangle$ is an eigenstate of $\hat{A}$ with eigenvalue $a_\phi$, the term is obviously zero: $$ \langle\psi|\hat{A}|\phi\rangle = \langle\psi|a_\phi|\phi\rangle =a_\phi \langle\psi|\phi\rangle = 0. $$ But this is only a special case, since $|\phi\rangle$ is not necessarily an eigenstate of $\hat{A}$ in general.

One of the simplest situations where this term does not vanish can occour when $|\psi\rangle$ and $|\phi\rangle$ are eigenstates of the harmonic oscillator, that we can denote $|n\rangle$ and $|n'\rangle$. As you probably know, these states are orthonormal: $$ \langle n|n'\rangle = \delta_{n,n'}. $$ If you consider the specific case $|\psi\rangle=|n\rangle$ and $|\phi\rangle=|n-1\rangle$, it immediately follows that $$ \langle n|n-1\rangle = 0. $$ Now if you take $\hat{A}= \hat{a}^\dagger$, for which you have $\hat{a}^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$ ($|n\rangle$ is not an eigenstate of $\hat{a}^\dagger$), you have ($n\geq1$) $$ \langle n|\hat{a}^\dagger|n-1\rangle = \langle n|\sqrt{n}|n\rangle = \sqrt{n} \langle n|n\rangle = \sqrt{n}\delta_{n,n} = \sqrt{n} \neq 0. $$

As I already stated, all depends on the action of $\hat{A}$ on your state.

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    $\begingroup$ Nice explanation, but perhaps more importantly, nice example! $\endgroup$ Nov 20, 2015 at 10:28

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