Implications of orthonormal wavefunctions 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?
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.