# Is this a sufficient condition for a state to be an eigenstate?

$\renewcommand{ket}{|#1\rangle} \renewcommand{bra}{\langle#1|} \renewcommand{braket}{\langle#1|#2\rangle} \renewcommand{bbraket}{\langle#1|#2|#3\rangle}$ I'm trying to wrap my head around quantum eigenstates, eigenvectors, and braket notation. I'm still fairly new to this.

Let $\Psi=\{\ket{\psi_n}\,|\,n=1\dots N\}$ be a complete set of all orthonormal eigenstates of operator $A$, with corresponding eigenvalues $\{a_n|n=1\dots N\}$, where $a_m = a_n$ iff $m=n$. Suppose a set of states is found $\Phi=\{\ket{\phi_m}\,|\,m=1\dots M\}$ such that $\bbraket{\phi_n}{A}{\phi_m} = a_n \delta_{nm}$ for all $\ket{\phi_n},\ket{\phi_m}\in \Phi$. If $N=M$, can we say that $\Psi=\Phi$? More generally, if $M<N$, must $\Phi\subseteq \Psi$?

EDIT

The above is an attempt to make precise a question I was given as a practice question in preparation for an upcoming exam. At this point, the question as stated above resembles the original wording very little. In case I've missed something of importance, here is the original wording of the question, verbatim (except for trivial notation changes to be consistent with the above):

Suppose $A$ is a Quantum Mechanical operator, with eigenstate $\psi_n$ and corresponding eigenvalues $a_n$, i.e. $A\ket{\psi_n} = a_n\ket{\psi_n}$ and $\bra{\psi_n}A^\dagger = a_n^\ast\bra{\psi_n}$, with $\braket{\psi_n}{\psi_m} = \delta_{nm}$. If $\bbraket{\phi_n}{A}{\phi_m} = a_n\delta_{nm}$, are the states $\ket{\phi_n}$ eigenstates of $A$?

• I think you're missing something: what is $\lvert\phi_m\rangle$? You introduced an arbitrary state $\lvert\phi_n\rangle$, but not another state $\lvert\phi_m\rangle$ that you can use to compute the matrix element $\langle\phi_n\rvert A\lvert\phi_m\rangle$. Perhaps you meant a set of states? And then is the goal to prove that all of these states are eigenstates? Or just one? – David Z Jun 21 '15 at 14:35
• Thank you, I have edited the question. I guess another way to ask my question is: does the set of eigenstates $|\phi_n\rangle$ that all satisfy the above criterion necessarily coincide with the given eigenstates $|\psi_n\rangle$? – robotopia Jun 21 '15 at 14:47
• It's not quite possible to answer that question as asked, because the criterion you propose is actually one that applies to pairs of states $\lvert\phi_n\rangle$ and $\lvert\phi_m\rangle$, whereas the definition of an eigenstate is something that applies to single states. You can talk about larger sets of states, though. – David Z Jun 21 '15 at 15:15
• Would it make more sense if I asked this? Let $F=\{|\psi_n\rangle\,|\,n=1\dots N\}$ be the (complete) set of all orthonormal eigenstates of $A$, with corresponding eigenvalues $\{a_n\,|\,n=1\dots N\}$. Suppose a set of states is found $G=\{|\phi_m\rangle\,|\,m=1\dots M\}$ such that $\langle \phi_n|A|\phi_m \rangle = a_n\delta_{nm}$ for all $|\phi_n\rangle, |\phi_m\rangle \in G$. If $N=M$, can we say that $F=G$? – robotopia Jun 21 '15 at 15:42
• Yeah, that's a lot better. I figured that's more or less what you were trying to ask, but thought I should clarify to be sure. BTW I can tell you it's obviously not true if there are repeated eigenvalues, because if $a_{n_1} = a_{n_2}$ then $\alpha\lvert\psi_{n_1}\rangle + \beta\lvert\psi_{n_2}\rangle$ is also an eigenvector for any $\alpha,\beta$. But perhaps for distinct eigenvalues, it could work. P.S. People might say this is a math question, which is fine; we can always migrate it to Mathematics if that is the case, so don't worry about it. – David Z Jun 21 '15 at 16:06

$\renewcommand{ket}{|#1\rangle}$ Any linear combination of states in the set $\Phi=\{\ket{\phi_m}|m=1\ldots M\}$ is going to be an eigenstate of $A$. I know this because this statement: $\langle \phi_n|A|\phi_m \rangle = a_n \delta n_m$ is really saying that you can represent the operator $A$ as an $M \times M$ matrix in the $\Phi$ basis. in this basis $A$ becomes a matrix with Eigenvalues am down the diagonal, and the $\ket{\phi_m}$'s become column vectors with a 1 in the $m^{\text{th}}$ row and zeros everywhere else. from this you can see that $A\ket{\phi_m} = a_m\ket{\phi_m}$, where am is just a complex number, meaning that |ϕm⟩ is an eigenstate of $A$. if youre studying from griffiths, check out chapter 3 (formalism, dirac notation)

As far as $\Psi$ and $\Phi$ being the same set, I don't think this is necessarily true. If you take the parity operator as an example, any state that is antisymmetric under exchange or symmetric under exchange will be an eigenstate with the same eigenvalue, even if they are completely different states. for example, you could have a spin wavefunction of two electrons, or a spatial wavefunction of two electrons that have the same eigenvalue under parity (-1), and they could easily be some part of an orthonormal basis with the same dimension.

• Hi dead cat and welcome to Physics.SE! Please see this help post to learn how to write your equations in a way nicer way i.e. in $\LaTeX$, in order to improve legibility. Thanks! – Gonenc Jun 21 '15 at 22:28
• My gut tells me this is what the author was getting at with his original question, but I'm still not clear (for example) how this tallies with Timaeus' counter example below. – robotopia Jun 22 '15 at 9:16
• It is a bit misleading that the quantities $\langle\phi_n|A|\phi_m\rangle$ are called matrix elements, since they are not generally the entries of the matrix of $A$ in the basis $\phi_i$. This is only the case when they form an orthonormal basis. In Timaeus' counterexample the basis is not orthonormal. – doetoe Jun 22 '15 at 23:07
• @doetoe the Phi stationary states are definitely orthogonal, normalized, and eigenstates of A or the last equation (in the original wording of the question) would not be true – dead cat Jun 24 '15 at 18:08

When you have an observable, the eigenvectors with different eigenvalues are orthogonal, but your second set does not require that the vectors be orthogonal. And since the elements of the second set are not required to be normalized we learn very very little (almost nothing) about how $A$ scales any particular vector, or even whether it scales the vector or does something else.

And all your conjectures are false and they all have counterexamples in two dimensions (they are all true in one dimension).

Consider $A=\sigma_z=\left[\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right]$ with $\Psi_1=\left[\begin{matrix} 1 \\ 0 \end{matrix}\right]$ and $\Psi_2=\left[\begin{matrix} 0 \\ 1 \end{matrix}\right]$ but we can consider a $\Phi_1=\left[\begin{matrix} a \\ b \end{matrix}\right]$ and all we need is that $a^*a-b^*b=1.$ This is a counter example to your generalization for $M<N$ since you could have for instance $\Phi_1=\left[\begin{matrix} \sqrt 2 \\ 1 \end{matrix}\right].$

But this also has zero relationship with the cited problem. Your nongeneralized problem is also false and again for reasons related to lack of orthonormalization of $\Phi_k$ and that also have no bearing on the problem cited. For example consider $A=\left[\begin{matrix} 4 & 0 \\ 0 & 9 \end{matrix}\right]$ with $\Psi_1=\left[\begin{matrix} 1 \\ 0 \end{matrix}\right]$ and $\Psi_2=\left[\begin{matrix} 0 \\ 1 \end{matrix}\right]$ but we can consider $\Phi_1=\left[\begin{matrix} 0 \\ a\end{matrix}\right]$ and $\Phi_2=\left[\begin{matrix} b \\ 0 \end{matrix}\right]$ so now the requirements are that $9a^*a=4$ and $4b^*b=9$ so we can for instance use $\Phi_1=\left[\begin{matrix} 0 \\ 2/3\end{matrix}\right]$ and $\Phi_2=\left[\begin{matrix} 3/2 \\ 0 \end{matrix}\right]$ and note that this is a counter example to your question but satisfies the cited question. An even simpler counter example would be just to multiply $\Psi_1$ by $-1.$

Which means your question is wrong, but the cited question is still up in the air. The cited question makes no claim about nondegenerate spectra or about how many vectors you have. And it isn't even clear that the spectra has to be real because it just days operator, not observable. But we can find a counterexample where it is an observable and has a nondegenerate spectra.

If all you want to do is understand the notation think of kets as column vectors, operators as square matrices and bras as row vectors, then think of dagger as taking the transpose of the matrix and then the complex conjugate of evey element of the matrix. Think of the bra version of a ket as the dagger of the ket. That's what you need to see what is going on.

But back to your cited question. If you have a bunch of orthonormal $\Psi_k$ that are eigenvectors of $A$ then this doesn't tell us anything at all about what happens outside the (closure) of the linear span of those eigenvectors, and we can definitely find a counter example if the $\Psi_k$ live in some proper subspace of the Hilbert space. So either that was a counterexample we were supposed to find, or else let's assume we have lots and lots of eigenvectors, $\Psi_k,$ say enough to span a dense subset of the Hilbert space. We can then actually write $A=B+iC$ for some observables $B$ and $C$ ($C=0$ if A is an observable). And we can then do it so $B$ and $C$ have common eigenvectors with the real and imaginary parts of the $a_k$ corresponding to the eigenvalues of $\Psi_k$ with $B$ and $C$ respectively.

OK now we want to consider whether the $\Phi_k$ are eigenvectors of $A.$

This again has a counterexample in the land of 2x2 matrices. Let $A=\left[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right]$ with $\Psi_1=\left[\begin{matrix} 1 \\ 0 \end{matrix}\right]$ and $\Psi_2=\left[\begin{matrix} 0\\ 1 \end{matrix}\right]$ but we can consider $\Phi_1=\left[\begin{matrix} 1 \\ 1\end{matrix}\right]$ and $\Phi_2=\left[\begin{matrix} 0\\ 1 \end{matrix}\right]$ it satisfies all the conditions but $\Phi_1$ is not an eigenvector of $A.$

Each time I just thought square matrix when I saw an operator, thought column vector when I saw a ket, thought row vector when I saw a bra, and thought transpose and conjugate when I saw a dagger and thought dagger of when a saw a bra version of a ket.

There can be some differences but they really come down to just extra details about infinite dimensional spaces and that is just because of unfamiliarity with infinite dimensional spaces not because they are weird or different the nice infinite dimensional spaces are nice and the finite dimensional spaces just happen to already be nice so you aren't used to worry about whether something is nice.