# Unitary Transformation of Eigenstates

Suppose I have two operators, $A$ and $B$, with eigenstates $A \lvert a \rangle = a \lvert a \rangle$ and $B \lvert b \rangle = b \lvert b \rangle$, where $a$ and $b$ are all unique. Furthermore, suppose that $A$ and $B$ are related by a unitary transformation $$A = U B U^{-1}.$$ This is equivalent to saying that the eigenstates are related as $$\lvert a \rangle = U \lvert b \rangle.$$

Then it seems I can prove the following: since $$A \lvert a \rangle = a \lvert a \rangle,$$ I also have $$A U \lvert b \rangle = U B U^\dagger U \lvert b \rangle$$ by inserting the identity, so that $$A U \lvert b \rangle = U B \lvert b \rangle = b U \lvert b \rangle = b \lvert a \rangle.$$ Thus, $a = b$.

Doesn't this imply then that the eigenvalues for corresponding eigenstates of $A$ and $B$ are equal, and therefore-- by the assumption that they are unique-- that the unitary transformation doesn't actually do anything?

• As nicely summarized by @ZeroTheHero, when you write $A=UBU^{\dagger}$ you are performing a unitary basis change on operator $B$. $A$ is then the new representation of $B$ in this new set of basis. Eigenvalues are invariant under basis transformation . Thus, eigenvalues of $A$ and $B$ are the same, but their eigenstates are not. – Ptheguy Jan 5 '18 at 8:52
• Here is a good article if you're not sure why eigenvalues are invariant under basis transformation. (link). – Ptheguy Jan 5 '18 at 9:00
• It might help to choose as an example $A=x$ and $B=p$, where $U$ is the Fourier transform and takes $x$ to $p$. Here $x$ and $p$ have identical eigenvalue spectra, but that doesn't mean that the Fourier transform doesn't "do" anything. – Emilio Pisanty Jan 5 '18 at 10:03

I’m not sure what you mean by “$a$ and $b$ are unique” but clearly if $A=UBU^\dagger$ and $U$ is unitary, $A$ and $B$ have the same eigenvalues but it doesn’t mean $U$ doesn’t do anything.

For instance, the Pauli matrices $\sigma_{x,y,z}$ all have the same eigenvalues, are related by a unitary transformation $U$, but are certainly different. The transformation $U$ is a change of basis, so if $B$ is initially diagonal, say $$B=\sigma_z=\left(\begin{array}{cc} 1&0 \\ 0&-1\end{array}\right)$$ and $U=\left(\begin{array}{cc} \cos\theta/2&-\sin\theta/2\\ \sin\theta/2 &\cos\theta/2\end{array}\right)$ then $$U\sigma_zU^{\dagger}=\cos\theta\sigma_z+\sin\theta \sigma_x$$ still have eigenvalues $\pm 1$ but obviously $U$ has done something.
Of course the eigenstates are no longer $\left(\begin{array}{c}1\\ 0\end{array}\right)$ and $\left(\begin{array}{c}0\\ 1\end{array}\right)$.

I interpret your uniqueness claim as the requirement that

every eigenspace of either $A$ and $B$ has dimension $1$.

So, if $a$ is an eigenvalue of $A$ and $A|a\rangle =a|a\rangle$, for $|a\rangle \neq 0$, then every other eigenvector with the same eigenvalue $a$ is of the form $c|a\rangle$ for every $c\in \mathbb C$, $c\neq 0$. If we consider only normalized eigenvectors, $c$ is of the form $e^{i \theta}$ for every $\theta \in \mathbb R$.

Assuming that the spectra of the said operators are pure-point spectra, we have the spectral decompositions (the sum are understood in the strong operator topology, but here it is quite irrelevant and you can safely interpret all follows at algebraic level) $$A = \sum_{a \in {\cal A}} a |a\rangle\langle a| \tag{1}$$ and $$B = \sum_{b\in {\cal B}} b |b\rangle\langle b| \:.$$ where I used normalized eigenvalues and ${\cal A}=\sigma(A)$, ${\cal B}=\sigma(B)$ (up to accumulation points) are the spectra of the operators.

On the other hand $$A = UBU^\dagger$$ implies $$\sum_{a \in {\cal A}} a |a\rangle\langle a| = \sum_{b \in {\cal B}} b U|b\rangle\langle b|U^\dagger\:.$$ That is $$A= \sum_{b \in {\cal B}} b|\psi_b\rangle \langle \psi_b|\tag{2}$$ where $$|\psi_b\rangle := U|b\rangle\:.$$

The crucial result is now that

for a given self-adjoint operator (with point spectrum) the spectral decomposition is unique.

Thus, comparing (1) and (2) we conclude that

(i) ${\cal A}= {\cal B}\:,$

so that we can re-arrange the decomposition of $A$ like this $$A= \sum_{a \in {\cal A}} a|\psi_a\rangle \langle \psi_a|\:,$$

(ii) $|a\rangle \langle a| = |\psi_a\rangle \langle \psi_a|$ so that, since every vector is normalized $$|\psi_a\rangle = e^{i\theta_a}|a\rangle\quad \mbox{for some \theta_a \in \mathbb R}\:.$$ there is no way to determine the phases $e^{i\theta_a}$, since normalized eigenvectors are defined up to a phase, but we are free to fix all them $e^{i\theta_a}=1$.

I stress that (i) and (ii) are the maximum you can obtain from the initial information at your disposal that eigenspaces are one-dimensional and that the unitary equivalence $A=UBU^\dagger$ holds.

You see that $U$ has an action (it is false that "it doesn't actually do anything"). In fact, it changes eigenvectors, but it leaves fixed the spectrum of the operators.

Dropping the hypotheses of one-dimensional eigenspaces but keeping the request of pure point spectrum, (i) remains valid in view of the uniqueness of spectral decomposition, which now reads $$A = \sum_{a \in {\cal A}} a P_a$$ where $P_a = \sum_{k=1}^{\dim E_a} |a,k\rangle \langle a,k|$ is the orthogonal projector onto the eigenspace $E_a$ of $A$ with eigenvalue $a$ and the vectors $|a,k\rangle$, varying $k=1,\ldots, \dim E_a$, form an orthonormal basis of that eigenspace.

Well, a similarity transformation for an invertible (not necessary unitary) operator$^1$ $U$ does generically change the eigenspaces but does not change the eigenvalue spectrum $\{a_1, a_2,\ldots, \}=\{b_1,b_2,\ldots\}$. Hence it would be inconsistent to claim that all the eigenvalues $\{a_1,a_2, \ldots, b_1,b_2,\ldots\}$ for both $A$ and $B$ are different, if that's what OP means by that they are all unique. Whether the individual spectra are degenerate or not is irrelevant.

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$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.