In the proof of Wigner's theorem, the crucial role is played by the quantity $|\langle\phi|\psi\rangle|^2$ where $|\psi\rangle$ and $|\phi\rangle$ represent two distinct physical states (or more precisely, two distinct rays) of a system. Expanding the states in the orthonormal eigenbasis $\{|a_n\rangle\}$ of a Hermitian operator $\hat{A}$ as $$|\psi\rangle=\sum_n c_n|a_n\rangle, \quad |\phi\rangle=\sum_n d_n|a_n\rangle,$$ the object of interest can be written as, $$|\langle\psi|\phi\rangle|^2=\left|\sum_n d_n^*c_n\right|^2.$$ Wigner's theorem is based on the notion of symmetry transformation which changes the states from $|\psi\rangle\to|\psi^\prime\rangle$ and $|\phi\rangle\to|\phi^\prime\rangle$ keeping $|\langle\phi|\psi\rangle|^2$ unchanged. Well, we could define symmetry transformation in this abstract way and proceed from here.

But I would like to feel it in my bones. What does the object $|\langle\psi|\phi\rangle|^2$ mean, physically? Even if not measurable, I would at least like to understand, why we expect this quantity to remain unchanged under a rotation, translation, etc.

Please note that neither $\psi$ nor $\phi$ be eigenstates of any hermitian operator. If $|\phi\rangle=|a_m\rangle$, where $|a_m\rangle$ is an eigenstate of a hermitian operator $\hat{A}$ with eigenvalue $a_m$, then postulate of QM, the quantity $|\langle a_m|\psi\rangle|^2$ has a direct physical meaning. It is the probability that the outcome of measurement will of the observable $A$, will return a value $a_m$. But $|\langle\phi|\psi\rangle|^2$, for a general state $|\phi\rangle$, do not have any such interpretation, as far as I understand.

  • $\begingroup$ Note that the similar thing happens with vectors and tensors in non quantum physics. $\endgroup$
    – Mauricio
    Jun 1, 2023 at 18:34
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    $\begingroup$ Regarding the "please note" - every state $|\phi\rangle$ is an eigenstate of a Hermitian operator! One can define a Hermitian operator as $|\phi\rangle\langle \phi|$ and be done, or do something more general like $H=h_1|\phi\rangle\langle \phi|+h_2|\phi_\perp\rangle\langle \phi_\perp|$ where $\langle \phi_\perp|\phi\rangle=0$, etc. As with @Mauricio's comment, this is true for any vector $\endgroup$ Jun 1, 2023 at 19:56

5 Answers 5


If you're looking for an answer with words more than with equations: when you "compare" two things, a symmetry means that you shouldn't have to worry where you are when you compare them.

In quantum theory, comparing two states is done by $|\langle \phi|\psi\rangle|^2$. It tells us the probability that, after preparing state $|\psi\rangle$, we will measure it to be state $|\phi\rangle$; alternatively, it is like comparing how close the two vectors are in Hilbert space. Symmetries mean that we don't care where we are in Hilbert space, we just care how close together these two vectors are.


Consider a normalised wave-function $\vert \phi\rangle$.

We know that the expectation value of some observable $\langle \hat{A} \rangle =\langle \phi\vert\hat{A}\vert \phi\rangle$

Now let's find the expectation that it will be projected onto some state $\vert\psi\rangle$ that is finding out the expectation value of projection operator for $\vert\psi\rangle$ given by $P_{\psi}=\vert\psi\rangle\langle\psi\vert$

This gives us $\langle \phi\vert P_{\psi}\vert \phi\rangle=\langle \phi\vert\vert\psi\rangle\langle\psi\vert\vert \phi\rangle=\vert\langle\psi\vert\vert \phi\rangle\vert ^2$

You can interpret it as the expectation value of projection of state $\phi$ on some state $\psi$.

Now all the transformation of my frame of reference are just changes in my way of measurement and physics or nature should not care about how we measure it. There is some underlying symmetry if it is preserving something (noether theorem) and that's why such transformations are symmetry transformations.

Under all such transformation, the length and direction of my vector should not change (which has all the physics in them) just because I decided to look it from a different frame of reference, length(norm) is deducted from inner product of a vector with itself and direction can be deducted from inner product of vector with basis vectors,i.e. the inner product should not change under any such transformations, (this is not abstract but rather a necessity)

Inner product in case of Hilbert vector space is defined as $$(\psi,\phi)=\langle\psi\vert\vert \phi\rangle=\int^{\infty}_{-\infty}\psi^*\phi\ dq$$ where $q$ are my generalized coordinates of configuration space in which the said wave is there of equiaction surfaces. More on it here

The operators that preserve the inner product are Unitary operators.

That's why we generally deal with the Unitary operators in Q.M. as they preserve the inner product and as a consequence also preserve $\vert\langle\psi\vert\vert \phi\rangle\vert ^2$


Quick but loose thoughts:

  1. Yes, such inner products correspond to probabilities (or amplitudes) which can be measured if you can set up / measure states $\psi$ and $\phi$.

  2. More fundamentally, you can think of the symmetry (S) as transforming the inner product on the Hilbert space $$\langle \psi | \phi \rangle \equiv \langle \psi | \Omega | \phi \rangle \xrightarrow{S} \langle \psi | S^{-1} \cdot \Omega \cdot S | \phi \rangle$$ and you can decompose $\Omega$ in an eigenbasis of $S$ to see that the inner product must be perserved if $S$ is to be a symmetry of the Hilbert space.

  • $\begingroup$ Thanks, @Siva But I don't understand point 1 of your answer. Let me explain a bit. If $|a_n\rangle$ is an eigenstate of a hermitian operator with eigenvalue $a_n$, then $|\langle a_n|\psi\rangle|^2$ is the probability that the outcome of measurement will return a value $a_n$. This is a postulate of quantum mechanics. But $|\langle\phi|\psi\rangle|^2$ do not have any such interpretation unless $|\phi\rangle$ is an eigenstate of a hermitian operator pertaining to some physical observable. $\endgroup$ Jun 1, 2023 at 4:49
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    $\begingroup$ Yes, but, as stated by , $|\phi\rangle$ is always an eigenstate of the hermitian operator $P_{|\phi\rangle} = |\phi\rangle \langle \phi |$. $\endgroup$
    – Tarik
    Jun 1, 2023 at 14:25

$|\langle \phi|\psi\rangle |^2$ is the probability to measure a system in the state $|\phi\rangle$, if the system was prepared in a site $|\psi\rangle$ before measurement. This is a consequence of the axioms of quantum mechanics, namely :

If $|a_n\rangle$ is an eigenstate of a hermitian operator $\hat A$ with eigenvalue an, then $|\langle a_n|\psi\rangle |^2$ is the probability that the outcome of measuring $\hat A$ in the state $|\psi\rangle$ will return the value $a_n$.

To see that this, consider $P_\phi = |\phi\rangle\langle\phi|$. This is a self-adjoint operator and therefore an observable. $|\phi\rangle$ is its eigenstate for the eigenvalue $1$, corresponding to "the system is in the state $ \phi$. The other eigenstates, with eigenvalue $0$, correspond to "the system is not in the state $\phi$". Therefore, $|\langle \phi|\psi\rangle|^2$ is the probability that the system in state $|\psi\rangle$ will be found in state $|\phi\rangle $ instead.


Under a translation, the overlap is invariant $$\eqalign{ \langle\phi|\psi\rangle &=\int \phi^*(x)\psi(x)dx\cr &=\int \phi^*(x+a)\psi(x+a)dx\cr &= \langle\phi’|\psi’\rangle }$$ This can be easily generalized to rotation.


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