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I was wondering how to physically interpret the inner product between two quantum states. Say that we have an operator $A$ to which the eigenvector $\psi_A$ belongs, and an operator $B$ to which the eigenvector $\psi_B$ belongs. Then we can determine the inner product as;

$$\langle\psi_A||\psi_B\rangle=\langle\psi_A|\psi_B\rangle$$

I understand that the inner product describes the overlap of the two quantum state, but what exactly does this mean? If we set up a quantum-mechanical experiment, what does the overlap between eigenvectors belonging to different operators tell us? We can also write;

$$|\langle\psi_A|\psi_B\rangle|^2$$

Say that we want to measure a particle which starts in state $\psi_B$. As far as I am concerned the expression then describes the probability of measuring the particle in state $\psi_A$ (given that the particle started in state $\psi_B$). This has a very explicit physical description, but I do not understand how the inner product can be described in a similar explicit physical way.

$\textbf{Update:}$ Thank you for the many answers; they were all very insightful. It was hard to choose a specific answer as they all contributed with useful knowledge, but I picked the one I thought was most specific to my question.

$\textbf{Reedit:}$ After studying more quantum mechanics, I changed the answer to this question to explain the more general case, however, the previous answer (by catalogue_number) was still helpful.

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6 Answers 6

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Assume the the particle is in state $\psi_B\,.$ The projection $$ P_A=|\psi_A\rangle\langle\psi_A| $$ is a Hermitian operator having eigen values zero and one which can be used to measure if the particle is in state $\psi_A\,.$ This measurement outcome has the expected value $$ \langle \psi_B|P_A|\psi_B\rangle=|\langle\psi_A|\psi_B\rangle|^2\,. $$

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    $\begingroup$ Forgive me being terse. The discussion what is real in QM I better leave to others. Measurement outcomes are real. States? I have no idea. You may want to read stuff like https://arxiv.org/abs/1111.3328. $\endgroup$
    – Kurt G.
    Commented Jul 24, 2023 at 18:44
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Imagine you have a coherent beam of light / electrons / cold atoms incident on two slits, $A$ and $B$. First, you cover $B$ and measure the wave field $I_A(r) = \langle \psi_A (r)| \psi_A(r) \rangle$. You do the same with the other slit to get the intensity pattern $I_B(r)$.

Uncovering both slits reveals the intensity pattern $$I_{A+B} = I_A(r) + I_B(r) + 2\mathrm{Re}\langle\psi_A(r)|\psi_B(r)\rangle$$ which, using the so-called polarisation identity, reveals an interpretation of the overlap integral as $$\mathrm{Re}\langle\psi_A(r)|\psi_B(r)\rangle = \frac{1}{2}\left[I_{A+B}(r) -I_A(r)-I_B(r)\right]$$ which is to say, the extent to which the states interfere with one another. Note that when $A$ and $B$ are orthogonal, their intensities add 'classically': $$I_{A+B} = I_A + I_B$$

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  • $\begingroup$ Well, here $\psi_A$ and $\psi_B$ are part of the same state. I am thinking of a screen with two holes and $\psi_A$ and $\psi_B$ are the parts of the same wavefunction after the screen, due to the two holes. This is the reason why the scalar product is not affect by arbitrary pahses, it is the same state, you cannot change the phase of $\psi_A$ without changing the phase of $\psi_B$ and the changes cancel each other. $\endgroup$ Commented Jul 26, 2023 at 10:24
  • $\begingroup$ But all that cannot be applied to the general case where $\psi_A$ ad $\psi_B$ are unrelated states and, in this case, their scalar product has no direct physical meaning. In this sense I find this answer a bit misleading. $\endgroup$ Commented Jul 26, 2023 at 10:24
  • $\begingroup$ You're right, of course - in order for the product to have a physical meaning, there must be some notion of transport (e.g. Berry connection) that allows them to interfere. If they are two states in boxes at opposite ends of the universe, there can be no sensible physical meaning for $\langle A | B \rangle$ $\endgroup$ Commented Jul 26, 2023 at 12:45
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First of all $\langle \psi|\phi\rangle$ is not the inner product between two quantum states. It is between vectors. But pure states are represented by unit vectors up to a phase, therefore $$\langle \psi|\phi\rangle$$ has no direct physical meaning as it is not invariant under $\psi \to e^{ia}\psi$, $\phi \to e^{ib}\phi$ for $a,b \in \mathbb{R}$, though it enters several intermediate mathematical computations giving rise to known interference terms.

What is physically meaningful is instead $$|\langle \psi|\phi\rangle|\:.$$ It encompasses the same information as $$|\langle \psi|\phi\rangle|^2\:,$$ and this is the transition probability between the two states as you know.

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    $\begingroup$ The expression $\langle \psi |\phi\rangle$ is definitely the inner product between two vectors in a Hilbert space. The fact whether these represent something physical is a different question. $\endgroup$ Commented Jul 25, 2023 at 9:31
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    $\begingroup$ I do not understand the -1 what is wrong with my answer? $\endgroup$ Commented Jul 25, 2023 at 10:20
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    $\begingroup$ And I cannot understand also the +1 to @AccidentalTaylorExpansion's comment as it refers to something which is not against my answer. The question was about the physical meaning of $\langle\psi|\phi\rangle$, I argued that there is no direct physical meaning since that object is affected by ambiguities. The ambiguities are removed whan taking the absolute value,. What is wrong with it? $\endgroup$ Commented Jul 25, 2023 at 10:25
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    $\begingroup$ As it was written, the first sentence would have lead to confusion. After the edit I have retracted my downvote $\endgroup$ Commented Jul 25, 2023 at 13:16
  • $\begingroup$ Your answer is interesting: it suggests that vectors (or states) are first defined up to a phase, from which is follows that $\langle\psi\vert\phi\rangle$ has no meaning. Historically at least, it's the other way around: Born suggested the physics was in $\vert\langle\psi\vert\phi\rangle\vert^2$, from which it follows that the overall phase of a vector is irrelevant. $\endgroup$ Commented Jul 25, 2023 at 15:01
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The overlap itself means nothing. In general it is a complex amplitude so not expected to be connected to any measurable quantities, which must be real. There is only physics associated with the modulus squared of the overlap.

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$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\clr}[1]{\left\{#1\right\}} \newcommand{\vlr}[1]{\left\vert#1\right\vert} \newcommand{\vra}[1]{\left|#1\right\rangle} \newcommand{\lavra}[2]{\left\langle#1|#2\right\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\Vp}[1]{\vphantom{#1}} \newcommand{\tl}[1]{\tag{#1}\label{#1}}$

Although mathematical and not physical I like the interpretation of inner product given by Feynman in his lectures on Quantum Mechanics.

Consider that the Hilbert space of states is of finite dimension $\:n\:$ and let a complete orthonormal basis of states $\:\clr{\vra{\mb e_1},\vra{\mb e_2},\cdots,\vra{\mb e_k},\cdots,\vra{\mb e_n}}$. For normalized quantum states $\:\vra{\bl \psi_\mr A}\:$ and $\:\vra{\bl \psi_\mr B}\:$ we have \begin{align} \!\!\!\!\vra{\bl \psi_\mr A} & \e \mr a_1\vra{\mb e_1} \p \mr a_2 \vra{\mb e_2} \p \cdots \mr a_k \vra{\mb e_k}\p\cdots\mr a_n\vra{\mb e_n}\,, \:\:\sum\limits_\imath \vlr{\mr a_\imath}^2 \e 1\,, \:\: \mr a_\imath \bl\in \mathbb C \tl{01.a}\\ \!\!\!\!\vra{\bl \psi_\mr B} & \e \mr b_1\vra{\mb e_1} \p \mr b_2 \vra{\mb e_2} \p \cdots \mr b_k \vra{\mb e_k}\p\cdots\mr b_n\vra{\mb e_n}\,, \:\: \sum\limits_\jmath \vlr{\mr b_\jmath}^2 \e 1\,, \:\: \mr b_\jmath \bl\in \mathbb C \tl{01.b} \end{align} For the probability to find the particle in state $\:\vra{\bl \psi_\mr A}\:$ if it starts in state $\:\vra{\bl \psi_\mr B}\:$ we use probability amplitudes and the laws of combining them, see Figure-01.

The probability amplitude for the transition from the state $\:\vra{\bl \psi_\mr B}\:$ to the basic state $\:\vra{\mb e_1}\:$ is $\:\mr b_1$, while in turn the probability amplitude for the transition from the basic state $\:\vra{\mb e_1}\:$ to the state $\:\vra{\bl \psi_\mr A}\:$ is $\:\mr a^{\bl *}_1$,$\:\:$the complex conjugate of $\:\mr a_1$. The amplitude for the transition from the state $\:\vra{\bl \psi_\mr B}\:$ to state $\:\vra{\bl \psi_\mr A}\:$ passing through the intermediate basic state $\:\vra{\mb e_1}\:$ is the product of the amplitudes $\:\mr b_1\,\mr a^{\bl *}_1$. So the amplitude for the transition from the state $\:\vra{\bl \psi_\mr B}\:$ to state $\:\vra{\bl \psi_\mr A}\:$ passing through any intermediate basic state $\:\vra{\mb e_k}\:$ is $\:\mr b_k\,\mr a^{\bl *}_k$. \begin{equation} \vra{\bl \psi_\mr B}\stackrel{b_k}{\bl{\m\!\!\!\m\!\!\!\m\!\!\!\longrightarrow}}\vra{\mb e_k}\stackrel{\mr a^{\bl *}_k}{\bl{\m\!\!\!\m\!\!\!\m\!\!\!\longrightarrow}}\vra{\bl \psi_\mr A} \tl{02} \end{equation} Since passing through any basic state $\:\vra{\mb e_k}\,\plr{k\e 1,2,\cdots,n}\:$ is a complete set of independent possibilities, the probability amplitude for the transition from the state $\:\vra{\bl \psi_\mr B}\:$ to the basic state $\:\vra{\bl \psi_\mr A}\:$ is the sum of the amplitudes $\:\mr b_k\,\mr a^{\bl *}_k$ \begin{equation} \mr b_1\,\mr a^{\bl *}_1\p\mr b_2\,\mr a^{\bl *}_2\cdots\p\mr b_n\,\mr a^{\bl *}_n \e\sum\limits_{k\e 1}^{k\e n}\mr b_k\,\mr a^{\bl *}_k\e \begin{bmatrix} \mr a^{\bl *}_1 & \mr a^{\bl *}_2 & \cdots & \mr a^{\bl *}_n\Vp{\tfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \mr b_1 \Vp{\tfrac{a}{b}}\\ \mr b_2 \Vp{\tfrac{a}{b}}\\ \vdots \Vp{\tfrac{a}{b}}\\ \mr b_n \Vp{\tfrac{a}{b}} \end{bmatrix}\e \lavra{\bl \psi_\mr A}{\bl \psi_\mr B} \tl{03} \end{equation} Equation \eqref{03} is given schematically in Figure-02.

We know that the inner product $\:\lavra{\bl \psi_\mr A}{\bl \psi_\mr B}\:$ is invariant with respect to the choice of the complete basis $\:\clr{\vra{\mb e_1},\vra{\mb e_2},\cdots,\vra{\mb e_k},\cdots,\vra{\mb e_n}}$.

The probability for the transition $\:\vra{\bl \psi_\mr B}\bl{\longrightarrow}\vra{\bl \psi_\mr A}\:$ is \begin{equation} \mr P\plr{\vra{\bl \psi_\mr B}\bl{\longrightarrow}\vra{\bl \psi_\mr A}\vp} \e \vlr{\lavra{\bl \psi_\mr A}{\bl \psi_\mr B}\vp}^2 \tl{04} \end{equation}

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Think of the regular scalar product in $\mathbb{R}^{3}$, the scalar product measures how much the two vectors are aligned. Take the vectors (0,1) and (1,0). Their scalar product is 0, they are orthogonal. (0,1) and (0,-1) on the other hand, have scalar product -1 or (1,0) and (1,0) have scalar product 1. If both vectors are normalized, the closer their scalar product approaches 1, the more they align, visually speaking. The same is true for a scalar product in any Hilbert space. The scalar product tells you how much the respective vectors overlap, i.e. how much they are aligned in the same direction. If the scalar product, between two states is 0, they are orthogonal. If a particle is being found to be in state $|\psi_{A}\rangle$ which is an eigenstate of operator $\hat{A}$ with eigenvalue a, there is a 100% of for outcome $a$ when measuring $\hat{A}$. On the other hand, there is zero probability to find it having value $b$ for observable $\hat{B}$, if $|\psi_{B}\rangle$ is an eigenstate of $\hat{B}$ with eigenvalue $b$. They are not at all aligned. The real physical content is in the squared version of the scalar product as long as there is no time evolution.

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  • $\begingroup$ This answer is a little confused at the end - $\hat{A} \neq \hat{B}$ does not imply that there is no overlap between their eigenstates, for example $S^z$ and $S^x$ - $\langle\leftarrow |\uparrow\rangle = (1/\sqrt{2}, 1/\sqrt{2})\cdot (0,1) \neq 0$. $\endgroup$ Commented Jul 24, 2023 at 20:10

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