While analysing a problem in quantum Mechanics, I realized that I don't fully understand the physical meanings of certain integrals. I have been interpreting:

  • $\int \phi^\dagger \hat A \psi \:\mathrm dx$ as "(square root of) probability that a particle with state $|\psi\rangle$ will collapse to a state $|\phi\rangle$ when one tries to observe the observable corresponding to $\hat A$"

  • $\int \phi^\dagger \psi \:\mathrm dx$ as "(square root of) probability that a particle with state $|\psi\rangle$ will collapse to a state $|\phi\rangle$".

All integrals are over all space here, and all $\phi$s and $\psi$s are normalized.

Now, I realize that interpretation of the first integral doesn't really make sense when put side by side with the second one.

For example, when $|\phi\rangle$ is an eigenstate of $\hat A$, I get two different expressions for "the square root of probability that one will get the corresponding eigenvalue of $\psi$ when one tries to observe the observable corresponding to $\hat A$". As far as I can tell, $\int \phi^\dagger \hat A \psi \:\mathrm dx \neq\int \phi^\dagger \psi \:\mathrm dx $, even if $|\phi\rangle$ is an eigenstate of $\hat A$. I am inclined to believe that the second integral is the correct answer here (it comes naturally when you split $|\psi\rangle$ into a linear combination of basis vectors). But I am at a loss as to the interpretation of the first integral.

So, my question is, what are the physical interpretations of $\int \phi^\dagger \hat A \psi \:\mathrm dx$ and $\int \phi^\dagger \psi \:\mathrm dx$?

(While I am familiar with the bra-ket notation, I have not used it in this question as I don't want to confuse myself further. Feel free to use it in your answer, though)

  • $\begingroup$ Your interpretation of the second integral is the most correct. That is, the second integral is like a dot product between two vectors. It represents "how much" of the amplitude of $\psi$ is in $\phi$. The first integral isn't a probability. $\endgroup$
    – Kenshin
    Jan 27, 2013 at 13:22
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    $\begingroup$ You're close to the Fermi Golden Rule en.wikipedia.org/wiki/Fermi's_golden_rule $\endgroup$
    – J L
    Jan 27, 2013 at 13:26
  • $\begingroup$ @chris thought so, since I could derive it easily by splitting $|\psi\rangle$ into a sum of basis eigenstates. Maybe you could explain what the first integral is in an answer? :) $\endgroup$ Jan 27, 2013 at 13:42
  • $\begingroup$ I wish I could. I can't think of a simple interpretation of the first integral. I agree with Ondrej though that it is the overlap of $\phi$ and $A\psi$, but that's about it. $\endgroup$
    – Kenshin
    Jan 27, 2013 at 14:07

3 Answers 3


I'll start with the second one. $\int\phi^\ast\psi\,\mathrm{d}x$ is, as Chris says in the comments, the scalar (or dot) product of $\phi$ and $\psi$. In the Dirac notation, it is written as $\langle\phi|\psi\rangle$ and it gives the overlap of the two wavefunctions. In other words, it gives the probability amplitude (i.e., what you call square root of probability) that starting from $|\psi\rangle$ we will measure the state to be $|\phi\rangle$.

The first integral works in pretty much the same way; it's just the scalar product of $|\phi\rangle$ with $\hat{A}|\psi\rangle$. This means that you first apply $\hat{A}$ to the state $|\psi\rangle$ and then measure its overlap with $|\phi\rangle$. But this is not the same as calculating the amplitude probability of measuring $|\phi\rangle$ after the measurement of $\hat{A}$ is performed on $|\psi\rangle$. If you measured $\hat{A}$, that would be described by a set of projectors $\hat{\Pi}_i = |\chi_i\rangle\langle\chi_i|$, where $|\chi_i\rangle$ is an eigenvector of $\hat{A}$. The state $|\psi\rangle$ collapses to $|\chi_i\rangle$ with probability $p_i = |\langle\chi_i|\psi\rangle|^2$ which then collapses to $|\phi\rangle$ with probability $q_i = |\langle\phi|\chi_i\rangle|^2$. The overall probability is then $P = \sum_i p_i q_i$, which is not the same as $|\langle\phi|\hat{A}|\psi\rangle|^2$, as you can see if you write $\hat{A}|\psi\rangle = \sum_i \chi_i|\chi_i\rangle\langle\chi_i|\psi\rangle$, with $\chi_i$ being the eigenvalues of $\hat{A}$.


The interpretation of $\hat{A}|\psi\rangle$ can be divided into several cases:

  1. If $\hat{A}$ is unitary, $\hat{A}|\psi\rangle$ can be understood as a time evolution of $|\psi\rangle$ governed by a Hamiltonian $\hat{H}$ fulfilling $\hat{A} = \exp(i\hat{H}t)$.

  2. If $\hat{A}$ is not unitary, but is trace-decreasing (i.e., $\langle\psi|\hat{A}^\dagger\hat{A}|\psi\rangle \le \langle\psi|\psi\rangle$ for each $|\psi\rangle$) it can describe some probabilistic evolution $|\psi\rangle\to|\varphi_i\rangle$ with probability $p_i\le 1$.

  3. The general case cannot be, I believe, interpreted generally. It requires renormalization (as does case 2) but this time the norm cannot be interpreted as the probability. Thus, it depends on the specific operator used.

As daaxix pointed out in the comments, the expression $\langle\phi|\hat{A}|\psi\rangle$ can be understood as a form of generalized mean value. Assuming $\hat{A}$ is Hermitian and writing $|\psi\rangle = \sum_k c_k|\chi_k\rangle$, $|\phi\rangle = \sum_k d_k|\chi_k\rangle$, we have $$ \langle\phi|\hat{A}|\psi\rangle = \sum_{k,l} c_k d_l^\ast \langle\chi_l|\hat{A}|\chi_k\rangle.$$ Due to orthogonality of $|\chi_k\rangle$, we can write $\langle\chi_l|\hat{A}|\chi_k\rangle = \delta_{kl} \langle\chi_k|\hat{A}|\chi_k\rangle$, so we have $$ \langle\phi|\hat{A}|\psi\rangle = \sum_k c_k d_k^\ast \langle\chi_k|\hat{A}|\chi_k\rangle. $$ This gives a sum of the expectation values $\langle\chi_k|\hat{A}|\chi_k\rangle$ (i.e., the eigenvalues $\chi_k$), with weight coefficients given by the expansion of the vectors $|\psi\rangle$, $|\phi\rangle$, namely $c_k d_k^\ast$.

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    $\begingroup$ Thanks for your answer (I'll accept tomorrow after giving others a chance to answer). Is there any way to interpret $\hat A|\psi\rangle$? Even a slightly mathematical/complicated explanation would do (editing it into your answer would be nice) $\endgroup$ Jan 27, 2013 at 17:24
  • $\begingroup$ Not a completely general one, as far as I know. I'll try and add something more on the topic tomorrow, don't have much time now... $\endgroup$ Jan 27, 2013 at 17:28
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    $\begingroup$ @Ondejernotík: "Not a completely general one" -- that's intriguing, my interest is piqued :) No problem, add it whenever you have the time.. $\endgroup$ Jan 27, 2013 at 17:29
  • $\begingroup$ It's not a big deal, there are just several possibilities and I think it's better to treat them separately.. $\endgroup$ Jan 27, 2013 at 18:31
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    $\begingroup$ The interpretation of the first integral is exactly the sum $\sum C_k\langle \hat{A}\rangle$, or the sum of expectation values over a canonical orthogonal wavefunction basis, where both $\phi$ and $\psi$ can be represented in that orthogonal basis. So it is sort of like a weighted expectation value, weighted by the respective wavefunctions $\phi$ and $\psi$. $\endgroup$
    – daaxix
    Jan 28, 2013 at 0:23

Think about vectors in $\mathbb{R}^2$. The inner product over this space gives you a notion of an angle and orthogonality. For example $\hat x \cdot \hat y= 0$ means they are completely orthogonal and you cannot express one in terms of the other.

This is the same interpretation for $\langle f | g\rangle = \displaystyle \int_a ^b f^*g \ dx$ which is an inner product over the space $L^2(a,b)$ where the solutions to the Schrodinger operator live. Same interpretations for $ z=x \cdot A y$ where $x,y \in \mathbb{R}^2$ and $A \in M_2(\mathbb{R})$ apply for $\langle f |A| g\rangle$ with $A\in \mathcal{L}(L^2)$ a linear operator over $L^2$.

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    $\begingroup$ You answer is basically the same as Ondrej. So what is the meaning of $A|g>$? It is clearly some kind of rotation. But after that why do you want to project it onto the $<f|$, what is the meaning then? $\endgroup$
    – unsym
    Jan 27, 2013 at 20:26
  • $\begingroup$ it need not be a "rotation". if by "meaning" you refer to the operator's counterpart in nature, well that will depend on the physical problem you are working on. $\endgroup$
    – yca
    Jan 27, 2013 at 20:49

The mathematical operation of applying $\hat{A}$ to a ket $| \phi \rangle$, is a generalized rotation in Hilbert space, that results in another ket $\hat{A}| \phi \rangle$ which may be useful in further calculations, but is not in general the result of measuring the physical quantity $A$.

Therefore, the quantity $\langle \psi |A| \phi \rangle$ has no simple, general interpretation. However, in many concrete problems, a quantity with physical meaning appears in that form, specially when dealing with continuous states. For example, in particle scattering, there is the scattering matrix $S$. The quantity $\langle \beta |S| \alpha \rangle$ represents the amplitude for the free-particle state $| \beta \rangle$ to be found from the initial state $| \alpha \rangle$ after the scattering has taken place. This is used to compute scattering cross sections.

Another example is that of the atomic transitions that account for the spectral lines. The different mechanisms involved (being the electric dipole the most usual) are represented by operators whose matrix elements $\langle k | E | n \rangle$ are directly related to the transition probabilities.

(Apart from that, you surely know that the special case $\langle \psi |A| \psi \rangle$ is the expectation value for the operator $\hat{A}$ for the state represented by $| \psi \rangle$, except by a normalization factor $\langle \psi |\psi \rangle$)

As for the other integral $\langle \phi | \psi \rangle$, your interpretation is essentially correct, except that it is not the square root of a probability, but rather a complex number (the difference matters when you have to add two amplitudes, for example)


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