quantum mechanics - Meaning of $int phi^dagger hat A psi :mathrm dx$

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)

Answer

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 $\hat{A}$.

Edit:

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 $c_k d_k^\ast$.