I Physical meaning of ##\psi_1(x)\hat{A}\psi_2(x)##

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What is the physical meaning of ##\psi_1(x)\hat{A}\psi_2(x)##, where ##\hat{A}## is an observable and, ##\psi_1(x)## and ##\psi_2(x)## are arbitrary wavefunctions?
 
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vanhees71

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I don't see any physical meaning in this expression.
 

DrClaude

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Inside an integral?
 

vanhees71

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What should it mean there?
 

DrClaude

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It wouldn't lead necessarily to a "physical meaning," but I was hoping for more context. Looks more like a set-up for the matrix representation of ##\hat{A}##.
 

haushofer

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I think bra-ket notation is missing.
 

olgerm

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I think he means:
physical meaning of ##<\psi_1(x)|\hat{A}|\psi_2(x)>##, where ##\hat{A}## is an observable and, ##\psi_1(x)## and ##\psi_2(x)## are arbitrary wavefunctions.
Is it expected value of quantity A?
 

DrClaude

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I think he means:
physical meaning of ##<\psi_1(x)|\hat{A}|\psi_2(x)>##, where ##\hat{A}## is an observable and, ##\psi_1(x)## and ##\psi_2(x)## are arbitrary wavefunctions.
Sorry for being pedantic, but this mix of notation doesn't make sense. It should be
$$
\langle \psi_1 | \hat{A} | \psi_2 \rangle
$$
or (which is why I asked about an integral)
$$
\int \psi_1^*(x) \hat{A} \psi_2(x) \, dx
$$

Is it expected value of quantity A?
No. That would only be the case for ##\psi_1 = \psi_2##.
 
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If ##\hat A## is a perturbing Hamiltonian, a search in "Fermi's Golden Rule" may present useful information related to the expressions mentioned by DrClaude in #8. They may be related to transition probabilities.
 

vanhees71

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Well, if you want to calculate
$$\langle \psi_1|\hat{A}|\psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x_1 \int_{\mathbb{R}} \mathrm{d} x_1 \psi_1^*(x_1) A(x_1,x_2) \psi_2(x_2).$$
Now defining the operator in position representation as
$$\hat{A} \psi(x)=\int_{\mathbb{R}} \mathrm{d} x' \langle x|\hat{A}|x' \rangle \psi(x')$$
Sorry for being pedantic, but this mix of notation doesn't make sense. It should be
$$
\langle \psi_1 | \hat{A} | \psi_2 \rangle
$$
or (which is why I asked about an integral)
$$
\int \psi_1(x) \hat{A} \psi_2(x) \, dx
$$


No. That would only be the case for ##\psi_1 = \psi_2##.
NO! It's
$$
\int \psi_1^*(x) \hat{A} \psi_2(x) \, dx
$$
Then it's a matrix element of ##\hat{A}##. The complex conjugation is of utmost importance!
 

DrClaude

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NO! It's
$$
\int \psi_1^*(x) \hat{A} \psi_2(x) \, dx
$$
Then it's a matrix element of ##\hat{A}##. The complex conjugation is of utmost importance!
Of course. I'll fix my typo.
 

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