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

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

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

#### DrClaude

Mentor
Inside an integral?

#### vanhees71

Gold Member
What should it mean there?

#### DrClaude

Mentor
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

I think bra-ket notation is missing.

#### olgerm

Gold Member
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

Mentor
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
$$\langle \psi_1 | \hat{A} | \psi_2 \rangle$$
$$\int \psi_1^*(x) \hat{A} \psi_2(x) \, dx$$

Is it expected value of quantity A?
No. That would only be the case for $\psi_1 = \psi_2$.

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#### DaTario

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

Gold Member
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
$$\langle \psi_1 | \hat{A} | \psi_2 \rangle$$
$$\int \psi_1(x) \hat{A} \psi_2(x) \, dx$$

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

Mentor
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.