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

• I
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?

Last edited:
olgerm

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

Mentor
Inside an integral?

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What should it mean there?

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}##.

I think bra-ket notation is missing.

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?

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##.

Last edited:
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.

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

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!

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.

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}##.
I will put more context to make my point more clear.
$$\hat{H}=\dfrac{p^2}{2m}-\dfrac{\partial_z^2}{2m}+V(z) +\gamma V'(z)(\hat{z} \times \vec{p})$$
The eigenfunctions of the Hamiltonian ##\hat{H}## have the following form:
##\vec{\varphi}_{k\sigma}(z)=\begin{cases}
\end{cases}##

Where ##V(z)=V\theta(z)##,##\xi_{\pm}## are the eigenfunctions of ##\sigma_x## and the terms ##(r_x\sigma_x+r_y\sigma_y)e^{-ikz}\xi_{\sigma}## and ##(t_x\sigma_x+t_y\sigma_y)e^{ik'z}\xi_{\sigma}## are the terms added due to the presence of Rashba spin-orbit coupling term at z=0.
The presence of the Rashba spin-orbit coupling term at the interface induces lateral spin and charge currents flowing along the x-y plane.
The interference between the spin-orbit coupling part of the wavefunction such as ##(r_x\sigma_x +r_y\sigma_y)\xi_{\sigma}## with the part not related to the spin-orbit coupling term is the fundamental reason behind the induction of the lateral charge and spin currents.
I speculated that the quantum overlap between ##e^{ikz}\xi_{\sigma}## and ##r_x\sigma_x\xi_{\sigma}## or ##r_y\sigma_y\xi_y## induces a charge current flowing along the x or y directions, respectively, as ##r_x \propto p_x## and ##r_y \propto p_y##.
In the case of spin currents, I speculated that instead of quantum overlapping the non-zero value, for example, of the integral below:
$$\int (e^{-ikz}\xi_{\sigma}^{\dagger})\sigma_{x/y/z}(r_{x/y}\sigma_{x/y}\xi_{\sigma})dz$$
induces a spin current flowing along the x/y direction and spin-polarized along the x/y/z-axis.
I want to know what physical meaning may this integral give.