- #1

amjad-sh

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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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- #1

amjad-sh

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- #2

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

- #3

DrClaude

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Inside an integral?

- #4

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What should it mean there?

- #5

DrClaude

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- #6

haushofer

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I think bra-ket notation is missing.

- #7

olgerm

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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?

- #8

DrClaude

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Sorry for being pedantic, but this mix of notation doesn't make sense. It should beI 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.

$$

\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##.Is it expected value of quantity A?

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- #9

DaTario

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- #10

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$$\langle \psi_1|\hat{A}|\psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x_1 \int_{\mathbb{R}} \mathrm{d} x_1 \psi_1^*(x_1) A(x_1,x_2) \psi_2(x_2).$$

Now defining the operator in position representation as

$$\hat{A} \psi(x)=\int_{\mathbb{R}} \mathrm{d} x' \langle x|\hat{A}|x' \rangle \psi(x')$$

NO! It'sSorry 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##.

$$

\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!

- #11

DrClaude

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Of course. I'll fix my typo.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!

- #12

amjad-sh

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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}

(e^{ikz}+(r_0+r_x\sigma_x+r_y\sigma_y)e^{-ikz})\xi_{\sigma} \quad\text{if} \quad z<0\\

(t_0+t_x\sigma_x+t_y\sigma_y)e^{ik'z}\xi_{\sigma} \quad \text{if} \quad z>0

\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.

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