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

  • I
  • Thread starter amjad-sh
  • Start date
  • Tags
    Physical
In summary, the physical meaning of the integral is that it gives the charge current induced by spin-orbit coupling in the case of a Hamiltonian with Rashba spin-orbit coupling.
  • #1
amjad-sh
246
13
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:
  • Like
Likes olgerm
Physics news on Phys.org
  • #2
I don't see any physical meaning in this expression.
 
  • Like
Likes bhobba and amjad-sh
  • #3
Inside an integral?
 
  • #4
What should it mean there?
 
  • #5
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}##.
 
  • #7
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?
 
  • #8
olgerm said:
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)
$$
\int \psi_1^*(x) \hat{A} \psi_2(x) \, dx
$$

olgerm said:
Is it expected value of quantity A?
No. That would only be the case for ##\psi_1 = \psi_2##.
 
Last edited:
  • #9
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.
 
  • #10
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
$$\hat{A} \psi(x)=\int_{\mathbb{R}} \mathrm{d} x' \langle x|\hat{A}|x' \rangle \psi(x')$$
DrClaude said:
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!
 
  • #11
vanhees71 said:
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.
 
  • #12
DrClaude said:
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}
(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.
 

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

1. What is the physical meaning of ##\psi_1(x)\hat{A}\psi_2(x)##?

The physical meaning of ##\psi_1(x)\hat{A}\psi_2(x)## is the probability amplitude for a particle described by wavefunctions ##\psi_1(x)## and ##\psi_2(x)## to be in a state described by the operator ##\hat{A}##. This quantity is used in quantum mechanics to calculate the probabilities of different outcomes of measurements on a system.

2. How is ##\psi_1(x)\hat{A}\psi_2(x)## related to the uncertainty principle?

The uncertainty principle states that certain pairs of physical quantities, such as position and momentum, cannot be precisely measured at the same time. In the case of ##\psi_1(x)\hat{A}\psi_2(x)##, the product of the wavefunctions and the operator represents the simultaneous measurement of two quantities. Therefore, the uncertainty principle applies to this quantity as well.

3. Can ##\psi_1(x)\hat{A}\psi_2(x)## have a negative value?

Yes, ##\psi_1(x)\hat{A}\psi_2(x)## can have a negative value. This indicates that the particle described by the wavefunctions ##\psi_1(x)## and ##\psi_2(x)## has a negative probability of being in the state described by the operator ##\hat{A}##. However, the overall probability of the particle being in any state must still be positive.

4. How does the physical meaning of ##\psi_1(x)\hat{A}\psi_2(x)## change with different operators?

The physical meaning of ##\psi_1(x)\hat{A}\psi_2(x)## can change depending on the operator used. For example, if ##\hat{A}## represents the position of a particle, then ##\psi_1(x)\hat{A}\psi_2(x)## represents the probability amplitude for the particle to be at a specific position. However, if ##\hat{A}## represents the momentum of a particle, then ##\psi_1(x)\hat{A}\psi_2(x)## represents the probability amplitude for the particle to have a specific momentum.

5. What is the significance of the product of two wavefunctions and an operator in quantum mechanics?

The product of two wavefunctions and an operator, ##\psi_1(x)\hat{A}\psi_2(x)##, is a fundamental concept in quantum mechanics. It allows us to calculate the probabilities of different outcomes of measurements on a system described by the wavefunctions ##\psi_1(x)## and ##\psi_2(x)##. This quantity also takes into account the inherent uncertainty of quantum systems, as described by the uncertainty principle.

Similar threads

Replies
12
Views
1K
Replies
1
Views
881
Replies
19
Views
2K
Replies
12
Views
2K
  • Quantum Physics
Replies
6
Views
1K
Replies
6
Views
961
  • Quantum Physics
Replies
8
Views
2K
  • Quantum Physics
3
Replies
87
Views
9K
Replies
5
Views
977
  • Quantum Physics
Replies
1
Views
781
Back
Top