Evaluate Wigner Weyl Transforms for xp+px/2

In summary, the problem is that when multiplying operators, the eigenvalues of the products will not necessarily be the same as the eigenvalues of the operands. You need to be careful when calculating the matrix elements of the transform.
  • #1
aim1732
430
2
For Wigner transforming the function of operators x and p : (xp+px)/2 we need to evaluate something like:

g(x,p) = ∫dy <x - y/2 | (xp+px)/2 | x+y/2> e(ipy/h)
where h is h/2π.

Now I am not sure how to evaluate <x - y/2 | (xp+px)/2 | x+y/2> . I mean what I did was think of |x+y/2> as a delta function whose eigenvalue is x+y/2 and the basis to use is (from the bra) x-y/2.But that gives

∫(xp+px)/2 * δ(-y) e(ipy/h)
which comes out to be a constant where I took x=x and p=(h/i)∂/∂x.
I was expecting g(x,p)=xp

Actually I realize its quite a stupid doubt, rather a problem of me not understanding notations.I would be grateful if somebody gets me out of this mess.
 
Physics news on Phys.org
  • #2
First of all let's calculate the matrix elements. For the first expression you have
[tex]\langle x_1|\hat{x} \hat{p}|x_2 \rangle=x_1 \langle x_1|\hat{p} x_2 \rangle = -\mathrm{i} x_1 \partial_{x_1} \langle x_1 \rangle x_2 = -\mathrm{i} x_1 \partial_{x_1} \delta(x_1-x_2).[/tex]
The other term is
[tex]\langle x_1 |\hat{p} \hat{x} |x_2 \rangle = x_2 \langle x_1|\hat{p}| x_2 \rangle = -\mathrm{i} x_2 \partial_{x_1} \delta(x_1-x_2).[/tex]
Now you set
[tex]x=\frac{x_1+x_2}{2}, \quad y=x_2-x_1[/tex]
which means
[tex]x_1=x-y/2, \quad x_2=x+y/2.[/tex]
Then we have
[tex]\partial_{x_1}=\frac{\partial x}{\partial x_1} \partial_x+\frac{\partial y}{\partial x_1} \partial_y=\frac{1}{2}\partial_x-\partial_y[/tex].
From this we get (in your convention for the Fourier transform, which differs from what I'm used to, but anyway):
[tex]g(x,p)=\frac{1}{2} \int \mathrm{d} y \exp(\mathrm{i} p y) [+\mathrm{i} (x-y/2) \partial_y \delta(y)+\mathrm{i} (x+y/2) \partial_y \delta(y)]=\mathrm{i} x \int \mathrm{d} y \exp(\mathrm{i} p y) \partial_y \delta(y) =x p,[/tex]
as you expected.

The difficulty is that one has to be careful with the expression of the matrix elements with help of the operators in the position representation and the resulting distributions. That's why we had to evaluate the matrix elements first in the original arguments [itex]x_1[/itex] and [itex]x_2[/itex] and then transform into the "macroscopic position" [itex]x[/itex] and the "relative postion" [itex]y[/itex] variables afterwards.
 
Last edited:
  • #3
Thanks a lot for the reply.I realized I should have seen something like this.My original mistake was in assuming that the operators themselves have an inherent basis when in fact they do not.
However my original (revised) attempt involved working with the key idea that the operators 'get' their basis from the left hand side term of the matrix element, as in <u|xp|v> will be written as [u*(h/i)∂u] δ(u-v).That worked out gives the result-xp but when I showed that to my professor he told me that it was not right.Specifically I was not justified in writing the matrix element like that because <u|p|v> was p δ(u-v) but terms involving products with other operators would not be necessarily so.He then did it the same way a you did.
Am I wrong?And why exactly?
 
  • #4
I don't see what's the difference between my derivation and yours. I only used other names for the eigenvalues, or do I miss something?
 
  • #5
The point of difference between the two is whether we can write

<u|xp|v> = [u*(h/i)∂u] δ(u-v)
 
  • #6
If [itex]|u \rangle[/itex] and [itex]|v \rangle[/itex] are position eigenvectors, it's correct. I used this myself in my derivation. I've only called these vectors [itex]|x_1 \rangle[/itex] and [itex]|x_2 \rangle[/itex].
 
  • #7
And the position eigenvectors figure only w.r.t the delta functions used so that you can write any two operators multiplied with each other and appearing in the <u|OPERATOR1*OPERATOR2|v> as OPERATOR1*OPERATOR2* [Eigenvector of the type |u> and |v> with basis u and eigenvalue v] where the basis of the operators are themselves u?
 

1. What is a Wigner Weyl Transform?

The Wigner Weyl Transform is a mathematical tool used in quantum mechanics to represent operators in phase space. It allows for the translation between position and momentum representations of operators, and is particularly useful in studying the dynamics of quantum systems.

2. How is the Wigner Weyl Transform related to xp+px/2?

The Wigner Weyl Transform is a specific way of representing operators in quantum mechanics. The expression xp+px/2 is a particular form of this representation, known as the Weyl symbol. It is used to describe the position and momentum operators in a quantum system.

3. What is the significance of the Wigner Weyl Transform?

The Wigner Weyl Transform is important in quantum mechanics because it allows for the study of quantum systems in phase space, which is a more intuitive and classical way of understanding their behavior. It also provides a useful way of analyzing the dynamics of quantum systems, particularly in terms of phase space trajectories.

4. How is the Wigner Weyl Transform evaluated for xp+px/2?

The evaluation of the Wigner Weyl Transform for xp+px/2 involves applying a specific mathematical formula to the expression. This formula involves the use of the Fourier transform and the Weyl symbol, and results in a phase space representation of the operator.

5. Where is the Wigner Weyl Transform used in scientific research?

The Wigner Weyl Transform is used in a wide range of scientific research, particularly in the fields of quantum mechanics and quantum information. It has applications in studying the dynamics of quantum systems, as well as in developing new quantum algorithms and protocols for quantum communication and computation.

Similar threads

Replies
17
Views
1K
  • Quantum Physics
Replies
4
Views
750
  • Quantum Physics
Replies
27
Views
3K
  • Advanced Physics Homework Help
Replies
2
Views
2K
Replies
6
Views
1K
Replies
19
Views
2K
  • Quantum Physics
Replies
14
Views
4K
Replies
4
Views
3K
  • Quantum Physics
Replies
2
Views
1K
Replies
4
Views
2K
Back
Top