What is the solution to [x,p2] and [x,p2]ψ(x) in quantum mechanics?

In summary, the conversation discusses using the results of a previous problem to find [x,p2 ] and determine [x,p2 ]\psi(x). The canonical commutation relation [x,p]=i\hbar is used to evaluate the expression, resulting in a complex wave function in momentum space. The final solution leads to the position space representation of the momentum operator px = i \hbar \partial/\partialx.
  • #1
atomicpedals
209
7

Homework Statement



Using the results of the previous problem, find [x,p2 ] and from that determine [x,p2 ][itex]\psi[/itex](x)

Homework Equations



The solution to the previous problem was [A,BC]=[A,B]C+B[A,C]

The Attempt at a Solution



As I'm suppose to use the results of the previous problem I think what I need to do is

[x,pp]=[x,p]p+p[x,p]

But that doesn't really seem right to me... shouldn't this lead me to an identity (it doesn't say but usually in a QM problem p is momentum and this would seem to be leading to a wave equation in momentum space). If I really am approaching this the correct way, then how should I treat [x,p2 ][itex]\psi[/itex](x)?
 
Physics news on Phys.org
  • #2
What is [x,p]?
 
  • #3
Yes, you are on the right track. I should have started my previous post with this.
 
  • #4
My first instinct is to answer [x,p]=-[p,x]...but I don't think that was what you were asking?
 
  • #5
atomicpedals said:
My first instinct is to answer [x,p]=-[p,x]...but I don't think that was what you were asking?

While this true, you need to evaluate [x,p] explicitly. The answer is something simple, and will have been covered earlier in your text/course.
 
  • #6
Am I correct in thinking that x is position and p momentum?
 
  • #7
The canonical commutation? [x,p]=i [itex]\hbar[/itex]
 
  • #8
Yes; now, use this in the expression at the end of your first post.
 
  • #9
(i [itex]\hbar[/itex])p+p(i [itex]\hbar[/itex]) ?
 
  • #10
Right, but remember that [itex]\left[ x , p \right] = i\hbar[/itex] really means [itex]\left[ x , p \right] = i\hbar I[/itex], where [itex]I[/itex] is the identity operator.
 
  • #11
So in the first part I then end up with a result of

(i [itex]\hbar[/itex] I)p + p(i [itex]\hbar[/itex] I)

Do the standard rules of algebra apply here allowing me to combine this?

Then for the second part is my result just the combined result with the [itex]\psi[/itex](x)? (a complex wave equation in momentum space?)
 
  • #12
atomicpedals said:
So in the first part I then end up with a result of

(i [itex]\hbar[/itex] I)p + p(i [itex]\hbar[/itex] I)

Do the standard rules of algebra apply here allowing me to combine this?

Yes, [itex]i\hbar I[/itex] commutes with all operators, and [itex]IA = A[/itex] for all operators [itex]A[/itex].
atomicpedals said:
Then for the second part is my result just the combined result with the [itex]\psi[/itex](x)? (a complex wave equation in momentum space?)

[itex]\psi \left( x \right)[/itex] is a complex wave function (not equation) in position space (since [itex]\psi[/itex] is a function of position [itex]x[/itex]). What is the position space representation of the momentum operator, i.e., [itex]p = ?[/itex] This also will have been covered earlier in your text/course.
 
  • #13
I think I'm starting to grasp this. It seems like the final solution is heading towards something similar to px = i [itex]\hbar[/itex] [itex]\partial[/itex]/[itex]\partial[/itex]x .
 

FAQ: What is the solution to [x,p2] and [x,p2]ψ(x) in quantum mechanics?

What is a commutator in quantum mechanics?

A commutator is an operation used in quantum mechanics to determine the relationship between two operators. It is defined as the difference between the product of two operators in one order and the product in the opposite order. In other words, A*B - B*A. The commutator can provide information about the uncertainty and compatibility of two operators.

How do commutators relate to the uncertainty principle?

According to the Heisenberg uncertainty principle, it is impossible to simultaneously know the position and momentum of a particle with 100% accuracy. The commutator between the position and momentum operators is directly related to the uncertainty principle, as it quantifies the degree of incompatibility between the two operators.

What is the significance of the commutator in quantum mechanics?

The commutator plays a crucial role in quantum mechanics as it allows us to calculate the commutator of any two operators and determine their compatibility. It also helps in understanding the uncertainty principle and the fundamental limitations of measuring certain physical quantities simultaneously.

How is the wave function related to commutators?

The wave function is a mathematical representation of a quantum system, and it can be described using operators. Commutators are used to determine the compatibility of these operators, which in turn affects the behavior of the wave function. In simple terms, the commutator of two operators tells us how they will interact and affect the wave function.

Can commutators be used to solve problems in quantum mechanics?

Yes, commutators are an essential tool in solving problems in quantum mechanics. They help in understanding the behavior of operators and their relationship with each other. By calculating commutators, we can determine the uncertainty of certain physical quantities and make predictions about the behavior of quantum systems.

Similar threads

Back
Top