On commuting of position and momentum operators

In summary, the conversation discusses proving the commutation relations for position and momentum operators in quantum mechanics. The equations and attempt at a solution involve using a wave function and product rule to show that the operators can be expressed in component form and ultimately relate to the identity operator.
  • #1
dingo_d
211
0

Homework Statement



I have to proove:

[tex][\hat{y},\hat{p}_y]=[\hat{z},\hat{p}_z]=i\hbar\hat{I}[/tex]


Homework Equations



[tex][\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}[/tex]

The Attempt at a Solution



Ok so I know that

[tex][\hat{y},\hat{p}_y]=\hat{y}\hat{p}_y-\hat{p}_y\hat{y}=y\left(-i\hbar\frac{\partial}{\partial y}\right)-(-i\hbar\frac{\partial y}{\partial y})=i\hbar[/tex]

Analogus for z component. But how to show that it's [tex]i\hbar\hat{I}[/tex]?

Since they're operators they can be expressed in component form - they correspond to some kind of matrix, right?

I don't see how to get that [tex]\hat{I}[/tex] :\
 
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  • #2
A few hints:

1. Remember that [,] is an operator and thus it should act on an arbitrary function, in QM its the wave function.

2. Remember that (d/dy)y f= f+ydf/dy (why?).
 
  • #3
Ok so I can set a wave function [tex]\psi (x)[/tex] and say: [tex]\psi (x)=\langle x|\psi\rangle[/tex]. And I can use product rule.

So [tex][\hat{y},\hat{p}_y]\psi(x)=i\hbar\psi(x)[/tex], but how to relate that with identity operator? :\ I mean, acting with identity operator won't change my wave function...
 
  • #4
dingo_d said:
Ok so I can set a wave function [tex]\psi (x)[/tex] and say: [tex]\psi (x)=\langle x|\psi\rangle[/tex]. And I can use product rule.

So [tex][\hat{y},\hat{p}_y]\psi(x)=i\hbar\psi(x)[/tex], but how to relate that with identity operator? :\ I mean, acting with identity operator won't change my wave function...

When you write

[tex][\hat{y},\hat{p}_y]\psi(x)=i\hbar\psi(x)[/tex]

that makes mathematical sense because the left and right-hand sides are both wavefunctions. If we want to write

[tex][\hat{y},\hat{p}_y]=?[/tex]

we have an operator on the left-hand side, so we should have an operator on the right-hand side. Since you already derived the result for the first equation, just note that it is equivalent to

[tex][\hat{y},\hat{p}_y]\psi(x)=i\hbar \hat{I}\psi(x)[/tex]

and you will have the correct value for the operator statement.
 
  • #5
I see, it's like when I have [tex]\vec{F}=m\vec{a}[/tex] - if I have a vector on the one side I must have vector on the other side...

(we said sth about it on my group theory class, but as a side note...)
 

Related to On commuting of position and momentum operators

1. What is meant by commuting of position and momentum operators?

The commuting of position and momentum operators refers to the mathematical property where the operators representing these physical quantities in quantum mechanics commute, or can be applied in any order, without changing the result of the operation.

2. Why is the commuting of position and momentum operators important in quantum mechanics?

This property is important because it allows for the determination of simultaneous values for position and momentum, known as the Heisenberg uncertainty principle. It also allows for the prediction of future states of a system based on its current state.

3. Can all operators in quantum mechanics commute?

No, not all operators in quantum mechanics commute. In fact, there are only certain pairs of operators that commute, such as position and momentum, or energy and time.

4. How do the operators for position and momentum commute?

The operators for position and momentum commute because they are represented by different mathematical operations. The position operator is represented by multiplication by the position variable, while the momentum operator is represented by differentiation with respect to position.

5. What are the consequences of non-commuting operators in quantum mechanics?

The consequences of non-commuting operators include the uncertainty principle, which states that the more precisely one quantity (such as position) is known, the less precisely the other quantity (such as momentum) can be known. This also affects the predictability of future states and the ability to make simultaneous measurements of different physical quantities.

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