# Commutator Troubles

1. Nov 10, 2004

### joycey

I'm having difficulty trying to figure out which of the following is the correct method to properly evaluate the effect of the operators on f(x).

Given that,
$$\hat{A}f(x)=<x|\hat{A}|f>$$

If the polarity operator, $$\hat{U_p}$$, and the translation operator, $$\hat{U_t}(a)$$, act as
$$\hat{U_p}f(x)=f(-x)$$
$$\hat{U_t}(a)f(x)=f(x-a)$$

Which of the following is the correct method of evaluating the commutator $$[\hat{U_p},\hat{U_t}(a)]f(x)$$.

$$\begin{array}{rl} \hat{U_p}\hat{U_t}(a)f(x)&=\hat{U_p}f(x-a)\\ &=f(-x+a)$$

or

$$\begin{array}{rl} \hat{U_p}\hat{U_t}(a)f(x)&=<x|\hat{U_p}\hat{U_t}(a)|f>\\ &=<-x|\hat{U_t}(a)|f>\\ &=<-x-a|f>\\ &=f(-x-a)$$

Why do I get a different result? The order of the operators acting has obviously changed, but which is the correct order? I am tempted to believe the second case, but I can't really see the difference.

2. Nov 10, 2004

### marlon

They are both correct. What you have proven here is that the commutator of the two operators is NOT 0.

marlon

3. Nov 10, 2004

### joycey

Hey Marlon,

Thanks for the answer, but that's not what I was trying to get at. I realize that either way I evaluate these the commutation relation will be non-zero, I just get a different relation depending on the method. Which is what I'm wondering here.

$$\begin{array}{rl} [\hat{U_p},\hat{U_t}(a)]f(x)&=<x|\hat{U_p}\hat{U_t}(a)|f>-<x|\hat{U_t}(a)\hat{U_p}|f>\\ &=<-x|\hat{U_t}(a)|f>-<x-a|\hat{U_p}|f>\\ &=<-x-a|f>-<-x+a|f>\\ &=f(-x-a)-f(-x+a) \end{array}$$

and

$$\begin{array}{rl} [\hat{U_p}\hat{U_t}(a)]f(x)&=\hat{U_p}\hat{U_t}(a)f(x)-\hat{U_t}(a)\hat{U_p}f(x)\\ &=\hat{U_p}f(x-a)-\hat{U_t}(a)f(-x)\\ &=f(-x+a)-f(-x-a) \end{array}$$

These commutation relations are clearly not equivalent. I am trying to figure out which one is the correct interpretation. I can not seem to see a flaw in either, but there must be some subtle difference between the way in which I have evaluated them. Basically, the order of the operators is reversed between the two, but I am unsure which is the correct order.

4. Nov 11, 2004

### James R

You say:

$$<-x|\hat{U_t}(a)|f> = <-x-a|f>$$

But

$$\hat{U_t}(a)|x> = |x - a>$$?

So, I think

$$\hat{U_t}(a)|-x> = |-(x-a)> = |-x+a>$$

Is that right? If so, that's your mistake.

5. Nov 11, 2004

### joycey

In fact, $$\hat{U_t}(a)|x> = |x+a>$$

$$\hat{U_t}(a)$$ represents the translation operator, translating the current x-coordinate.

$$<x|\hat{U_t}(a) = <x|\hat{U_t}^\dagger(-a) = <x-a|$$

hence, my statement that

$$\hat{U_t}(a)f(x) = <x|\hat{U_t}(a)|f> = <x-a|f> = f(x-a)$$

So, if the current x coordinate is -x then the new coordinate should be -x+a, not -x-a. Following that method, I get the following though, which still doesn't match.

$$\begin{array}{rl} [\hat{U_p},\hat{U_t}(a)]f(x)&=<x|\hat{U_p}\hat{U_t}(a)|f>-<x|\hat{U_t}(a)\hat{U_p}|f>\\ &=<-x|\hat{U_t}(a)|f>-<x-a|\hat{U_t}(a)|f>\\ &=<-(x-a)|f>-<-(x-a)|f>\\ &=f(-x+a)-f(-x+a) \end{array}$$

6. Nov 11, 2004

### ZapperZ

Staff Emeritus
Have we established the validity that if A and B are each Hermitian, that AB and BA are also hermitian? If not, then even if you can write <i|A|j> and <i|B|j>, it doesn't necessarily follow that you can write <i|AB|j> etc. This is only valid for a hermitian operator, is it not? That might explain why it "doesn't match".

Zz.

Edit: Er.. never mind. I should have proved it to myself before I posted this. Carry on....

Last edited: Nov 11, 2004
7. Nov 11, 2004

### reilly

The problem lies in assuming that A f(x) and <x|A|f> are the same thing. They are not.
In fact, <x|a|f> = integral/sum over y of <x|A|y><y|f>. So your second method is incorrect, as far as evaluating Up Ut f is concerned.
Regards,
Reilly Atkinson

8. Nov 11, 2004

### joycey

I just want to ensure that I understand you correctly. Especially since I wrote out the statements in two different orders.

To recap how I interpret things,

$$f(x)=<x|f>$$

$$\begin{array}{rl} \hat{A}f(x)&=<x|\hat{A}|f>\\ &=\int dx' <x|\hat{A}|x'><x|f> \end{array}$$

Now, if I let $$\hat{A}=\hat{U_p}\hat{U_t}(a)$$ then,

$$\begin{array}{rl} \hat{U_p}\hat{U_t}(a)f(x)&=\int dx' <x|\hat{U_p}\hat{U_t}(a)|x'><x'|f>\\ &=\int dx' <x|\hat{U_p}|x'+a><x'|f>\\ &=\int dx' <x|-x'-a><x'|f>\\ &=\int dx' \delta(x+x'+a) <x'|f>\\ &=f(-x-a)$$

since, $$x'=-x-a$$.

Thanks for your help.

9. Nov 11, 2004

### lalbatros

This step is not correct I think:

$$\begin{array}{rl}\hat{U_p}\hat{U_t}(a)f(x)&=\hat{U_p}f(x-a)\\&=f(-x+a)$$​

indeed:

$$\begin{array}{rl}\hat{U_p}\hat{U_t}(a)f(x)&=\hat{U_p}f(x-a))\\&=\hat{U_p}g(x)\\&=g(-x)\\&=f(-x-a)$$​

where g(x) = f(x-a)
and where I assume the rules of the game are as above since the definitions are valid for any functions

10. Nov 11, 2004

### joycey

lalbatros,

I think you made a mistake in going from $$g(-x')$$ to $$f(-x-a)$$.

If we define $$x' \equiv x-a$$, as you have done to go from $$f(x-a) \Rightarrow g(x')$$, then shouldn't we find that $$-x'=-(x-a)=-x+a$$?

11. Nov 12, 2004

### lalbatros

Hi

I did not define x' = x-a and g(x') = f(x-a)

I defined g(x) = f(x-a) , g is another function of x

and since the operators apply in the same way for all functions of x, the result follows.

To summarize: check the definition of the parity operator, it applies to any function of course

12. Nov 12, 2004

### joycey

lalbatros,

To manipulate it in such a manner, you assume the particle is at x, not x+a. The parity operator should effectively reflect around the axis, meaning something that is at x+a must be x+a to the left, or, at -x-a.

13. Nov 12, 2004

### lalbatros

The parity operator is simply:

"replace x by -x"​

14. Nov 12, 2004

### joycey

By that definition (very different than the one I have learned) these operators do commute. Something I am quite certain they should not. Your definition leads to,

$$\begin{array}{rl} [\hat{U_t}(a),\hat{U_p}]f(x)&=\hat{U_t}(a)\hat{U_p}f(x)-\hat{U_p}\hat{U_t}(a)f(x)\\ &=\hat{U_t}(a)f(-x)-\hat{U_p}f(x-a)\\ &=f(-x-a)-f(-x-a)\\ &=0 \end{array}$$

15. Nov 12, 2004

### dextercioby

Why the polemics,this the correct version,in agreement with QM principles and position/momentum representations of vectors and operators...

16. Nov 12, 2004

### lalbatros

Read a definition here:

Other starting points are possible of course. You can choose different representations.
But in the choosen way, all this is quite simple.

I did not make any comment on their commutator.
You can check that they do not commute:

UaUp f(x) = Ua f(-x) = f(-(x-a)) = f(-x+a)
UpUa f(x) = Up f(x-a) = f(-x-a)​

Interpretation:
plane waves are eigenfunctions of the translation operator Ua,
but they are obviously not eigenfunctions of the parity operator (except for a zero wavevector).​

ps: the reason of my posts is that the formula for UpUa f(x) in the first post was not correct (see above here), not taste for polemic!

Last edited: Nov 12, 2004
17. Nov 12, 2004

### reilly

Note, a la lalbatros, Up Ut(a) = Up Ut(a) U(-1)p Up = Ut(-a) Up, where, in spite of cumbersome notation, U(-1)p is the inverse of Up, the parity transformation. Indeed, parity and translations do not commute.
Regards,
Reilly Atkinson

18. Nov 12, 2004

### lalbatros

Reilly,

Thanks for your summary:

Up Ut(a) = Ut(-a) Up​

This tells it crystal clear.