# Homework Help: Commutation relation problem

1. Sep 12, 2009

### noblegas

1. The problem statement, all variables and given/known data

Suppose the operators P and Q satisfy the commutation relation

[P,Q]=a, where a is a constant(a number, not an operator).

a)Reduced the commutator

[P,Q^n] where Q^n means the product of n Qs, to the simplest possible form.

b) Reduce the commutator

[P,e^iQ] to the simplest form.(The operator function e^iQ is defined by the equation, U*$$\varphi$$=$$\sum$$(iQ)^n/(n!), U=e^iQ

2. Relevant equations

3. The attempt at a solution

a) do they want me to reduced both of my commutation relations to constant like the constant a? [P,Q^n]=d/dx*(P*Q^n-Q^n*P)*phi =0? zero is a constant.

b) [P, e^iQ]=d/dx*(P*e^iQ-e^iQ*P)*phi

2. Sep 12, 2009

### javierR

I think what they want is for you to use the commutator [P,Q]=a to re-write [P,Q^n]=P*Q^n - Q^n*P. Leaving the second term alone, you can move the P in the first term over space by space to end up with [P,Q^n]=Q^n*P-Q^n*P+... where ... is what's left over. Just to illustrate the first move: P*Q^n = Q*P*Q^(n-1) + aQ^(n-1) where I've used the [P,Q] relation: P*Q=Q*P+a. Keep doing this to find out what [P,Q^n] becomes.

Do a similar thing for the the exp function, replacing exp[iQ] by the series expansion.

3. Sep 12, 2009

### noblegas

[P,Q^n]=Q^n*P-Q^n*P=0; But generally , matrices don't commute. How do you know P*Q^n = Q*P*Q^(n-1) + aQ^(n-1) ? I understand how you got the first term, I don't understand how you you got the second term?

never mind , got it. since PQ-QP=a, then Q^(n-1)*PQ-QP(Q^(n-1)=Q^(n-1)(PQ-QP)=Q^n-1*a

Last edited: Sep 12, 2009
4. Sep 12, 2009

### noblegas

since, e^iQ=U and U*phi= $$sum$$(iQ)^n/n!

then [P,e^IQ]= P(iQ)^n*phi/n!-(iQ)^n*phi/n*P=PQ^1*Q^(i)*Q^(n-1)-Q^1*Q^(i)*Q^(n-1)*P/(n!); correct?

5. Sep 12, 2009

### gabbagabbahey

Right, generally operators don't commute....so why are you setting this commutator equal to zero?

Also, the order in which you multiply the operators together is important! Otherwise all commutators would be zero.

$$[P,Q^n]=PQ^n-Q^nP\neq Q^n*P-Q^n*P$$

No, order is important

$$Q^{n-1}PQ-QPQ^{n-1}\neq Q^{n-1}(PQ-QP)[/itex] but [tex]PQQ^{n-1}-QPQ^{n-1}=(PQ-QP)Q^{n-1}=aQ^{n-1}\implies PQQ^{n-1}=aQ^{n-1}+QPQ^{n-1}$$

And, even more useful to you:

$$[P,Q]=PQ-QP=a\implies PQ=a+QP$$

...what do you get for your completely reduced $[P,Q^n]$?

6. Sep 13, 2009

### noblegas

I get an expression that reduces to $$Q^(n-1)QP=QPQ^(n-1)$$ =>the expression is zero when I move one of the terms on the left or right side? Is the second part to my problem with the imaginary term correct?

7. Sep 13, 2009

### gabbagabbahey

Huh?!...How on Earth do you get that?

Let's see here...using $PQ=a+QP$, we have:

\begin{aligned} \left[P,Q^n\right] & = PQ^n-Q^nP \\ & = PQQ^{n-1}-Q^nP \\ & = (a+QP)Q^{n-1}-Q^nP \\ & = aQ^{n-1}+QPQ^{n-1}-Q^nP \\ & = aQ^{n-1}+QPQQ^{n-2}-Q^nP \\ & = aQ^{n-1}+Q(a+QP)Q^{n-2}-Q^nP\end{aligned}

and so on.....that doesn't look like its zero to me!

Last edited: Sep 13, 2009
8. Sep 13, 2009

### noblegas

I got the first part of the problem. Is my approach to the second part of the problem correct?

9. Sep 13, 2009

### gabbagabbahey

You say you got the first part, but you haven't yet posted a result that makes any sense....what was your final result?

The second part of the problem uses the result of the first part, so you'd better get that correctly before attempting it.

10. Sep 13, 2009

### noblegas

$$2aQ^(n-1)+[P,Q^n]$$; I can get my latex to work properly , the Q term is suppose to be Q raise to the n-1 power .

11. Sep 14, 2009

### gabbagabbahey

So you are telling me that you get $[P,Q^n]=2aQ^{n-1}+[P,Q^n]$?

Do you think $Q(a+QP)Q^{n-2}=aQ^{n-1}+PQ^n$?....If so, remember my earlier comment (it is the end all and be all of operator rules)-the order in which you multiply operators together is important

12. Sep 14, 2009

### noblegas

I thought Law of association applied to matrices? $$aQ^{n-1}+(QaQ^n)(Q^-2)+QQPQ^nQ^-2=aQ^n-1+(QaQ^n)Q^-1Q^-1+QQPQ^nQ^-2=2*a*Q^{n-1}+Q^2PQ^nQ^{-2}$$. Matrices and constants can commute , but two or more matrices cannot commute.

Last edited: Sep 14, 2009
13. Sep 14, 2009

### gabbagabbahey

Right, so why did you tell me your final result was $2aQ^{n-1}+[P,Q^n]$?

14. Sep 14, 2009

### noblegas

i incorrectly assumed that matrices commute,but only commute if P and Q share the same eigenfunctions. Do you think I should continue to simplify my new expression?

15. Sep 14, 2009

### Preno

Noblegas, are you familiar the Leibniz rule for commutators? $[A,BC]=B[A,C]+[A,B]C$ (try writing out the commutators in full to verify this). You can easily remember this because it has the same structure as the Leibniz rule for the derivative of a product.

Apply this to your problem and you get a recursive formula for $[P,Q^n]$.

16. Sep 14, 2009

### gabbagabbahey

Yes, you should continue to simplify it...you now have $[P,Q^n]=2aQ^{n-1}+Q^2PQ^{n-2}-Q^nP$ (assuming $n\geq 2$ ), right? Keep simplifying the middle term...you should see a pattern...

17. Sep 14, 2009

### noblegas

I let QP=-a+PQ and therefore I got an expression that looks like this:

$$2aQ^{n-1}-aQ^{n-1}+PQ^{n-1}-Q^{n}P=aQ^{n-1}+PQ^{n-1}-Q^{n}P$$

18. Sep 14, 2009

### gabbagabbahey

But $QP=+a+PQ\neq-a+PQ$.

And concentrate on the middle term...$Q^2PQ^{n-2}=$____?

Alternatively (since you don't seem to be grasping this at all), you can try Preno's suggestion:

$$[A,BC]=B[A,C]+[A,B]C\implies [P,Q^n]=[P,QQ^{n-1}]=Q[P,Q^{n-1}]+[P,Q]Q^{n-1}=Q[P,Q^{n-1}]+aQ^{n-1}$$

then use the same method to simplify $Q[P,Q^{n-1}]$ and then $Q^2[P,Q^{n-2}]$ and then $Q^3[P,Q^{n-3}]$ all the way down to $Q^n[P,Q^0]$.

Last edited: Sep 14, 2009
19. Sep 16, 2009

### noblegas

sorry to rehash this thread, my final solution was $$[P,Q^{n}]=naQ^{n-1}$$ . I am having trouble relating this solution to $$[P,e^{iQ}] =Pe^{iQ}-e^{iQ}P$$. $$U*\varphi={iQ}^n/n!$$ which means $$[P,e^{iQ}] =P(iQ)^n/n!-(iQ)^n/n!*P=i^{n}/n!*(PQ^n-Q^nP)=i^{n}/n!*(naQ^{n-1})$$? please look at the latex code if you have trouble reading the output of my latex code.

Last edited: Sep 16, 2009
20. Sep 16, 2009

### gabbagabbahey

What are the $U$ and $\varphi$ supposed to represent?

The definition you want use is $$e^{iQ}=\sum_{n=0}^{\infty}\frac{Q^n}{n!}$$

21. Sep 16, 2009

### noblegas

According to my book , $$U*\varphi=\sum (iQ)^{n}/n!$$ and $$U=e^{iQ}$$* $$\varphi$$ represents an eigenfunction. read as e^(iQ)^n*phi

Last edited: Sep 16, 2009