Raising and lowering operators & commutation

1. Jun 2, 2009

Werbel22

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

Show [a+,a-] = -1,

Where a+ = 1/((2)^0.5)(X-iP)
and a- = 1/((2)^0.5)(X+iP)

and

X = ((mw/hbar)^0.5)x

P = (-i(hbar/mw)^0.5)(d/dx)

2. The attempt at a solution

It would take forever to write it all up, but in summary:

I said:

[a+,a-] = (a+a- - a-a+)

then subbed in equations to finally get:

-((hbar/mw)^0.5)(d/dx)((mw/hbar)^0.5)x +((hbar/mw)^0.5)(d/dx)((mw/hbar)^0.5)x

which cancels out to zero, I don't get why! I made sure I did them in the right order, is there something I'm missing? Is it not just a matter of subbing these formulae in?

I put lambda as the wavefunction in when I was doing it on paper, so that the (d/dx) had something to operate on, but it didn't seem to help as it still gives zero.

2. Jun 3, 2009

nicksauce

Ignoring the constant factors you'll have something like have
[a+,a-] = [x - ip, x+ip] = [x,x] + [x,ip] - i[p,x] + [p,p] = i[x,p] - i[p,x] = 2i[x,p]

Now what is [x,p] ?

3. Jun 3, 2009

Werbel22

[X,P] = xp - px ?

When I sub in the equations for those now, I get zero for [x,p], which gives zero for [a+,a-].

I got down to:

[x,p] = (-i(d/dx)(x)) - (-i(d/dx))x
= -i + i = 0

:S I don't get why I can't get the answer.

4. Jun 3, 2009

Pengwuino

Also, how did you compute [a+,a-]? Remember, operators do not normally commute, so you must be careful about what you switch around. Also, you typically can't determine a commutation rule without having a "test function" for your operators to act on. d/dx makes no sense by itself unless it acts on something for example.

5. Jun 3, 2009

Werbel22

Yeah, but I thought in the example d/dx was operating on x, just to give 1. When I put it in, in my original post, it didn't make a difference.

I know operators don't commute, but when I sub IN the actual operator (say into xp - px) it isn't an operator anymore, it's just some variables/constants, so I have just numbers and don't get how THEY can't commute :S

6. Jun 3, 2009

nicksauce

You really need to be more careful with this.
Consider xp operating on f(x). This is x * -ih d/dx f(x) = -ih x * f ' (x)
Now consider px operating on f(x). This is -ih * d/dx (x f(x)) = -ih (f(x) + x*f '(x)) (by the product rule)

7. Jun 3, 2009

Werbel22

I'm still stuck. I understand that last post, and it has helped.

But regardless if I take that into account, I don't see how having a f'(x) helps me, I tried doing it that way, and seeing I end up with a combination of f(x) and f'(x) in my final answer I don't see how that will give 1 :(

8. Jun 3, 2009

nicksauce

Well if you understand the last post, you should be able to calculate [x,p]. The "f" is a test function, and should not be included in the final commutator.

$$[x,p]\psi = xp\psi - px\psi$$
$$[x,p]\psi = -i\hbar x\psi'(x) - ( -i\hbar x\Psi'(x) -i\hbar\psi(x))$$
$$[x,p]\psi = i\hbar\psi(x)$$

Which means that, after "factoring out" the test function,
$$[x,p] = i\hbar$$
This is known as the canonical commutation relation, and is one of the most important equations in quantum mechanics, if not the most. Don't forget it!

Now remember in the first post I showed that
$$[a^+, a^-] \sim [x, p]$$
up to some constant. I hope you'll be able to see why getting 1 as the answer is plausible.