# Quantum Mechanics: Show that this identity is true

1. Sep 3, 2013

### Xyius

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

$$e^{L}a e^{-L}=a+[L,a]+\frac{1}{2!}[L,[L,a]]+\frac{1}{3!}[L[L[L,a]]]+....$$

Where L and a are operators.

3. The attempt at a solution

Right now I am writing the exponentials as their Taylor expansions, and then expanding the RHS of the above equation to see if they are equal. I am having difficulties however, because I cannot seem to make them equal!

2. Sep 3, 2013

### D H

Staff Emeritus
The Taylor expansion is the way to go, as that is how the operator eL is defined. Next you'll have to expand that product on left hand side, and finally you'll have to compare like terms on the left and right hand sides.

Where did you start having problems?

3. Sep 3, 2013

### Xyius

I just cannot seem to make the two sides equal. Maybe I am screwing up the math. I will scan my work and upload it later today.

4. Sep 3, 2013

### TSny

A standard method is sketched out in the "important lemma" here .

5. Sep 3, 2013

### Xyius

I saw that last night and (this is going to sound really lazy) I really didn't feel like going through all that work. I don't think my professor is expecting we do that since the question is not worth that much with respect to the other questions on my homework.

6. Sep 3, 2013

### D H

Staff Emeritus
What level class is this? If it's a typical undergrad class, the lazy physics math approach will probably be acceptable. If it's an honors class or a graduate class, you might need a bit more rigor. The lazy approach is to simply expand the left hand side as if those operators acted like variables (but in a non-commutative algebra) and collect terms. In other words, just do

(1+L+L2/2+...) a (1-L+L2/2+...) = a+(La-aL)+(L2a/2-LaL+aL2/2)+...

There's a slight problem here: How do you know that this expansion is valid thing to do?

Last edited: Sep 3, 2013
7. Sep 3, 2013

### Xyius

It is a graduate level course. I really don't know what my professor expects to be honest.

8. Sep 3, 2013

### D H

Staff Emeritus
In your favor, even graduate level physics classes are chock full of "physics math", stuff that makes mathematicians cringe. To a mathematician, the difference between "physics math" and "engineering math" is a difference of degree rather than of kind.

9. Sep 3, 2013

### Xyius

Haha! Yes I have actually heard this a few times in the past. :]

10. Sep 3, 2013

### vanhees71

If I remember right, another trick is to define
$$a(\lambda)=\exp(\lambda L)a \exp(-\lambda L),$$
derive an operator valued initial-value ODE problem and solve by formal iteration in powers of $\lambda$ and then setting $\lambda=1$.

As a physicist, of course you forget about all the subtle problems of convergence of the series, etc. ;-).

11. Sep 3, 2013

### D H

Staff Emeritus
That's what Xyius doesn't want to do. I think he wants to forget all those subtleties and use

eLae-L = (1+L+L2/2+...) a (1-L+L2/2+...) = a+(La-aL)+(L2a/2-LaL+aL2/2)+...

Now it's just a matter of comparing terms. That initial term, a, That's the initial term on the right hand side. The next term, (La-aL), that's just the bracket [L,a]. The next term after that is 1/2[L,[L,a]]. And so on. Just collect terms that involve n L's into one term and show that each one those collected terms is equal to 1/n![L,[...,[L,a]...]]. Done and done! Whether this approach is valid, that's physics math for ya. You forget all those subtleties and pretend it just works.

Last edited: Sep 3, 2013
12. Sep 3, 2013

### TSny

The derivation outlined in the link I gave, which is essentially vanhees71's trick, is really not too bad to carry out. It's worth seeing.

Using vanhees71's notation, note that $\lambda$ is just a real number parameter. What do you get for $a(\lambda)$ when $\lambda = 0$ and when $\lambda = 1$?

Next, take the derivative of $a(\lambda)$ with respect to $\lambda$ and see if you can write the result $a'(\lambda)$ in terms of $L$ and $a(\lambda)$.

13. Sep 4, 2013

### Xyius

Okay so I started out with.. (I switched the notation to R instead of a)

$$R(\lambda)=e^{\lambda L} a e^{-\lambda L}$$

Next I took the derivative..

$$\frac{dR(\lambda)}{d\lambda}=Le^{\lambda L} a e^{-\lambda L}-e^{\lambda L} L e^{-\lambda L}$$

$$\frac{dR(\lambda)}{d\lambda}=LR(\lambda)-e^{\lambda L}Le^{-\lambda L}e^{\lambda L} a e^{-\lambda L}$$

$$\frac{dR(\lambda)}{d\lambda}=LR(\lambda)-\left( e^{\lambda L}Le^{-\lambda L} \right) R(\lambda)$$

$$\frac{dR(\lambda)}{d\lambda}=\left(L-e^{\lambda L}Le^{-\lambda L} \right) R(\lambda)$$

I am not sure where to go from here, but I see that I now have a differential equation...

14. Sep 4, 2013

### vanhees71

There is something wrong. Note that
$$\frac{\mathrm{d}}{\mathrm{d} \lambda} R(\lambda)=L \exp(\lambda L) a \exp(-\lambda L) - \exp(\lambda L) a \exp(-\lambda L) L.$$
Now this you can easily transform into an differential equation for $R(\lambda)$, which you can then formally solve by iteration of the corresponding integral equation, where you build in the initial value $R(0)=a$.

15. Sep 4, 2013

### D H

Staff Emeritus
The goal is to develop R(1), which is obviously $e^Lae^{-L}$ -- exactly what you want. So how to express this in a different way? Simple! Expand R(λ) as a Taylor series about λ=0 and evaluate at λ=1:
$$R(1) = R(\lambda)|_{\lambda=0} + \left. \frac{dR(\lambda)}{d\lambda} \right|_{\lambda=0}*1 + \left. \frac 1{2!} \frac{d^2R(\lambda)}{d\lambda^2}\right|_{\lambda=0}*1^2 + \cdots$$

Aside: When I wrote "Simple" I did that with my tongue deeply embedded in my cheek.

The zeroth term is R(0)=$e^{\lambda L}ae^{-\lambda L}$ evaluated at λ=0, or just a. The first term, well, just evaluate what you just derived at λ=0 (after fixing the problem pointed out by vanhees). You should get the Lie bracket. You might want to generate a recursive expression for the nth derivative of R wrt lambda.

16. Sep 4, 2013

### TSny

That should be $$\frac{dR(\lambda)}{d\lambda}=Le^{\lambda L} a e^{-\lambda L}-e^{\lambda L} a e^{-\lambda L}L = LR(\lambda)-R(\lambda)L = [L,R(\lambda)]$$
Next, try $\frac{d^2R(\lambda)}{d\lambda^2} = \frac{d}{d \lambda}[L, R(\lambda)]$ and see if you can write it in terms of commutators.

Then note the pattern (which you can easily prove by induction) and plug into DH's taylor expansion.

17. Sep 4, 2013

### Xyius

Awesome! Thanks everyone, I got it!

EDIT: The non-lazy way too!

18. Sep 4, 2013

### D H

Staff Emeritus
Most likely the non-lazy but still "physics math" way. You are working in operator space here, non-commutative operators at that. How do you know that that series converges? You probably just assumed that it does and bulldozed your way through. That's the physics math way, after all.

Physics math drives my son crazy. He's in a PhD math program but has to deal with "physics math" far more often than he'd like. I showed him a PhD thesis that I was basing some work on recently for a numerical integration technique. Something in it bothered me. The author developed an expectation operator E in terms of the backward difference operator ∇, writing $E=\frac 1{1-\nabla}$ (it should have been $E=(\mathop{id}-\nabla)^{-1}$, where id is the identity operator and the superscript -1 denotes the inverse function). To make matters worse, he then expanded that just as one would expand 1/(1-x), and then proceeded to build a differential and integral operator using the same haphazard math. "This looks a bit sloppy", I said. He turned a bit pale. "This is a PhD thesis? Arghhh!"

19. Sep 4, 2013

### vanhees71

Don't worry, this robust physicist's way to do math works better than you might think. The series expansion the physicist did is known to mathematicians as the Neumann series. Of course, the mathematician has to worry about all the formalities like convergence (in which sense) etc. The physicist is more interested in the physics. Of course, also the physicist has to worry about convergence, when he applies this sloppy calculations to real problems, particularly in numerics.

20. Sep 4, 2013

### D H

Staff Emeritus
My son recognized it as the Neumann series right off the bat, also. He then quickly came up with a scheme where divergence was guaranteed. Fortunately, I most likely wouldn't be using the technique there because this requires a sampling rate that is far lower than the Nyquist frequency for the typical usage.

Every reasonably accurate numerical integration has some step size you don't want to cross. This is yet another another example of that. For this technique, cross that line and things do just go south. They quickly go so far south that south turns into north, then south, then north again. Divergence, when it does happen is rather catastrophic.