Difficult quantum mechanics question

1. Oct 22, 2007

guruoleg

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

I need to prove:

$$a'(z)=S(z)aS^\dagger(z)$$

Where $$S(z)$$ is the squeeze operator and $$a'(z)$$ is the pseudo-lowering operator.

2. Relevant equations

$$S(z)=e^{\frac{1}{2}({z^\ast}a^2-za^{\dagger 2})}$$

$$e^x=\sum_n{\frac{x^n}{n!}}$$ ; I don't think 2 simultaneous Taylor series expansions will get you very far here...

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$$a'={\mu}a+{\nu}a^\dagger$$

$$\mu=cosh(r)$$

$$\nu=sinh(r)e^{i\theta}$$

$$cosh(r)=F(\frac{1}{2}},\frac{x^2}{4}})$$

$$sinh(r)=xF(\frac{3}{2}},\frac{x^2}{4}})$$

where $$F(a,x)$$ is a power series expansion

3. The attempt at a solution

From the second set of equations (which operate on the left side of the original problem), I can make a'(z) into something that looks like a Taylor series expansion. I think you should use the operator expansion theorem now... except I don't know what it is, concretely, or how to use it. Please explain briefly or even better link to a site that provides an explanation (I looked hard: apparently QM isn't that popular).

This is due on Wed and this site is a last resort. Please respond soon.

2. Oct 23, 2007

µ³

How are tau and theta defined?

3. Oct 23, 2007

guruoleg

theta is the angle that the squeeze state makes with one of the quadrature axes... I think it can be explained away by saying its absolute value is zero, but it's probably not that important. There is no tau; if you mean the 'r' in cosh(r), it is the displacement of the squeeze state or something, but it doesn't really matter because it goes away in the next step.

4. Oct 23, 2007

µ³

Well, here's some things that might help:
$$S(z)S^{\dagger} (z)= I^{\hat}$$
so S(z) is unitary
$$e^{A+B}=e^A e^Be^{-1/2[A,B]}=e^B e^A e^{1/2 [A,B]}$$
provided that
$$[A,[A,B]]=[B,[A,B]]=0$$
and given the eigen value equation
$$\hat{A}|a>=a|a>$$
Then
$$e^{\hat{A}}|a> = e^a |a>$$
I tried it for a while and then I realized I had my own home work to do. Best of luck!

$$\langle\beta|\alpha\rangle=e^{-{1\over2}(|\beta|^2+|\alpha|^2-2\beta^*\alpha)}$$
for where $|\alpha\rangle$ and $\beta\rangle$are eigenvectors of the $\hat{a}$ operator

So try inner producting the original expression with arbitrary state vectors beta and alpha and see what you get.

Last edited: Oct 23, 2007
5. Oct 23, 2007

meopemuk

Did you try this formula?

$$e^A B e^{-A} = B + [A,B] +\frac{1}{2!}[A,[A,B]] + \frac{1}{3!}[A,[A,[A,B]]] + \ldots$$

Eugene.

6. Oct 23, 2007

µ³

Ohhhhhhhh. Don't listen to me. Listen to that guy /\/\/\

7. Oct 23, 2007

guruoleg

YES!!! That's the operator expansion theorem that I couldn't find! Thank you so much meopemuk. And I truly appreciate your efforts $$\mu 3$$. Meopemuk, do you know this equation off the top of your head or can you recommend a book? My quantum optics prof. assumes we know QM but we're all engineers and very much do not.

Thanks again and I'm very open to reading recs.

8. Oct 26, 2007

meopemuk

Hi guruoleg,

this is probably the single most useful formula in quantum mechanics, because expressions like $\exp(A)B\exp(-A)$ are all over the place. It can be proven by power expansion of $\exp(A)$, which is kind of tedious. A more intelligent proof goes like this:

Take the derivative of the left hand side

$$\frac{d}{dx}B(x) = \frac{d}{dx} e^{Ax}Be^{-Ax} = Ae^{Ax}Be^{-Ax} - e^{Ax}Be^{-Ax}A = [A,e^{Ax}Be^{-Ax} ]$$

So, $B(x)$ is a solution of the differential equation

$$\frac{d}{dx}B(x) = [A,B(x) ]$$

Next notice that the right hand side

$$B + x[A,B] +\frac{x^2}{2!}[A,[A,B]] + \frac{x^3}{3!}[A,[A,[A,B]]] + \ldots$$

is also a solution of this differential equation. We still need to show that both sides are equal at x=0, which is trivial.

Eugene.

Last edited: Oct 26, 2007
9. Oct 27, 2007

PhysiSmo

Won't simple Taylor expansion do the trick?

i.e.,first order, $$e^{A}=1+A+...$$

$$e^{-A}=1-A+...$$

and

$$e^A B e^{-A}=B+AB-BA-ABA+...=B+[A,B]+...$$

Edit: Lol, I just saw it's already been written!

10. Oct 27, 2007

f95toli

This expanision is known as the Baker-Hausdorff formula and as Meopemuk has already pointed out it is extremely useful. You should be able to find it in any QM book.

11. Nov 4, 2007

guruoleg

Thank you for all your help :) I figured the problem out.

^-^