Making sense of perturbation (QM)

[mooeypoo]

Hey guys.

 

I'm going over my notes in class and I probably missed something the professor said, because one of the steps makes no sense to me. I was hoping someone could give me a hint on what's going on.

 

We're doing basic non-degenerate perturbation at the moment, and in this case we examined the first-order correction for the harminic oscillator with H1 defined as [math]\lambda x^4[/math]

 

So, H0 is obvious (the usual solution for the harmonic oscillator), but then we move on to try and get the H1 correction. My professor used a(dagger) and a for it.

 

[math]H_0=(a a^{\dagger}+\frac{1}{2})\hbar \omega[/math]

 

Ground States

a|0>=0

 

[math]N=a^{\dagger}a[/math]

 

And so N|n>=n|n>

 

And I also know that:

 

[math]a|n>=\sqrt{n}|n-1>[/math]

 

[math]a^{\dagger}|n>=\sqrt{n+1}|n+1>[/math]

 

And also

 

[math]a=x\sqrt{\frac{m\omega}{s\hbar}}+i\frac{p}{\sqrt{2m\hbar\omega}}[/math]

 

[math]a^{\dagger}=x\sqrt{\frac{m\omega}{s\hbar}}-i\frac{p}{\sqrt{2m\hbar\omega}}[/math]

 

And hence I can rewrite my x to say:

 

[math]x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger})[/math]

 

So far so good, we went on to try and find the first correction by stating that since H1 has the fourth power, it will look like:

[math]H_1=\lambda \left( \frac{\hbar}{2m\omega} \right)^2 (a+a^{\dagger})^4[/math]

 

And:

 

[math] (a+a^{\dagger})^4 = a^2a^{\dagger 2}+a a^{\dagger}aa^{\dagger}+a^{\dagger}a^{2}a^{\dagger}+a^{\dagger 2}a + aa^{\dagger 2} a+a^{\dagger}aa^{\dagger}a+ ... [/math]

[math]a^{2}a^{\dagger 2}+a^{\dagger 2}a^{2}+(N+1)^2+N(N+1)+(N+1)N+N^2=[/math]

[math]a^{2}a^{\dagger 2}+a^{\dagger 2}a^{2}+(2N+1)^2[/math]

 

Now, I'm having two problems of understanding here.. First, what are those ellipses sign up there in the first line of the solution? Why can we ignore the *rest* of this equation..? second, where did all the a/a-dagger multiplications go? I remember that there is an issue of orthogonality, but I'm not sure I understand why we seemed to have lost all those ..

 

I'm a bit confused here.

 

Okay, I worked out the question from scratch and I think I figured it out. If I set

[math]N=aa^{\dagger}[/math]

and

[math]N+1=a^{\dagger}a[/math]

Then it works out. I'm not quite sure why N+1 is the reversed action, though, I see it in my notes, but I'm not sure where it came from... help?

 

 

Also, I'm trying to solve the same question now but with a different definition of H1. This time,

[math]H_1=\lambdax^3[/math]

 

The strategy is the same, but I get stuck at the end because this time I don't have pairs of a/a-dagger... meh. Not sure how to proceed:

 

[math](a+a^{\dagger})^3=a^{\dagger 3}+a^{3}+(a^{\dagger}a^{\dagger}a+a^{\dagger}aa^{\dagger}+aa^{\dagger}a^{\dagger}+a^{\dagger}aa+aa^{\dagger}a+aaa^{\dagger})=[/math]

 

[math]=a^{\dagger 3}+a^{3}+[ a^{\dagger}(N+1)a^{\dagger}(N)+Na^{\dagger}+(N+1)a+a(N+1)+aN ]=[/math]

 

[math]=a^{\dagger 3}+a^{3}+[ a^{\dagger}(N(N+1))+Na^{\dagger}+(N+1)a+a((N+1)N) ]=[/math]

 

[math]=a^{\dagger 3}+a^{3}+\left[ (a^{\dagger}+a)(N^2+N)+Na^{\dagger}+(N+1)a \right] =[/math]

 

... and now what? I'm stuck. Help?

[mooeypoo]

Okay, I think I at least know where I get stuck -- in the bra-ket notation. I'm following my prof's method, and separated the above equation into 4 parts for ease of solving.

 

First and second parts should be relatively easy:

 

V1

m=n+3

[math]<n+3|a^{\dagger 3}|n>=\sqrt{(n+1)(n+2)(n+2)}[/math]

 

V2

m=n-3

 

[math]<n-3|a^{3}|n>=\sqrt{n(n-1)(n-2)}[/math]

 

These.. I think.. should be right. I'm very confused, though, so I hope I didn't mess it up.

 

My problem is (a) did I do the above right? and (B) the next part, that has Ns in it:

 

V3

m=n+1

[math]<n+1|a^{\dagger}(N+1)+a^{\dagger}N+Na|n>[/math]

 

V4

m=n-1

 

[math]<n-1|(N+1)a+Na+aN|n>[/math]