Question about quantum harmonic oscilator

[Miguel Paramo]

Hi, I am preparing for a quantum mechanics exam, and I have this problem that I can`t solve:

I have to find the complete energy eigenvalue spectrum of a hamiltonean of the form:

H = H0 + c

and also another of the form

H = H0 + \lambdax^{2}

Where in both cases, H0 is the hamiltonean of an harmonic oscilator, c and lambda are constants. The variable is x.

I have to find the exact eigenvalues, cannot use perturbation theory, but I can use the fact that I already know the eigenvalues for the harmonic oscilator.

Can anybody help?

Thant you.

 

[George Jones]

H = H0 + c

Let's do this one first. After we get this, we'll do the second one. Might have to take a supper break in the middle

Suppose |psi> is such that

H0 |psi> = E |psi>.

What does H |psi> equal?

 

[Miguel Paramo]

Thak you.

Well, I think that H|Psi> = (Ho+C)|Psi> = (E+C)|Psi>

The the eignevalues are just E+C, where E is the harmonic oscilator energy:

E = (n+1/2)\hbar\omega

If it is just that it was so easy, but the second part looks more complicated.

I made a mistake when I copy the problems, the hamiltonean should read:

H = H0 + \lambdax (instead of the x being squared).

Thank you!

 

[George Jones]

I made a mistake when I copy the problems, the hamiltonean should read:

H = H0 + \lambdax (instead of the x being squared).

Thank you!

I think I liked it better with x^2!

Maybe try and complete the square for the two terms

\frac{1}{2}m \omega^2 x^2 + \lambda x[/itex].

 

[Miguel Paramo]

OK, I completed squares so the potencial energy part of the hamiltoneal reads:

(1/2) mw^{2} [ x + (\lambda/ (m w^{2})) ]^{2} - (\lambda^{2} / 2m).

I think that I should now define y as y= x + (\lambda/ (m w^{2}))

So that the term reads

(1/2) mw^{2} y^{2} - (\lambda^{2} / 2m)

So, as a function of y, the hamiltonean is again an harmonic oscilator plus a constant, and I can solve it as the previous excercise.

My doubt is if the change of variable may alter the kinetic energy part of the hamiltonean, which depends on the second derivative of x. My guess is that not because the relation between x and y is linear.

Thank you for you help!

 

[George Jones]

\frac{d}{dx} = \frac{dy}{dx} \frac{d}{dy} = \frac{d}{dy},

so, as you say, I don't think the kinetic energy term changes.