# Question about quantum harmonic oscilator

1. Apr 26, 2008

### 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 + $$\lambda$$x$$^{2}$$

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.

2. Apr 26, 2008

### George Jones

Staff Emeritus
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?

3. Apr 26, 2008

### 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 + $$\lambda$$x (instead of the x being squared).

Thank you!

4. Apr 26, 2008

### George Jones

Staff Emeritus
I think I liked it better with $x^2$!

Maybe try and complete the square for the two terms

$$\frac{1}{2}m \omega^2 x^2 + \lambda x[/itex]. 5. Apr 26, 2008 ### Miguel Paramo OK, I completed squares so the potencial energy part of the hamiltoneal reads: (1/2) mw[tex]^{2}$$ [ x + ($$\lambda$$/ (m w$$^{2}$$)) ]$$^{2}$$ - ($$\lambda$$$$^{2}$$ / 2m).

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

(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!

6. Apr 26, 2008

### George Jones

Staff Emeritus
$$\frac{d}{dx} = \frac{dy}{dx} \frac{d}{dy} = \frac{d}{dy},$$

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