# Quantum mechanics problem

1. May 20, 2013

### EEnerd

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

in the 3rd problem consider a particle with mass m and of potential energy
V(x,y,z)=( (mω^2)/2) * [(1+ (2λ/3))*(x^2 +y^2) + (1- (4λ/3))*z^2]
where ω , λ are constants ω≥0 , 0≤λ≤3/4
what are the eigenstates of the Hamiltonion and the corresponding energies
whats the degeneracy of the ground state λ=0 and the first state λ=3/4

2. Relevant equations

3. The attempt at a solutionok if λ=0 i know that v= 1/2 ω(^2) m (x^2+y^2+z^2) which means the H=Hx+Hy+Hz, and that means E=(nx +ny+nz+3/2) hw and the ground state u have (1.0.0) and (0.1.0). and (0.0.1) and E= 5/2 hw
for λ=3/4 V=1/2 mw^2(3/2)(x^2+y^2) so
v=3/2 1/2 mw^2(x^2+y^2)
so H=Hx+Hy=(nx+ny+1)3/2 h w and the ground state degeneracy is (1.0) (0.1) ?! ( i know i made alot of mistakes :P, but i am EE major and this class is supposed to be introductory quantum)

i am not sure about the eigen states, but it looks like a harmonic oscillator so its ground state it should be ψ=N0e^-1/2 ζ ?!!

2. May 20, 2013

### Staff: Mentor

This is not clear. Can you check the exact formulation of the problem?

These are not the correct quantum numbers for the ground state.

Do you know what degeneracy mean? And don't forget about the $z$ axis.

What is ζ?

3. May 20, 2013

### EEnerd

ok the questions says find the eigenstates and the corresponding energies , and the degeneracy is when many states or multiple states share the same energy value right?!

4. May 20, 2013

### EEnerd

ok i see what i did, for ground state n=0 and that means (0,0,0) and E=3/2hw, and for the first state when y=3/4 i will get (0.1.0) , (1.0.0) ?!! and E will be 3hw?!

5. May 20, 2013

### Staff: Mentor

Yes, but when the question asks for the degeneracy of a state, the answer is a single number (or formula).

6. May 20, 2013

### EEnerd

oh so in ground state when we have n=0 we got degeneracy=0 or 1?! cause there is only one state !!
and when we have n=1 , we have degeneracy =3 ?

7. May 20, 2013

### EEnerd

Thanks alot for the help, the website is actually better than our school TAs, :)

8. May 20, 2013

### Staff: Mentor

You would call that a degeneracy of 1 (even though it is not degenerate!).

[/QUOTE]
Correct.