Harmonic Oscillator in 3D, different values on x, y and z

[Ofinns]

Hi,

For a harmonic oscillator in 3D the energy level becomes E_n = hw(n+3/2) (Note: h = h_bar and n = n_x+n_y+n_z) If I then want the 1st excited state it could be (1,0,0), (0,1,0) and (0,0,1) for x, y and z.

But what happens if for example y has a different value from the beginning? Like this: V(x,y,z) = 1/2mw²(x²+4y²+z²) and for this decide the energy level AND degeneracy for the 1st excited state. I can only find simple examples when x, y and z are equal and 1.

Best regards

 

[Demystifier]

In such a more general case you have
$$E_{n_1n_2n_3}=\hbar \omega_1 \left( n_1+\frac{1}{2} \right) + \hbar \omega_2 \left( n_2+\frac{1}{2} \right) + \hbar \omega_3 \left( n_3+\frac{1}{2} \right)$$

 

[Ofinns]

In such a more general case you have
$$E_{n_1n_2n_3}=\hbar \omega_1 \left( n_1+\frac{1}{2} \right) + \hbar \omega_2 \left( n_2+\frac{1}{2} \right) + \hbar \omega_3 \left( n_3+\frac{1}{2} \right)$$

Can you elaborate on that? Is 4y² just n₂ here? And in that case you will get three different energy values:

E₁₀₀ = 3hw₁/2
E₀₁₀ =6hw₂
E₀₀₁ =3hw₃/2

Which one is the 1st excited state? Is it E₀₁₀?

 

[Vanadium 50]

Can you elaborate on that? Is 4y2 just n2 here?

No, it enters in as the frequency.

 

[Demystifier]

Can you elaborate on that?

In your case
$$\omega_1=w$$
$$\omega_2=2w$$
$$\omega_3=w$$
Therefore
$$E_{000}=2\hbar w$$
$$E_{100}=E_{001}=3\hbar w$$
$$E_{010}=4\hbar w$$
Hence the first excited states are $E_{100}=E_{001}$.

 

[Ofinns]

Thank you, now I understand that part.

What will the degeneracy become for the 1st excited state then? Can I use the same formula g_n = 1/2(n+1)(n+2) for this case?

 

[Demystifier]

What will the degeneracy become for the 1st excited state then?

It's 2.

Can I use the same formula g_n = 1/2(n+1)(n+2) for this case?

No.

 

[Ofinns]

It's 2.

Why is it 2? What formula do you use to calculate that? (Sorry for all the questions..)

 

[Demystifier]

Why is it 2? What formula do you use to calculate that? (Sorry for all the questions..)

It follows from the last line of post #5. There you see that there are 2 "first excited states" with equal energies. Hence the degeneracy of first excited state is 2.

 

[Ofinns]

It follows from the last line of post #5. There you see that there are 2 "first excited states" with equal energies. Hence the degeneracy of first excited state is 2.

Oh! Thank you so much for the answers, this has been bugging me for a while now.

Best regards

 

[Ofinns]

In your case
$$\omega_1=w$$
$$\omega_2=2w$$
$$\omega_3=w$$

Late questions.. but why is w₂=2w and not 4w?

 

[Demystifier]

Late questions.. but why is w₂=2w and not 4w?

Because, by definition,
$$V(x)=\frac{1}{2}m\omega^2 x^2$$

 

[Ofinns]

Because, by definition,
$$V(x)=\frac{1}{2}m\omega^2 x^2$$

Right, of course. Thank you.