# Quantum mechanics exercise

1. Jun 1, 2010

### eoghan

1. The problem statement, all variables and given/known data
Let there be 3 particles with mass m moving in the 1D potential:
$$\frac{k}{2}[(x_1-x_2)^2 + (x_2-x_3)^2 + (x_1-x_3)^2]$$
where $$x_i$$ is the coordinate of the particle i.

1)Show that with the following coordinat change the Schroedinger equation is easy to solve:
$$y_1=x_1-x_2$$
$$y_2=\frac{1}{2}(x_1+x_2)-x_3$$
$$y_3=\frac{1}{3}(x_1+x_2+x_3)$$

2) Find the eigenstates and the energies of the equation you got in point 1)

2. Relevant equations

3. The attempt at a solution
$$x_1-x_2=y_1$$
$$x_2-x_3=y_2-\frac{1}{2}y_1$$
$$x_1-x_3=y_2+\frac{1}{2}y_1$$

$$V=\frac{k}{2}\left[\frac{3}{2}y_1^2+2y_2^2\right]$$
$$H=\frac{P_1^2}{2m}+\frac{P_2^2}{2m}+\frac{P_3^2}{2m}+V$$

So I have 2 independent harmonic oscillators with angular frequencies $$\sqrt{\frac{3k}{2m}}$$ and $$\sqrt{\frac{2k}{m}}$$
and a free particle whose eigenfunction is $$exp\left[\frac{i}{\hbar}\vec P\vec r\right]$$
So the eigenstates are the tensor product of the eigenstates of two harmonic oscillators and an exponential.
The energies are $$(a+\frac{1}{2})\hbar w_1+(b+\frac{1}{2})\hbar w_2 + E$$
where w1 and w2 are the two frequencies of the two harmonic oscillators and E is the energy of the free particle.

2. Jun 1, 2010

### vela

Staff Emeritus
You need to convert the momenta to the new coordinates as well.

3. Jun 1, 2010

### eoghan

ok, if I convert the momenta I get:
$$H=\frac{2P_1^2}{2m}+\frac{3}{4m}P_2^2+\frac{5}{18m}P_3^2+V$$

Where $$P_i$$ now refers to the momenta in the new basis.

Now the new Hamiltonian can be written as $$H=H_1+H_2+H_3$$
So I have the hamiltonian of 3 independent particles:
$$H_1=2\left[\frac{P_1^2}{2m}+\frac{k}{2}\left(\frac{3}{4}y_1^2\right)\right]$$
$$H_2=\frac{3}{2}\left[\frac{P_2^2}{2m}+\frac{k}{2}\left(\frac{9}{4}y_2^2\right)\right]$$
$$H_3=\frac{5}{9}\left(\frac{P_3^2}{2m}\right)$$

H1 and H2 are two independent harmonic oscillators with their own frequency, and H3 is an independent particle whose energy is multiplied by 5/9
Is this right?

Last edited: Jun 1, 2010
4. Jun 1, 2010

### vela

Staff Emeritus
I got the same potential in terms of the y's, but for the kinetic term, I found

$$\frac{1}{2m}\left(\frac{1}{2}p_1^2+\frac{2}{3}p_2^2+3 p_3^2\right)$$

Besides the constant factors, though, you're correct in that you get what looks like two oscillators and a free particle. The coordinate y3 is the center of mass of the three particles, so the "free particle" is actually the system taken as a whole.

5. Jun 2, 2010

### eoghan

Uhm... I calculate the momenta in the new coordinates again and I get:
$$\frac{1}{2m}\left(2p_1^2+\frac{3}{2}p_2^2+\frac{1}{3}p_3^2\right)$$
Anyway... in the end, the energy is the sum of 3 energies:
E1: energy of a harmonic oscillator with mass m/2 (or 2m according to your result) and frequency $$\left(\omega=\sqrt{\frac{m}{k}}\right)$$ w=3 (or 3/4 according to your results);
E2: the energy of the second harmonic oscillator
E3: the energy of a free particle of mass 3m, i.e. E3=$$\frac{P_3^2}{3m}$$

6. Jun 2, 2010

### vela

Staff Emeritus
Your answer is probably right. A mass of 3m for the whole system makes more sense than m/3.

EDIT: I redid the calculation and now get the same result you did.

Last edited: Jun 2, 2010
7. Jun 2, 2010

### eoghan

And if the 3 particles are bosons what is the ground state energy?
I guess it is:
3*(E1+E2+E3)
and the wavefunction is:
$$\frac{\psi_1(x_1)\psi_2(x_2)\psi_3(x_3)+\psi_1(x_1)\psi_2(x_3)\psi_3(x_2)+\psi_1(x_2)\psi_2(x_1)\psi_3(x_3)+\psi_1(x_2)\psi_2(x_3)\psi_3(x_1)+\psi_1(x_3)\psi_2(x_2)\psi_3(x_1)+\psi_1(x_3)\psi_2(x_1)\psi_3(x_2)}{\sqrt{6}}$$

But then, wouldn't the energy be 6*(E1+E2+E3)?

Where $$\psi_1$$ is the wavefunction of the first harmonic oscillator,
$$\psi_1$$ is the wavefunction of the second HO
and $$\psi_1$$ is the wavefunction of the free particle

8. Jun 2, 2010

### eoghan

Ah.. no...the energy is still 3*(E1+E2+E3) because I have to divide by 6

9. Jun 2, 2010

### vela

Staff Emeritus
What are E1, E2, and E3? The argument of the wavefunctions should be the y's, not the x's. The Hamiltonian isn't symmetric with respect to exchanging the y's, so if $\psi_1(y_1)\psi_2(y_2)\psi_3(y3)$ is a solution, $\psi_1(y_2)\psi_2(y_1)\psi_3(y3)$ won't generally be one.

10. Jun 3, 2010

### eoghan

E1: energy of a harmonic oscillator with mass m/2 (or 2m according to your result) and frequency $$\left(\omega=\sqrt{\frac{m}{k}}\right)$$ w=3;
E2: the energy of the second harmonic oscillator
E3: the energy of a free particle of mass 3m.

So the wave function of the ground state is simply $\psi_1(y_1)\psi_2(y_2)\psi_3(y_3)$?
Where psi1 is the wavefunction of a HO with frequency w=3, psi2 is the wavefunction of the second HO and psi3 is the wavefunction of a free particle

11. Jun 3, 2010

### vela

Staff Emeritus
So why would the energy be 3*(E1+E2+E3)? Wouldn't it be just E1+E2+E3?

12. Jun 3, 2010

### eoghan

Ah, yes, the energy is E1+E2+E3.
So, I don't care if the particles are bosons or fermions? In both cases the wave-function is
$\psi_1(y_1)\psi_2(y_2)\psi_3(y_3)$

13. Jun 3, 2010

### vela

Staff Emeritus
I'm not sure actually. I'll see if someone else can offer some help on this point.

14. Jun 3, 2010

### thegtoman

plug and chug

15. Jun 3, 2010

### diazona

Yeah, I think it doesn't matter. As far as I remember, the Pauli exclusion principle really comes from the antisymmetrization of the wavefunctions of identical fermionic particles, but in this case, the masses corresponding to y1 and y2 (and y3) are different, so the particles are distinguishable and we don't have that issue. And in any case, these are bosons you're talking about, so there's definitely no exclusion principle to worry about. Just drop everything in its lowest energy state and add them up.

16. Jun 4, 2010

### eoghan

Thanks to everybody for your help!

17. Jun 4, 2010

### vela

Staff Emeritus
I'm wondering about the ground state wavefunctions for the fermion and boson cases. Because of the symmetry of the original Hamiltonian, I'd expect the individual particle wavefunctions to be the same and the ground state would be symmetric and antisymmetric combinations depending on the type of particle. Since the ground state wavefunctions in terms of the the x's are different, I'd expect the ground state wavefunctions in terms of the y's to be different.

18. Jun 4, 2010

### diazona

Good point, I'll have to take a closer look at this...

19. Jun 5, 2010

### vela

Staff Emeritus
You can swap x2 and x3 in the original coordinate transformation to get

\begin{align*} y'_1 &= x_1-x_3 \\ y'_2 &= \frac{x_2+x_3}{2}-x_1 \\ y'_3 &= \frac{x_1+x_2+x_3}{3} \end{align*}

and find a solution. Other transformations derived from permuting the xi's will work as well, so the ground state solution would be

\begin{align*} \psi(x_1,x_2,x_3) = \frac{1}{\sqrt{3}} [ & \psi_1(x_1-x_2)\psi_2\left(\frac{x_1+x_2}{2} - x_3\right)\psi_3\left(\frac{x_1+x_2+x_3}{3}\right) + \\ & \psi_1(x_2-x_3)\psi_2\left(\frac{x_2+x_3}{2} - x_1\right)\psi_3\left(\frac{x_1+x_2+x_3}{3}\right) + \\ & \psi_1(x_3-x_1)\psi_2\left(\frac{x_3+x_1}{2} - x_2\right)\psi_3\left(\frac{x_1+x_2+x_3}{3}\right) ] \end{align*}

where $\psi_1$ and $\psi_2$ are the harmonic oscillator ground state wavefunctions (which are even so sign changes in the arguments don't matter) and $\psi_3$ is the free particle wavefunction. The ground state is symmetric so the spin part of the state would have to take care of the overall symmetry or asymmetry of the state.

Last edited: Jun 5, 2010
20. Jun 7, 2010

### eoghan

Ok, so I have to return back to the original coordinates x's and make the wave function symmetric with respect to the x's. In that case the spin part would have to be symmetric as well.
Is this right?