# Quantum mechanics exercise

by eoghan
Tags: exercise, mechanics, quantum
 P: 181 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.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,774 You need to convert the momenta to the new coordinates as well.
 P: 181 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?
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,774 Quantum mechanics exercise 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.
 P: 181 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=$$\frac{P_3^2}{3m}$$
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,774 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.
 P: 181 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: $$\frac{\psi_1(x_1)\psi_2(x_2)\psi_3(x_3)+\psi_1(x_1)\psi_2(x_3)\psi_3(x_ 2)+\psi_1(x_2)\psi_2(x_1)\psi_3(x_3)+\psi_1(x_2)\psi_2(x_3)\psi_3(x_1)+ \psi_1(x_3)\psi_2(x_2)\psi_3(x_1)+\psi_1(x_3)\psi_2(x_1)\psi_3(x_2)}{\s qrt{6}}$$ 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
 P: 181 Ah.. no...the energy is still 3*(E1+E2+E3) because I have to divide by 6
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 Quote by 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: $$\frac{\psi_1(x_1)\psi_2(x_2)\psi_3(x_3)+\psi_1(x_1)\psi_2(x_3)\psi_3(x_ 2)+\psi_1(x_2)\psi_2(x_1)\psi_3(x_3)+\psi_1(x_2)\psi_2(x_3)\psi_3(x_1)+ \psi_1(x_3)\psi_2(x_2)\psi_3(x_1)+\psi_1(x_3)\psi_2(x_1)\psi_3(x_2)}{\s qrt{6}}$$ 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
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.
 P: 181 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
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,774 So why would the energy be 3*(E1+E2+E3)? Wouldn't it be just E1+E2+E3?
 P: 181 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 $\psi_1(y_1)\psi_2(y_2)\psi_3(y_3)$
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,774 I'm not sure actually. I'll see if someone else can offer some help on this point.
 P: 1 plug and chug
HW Helper
P: 2,155
 Quote by 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 $\psi_1(y_1)\psi_2(y_2)\psi_3(y_3)$
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.
 P: 181 Thanks to everybody for your help!
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,774 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.
 HW Helper P: 2,155 Good point, I'll have to take a closer look at this...

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