- #1

Brian-san

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## Homework Statement

Some systems are adequately described by a one-dimensional potential in the form of an asymmetric double well. To good accuracy each can assumed to be harmonic with potential energies:

[tex]V_L(x)=\frac{1}{2}k_Lx^2, V_R(x)=\epsilon+\frac{1}{2}k_R(x-a)^2[/tex]

Here, [itex]\epsilon=V_R(a)>0[/itex]. [itex]N[/itex] classical particles of mass [itex]m[/itex] are brought into thermal equilibrium in this potential.

a) At temperature [itex]T[/itex] what is the average number of particles in each well?

b) What conditions need to be imposed on the parameters of the potential in order to have equal populations at temperature [itex]T^*[/itex]?

c) Calculate the difference, [itex]u_R-u_L[/itex], between the internal energy of two particles in the two wells. Explain why your result is not inconsistent with the way the particles are distributed between the wells for [itex]T\geq T^*[/itex] (when the conditions found in (b) hold.)

## Homework Equations

Relation between canonical and grand canonical partition functions:

[tex]\Xi=\sum_{N=0}^{\infty}z^NQ_N[/tex]

Canonical partition function for 1 particle/oscillator:

[tex]Q_1=\int e^{-\beta H(p,q)}dpdq[/tex]

Average number of particles:

[tex]\langle N\rangle=z\frac{\partial}{\partial z}ln\Xi[/tex]

## The Attempt at a Solution

Assuming the N oscillators are non interacting, [itex]Q_N=Q_1^N[/itex], so then

[tex]\Xi=\sum_{N=0}^{\infty}z^NQ_N=\sum_{N=0}^{\infty}(zQ_1)^N=\frac{1}{1-zQ_1}[/tex]

Then the average number of particles is simply

[tex]\langle N\rangle=z\frac{\partial}{\partial z}ln\Xi=-z\frac{\partial}{\partial z}ln(1-zQ_1)=\frac{zQ_1}{1-zQ_1}=\frac{1}{(zQ_1)^{-1}-1}[/tex]

For the left well,

[tex]Q_1=\frac{1}{h}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\beta(\frac{p^2}{2m}+\frac{k_Lx^2}{2})}dxdp=\frac{1}{h}\int_{-\infty}^{\infty}e^{-\frac{\beta p^2}{2m}}dp\int_{-\infty}^{\infty}e^{-\frac{\beta k_Lx^2}{2}}dx=\frac{1}{h}\sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi}{\beta k_L}}=\frac{2\pi}{h\beta}\sqrt{\frac{m}{k_L}}=\frac{k_BT}{\hbar\omega_L}[/tex]

using [itex]\beta=\frac{1}{k_BT}, k_L=m\omega_L^2[/itex]. Then I got

[tex]\langle N_L\rangle=\frac{1}{\frac{\hbar\omega_L}{k_BT}e^{-\frac{\mu}{k_BT}}-1}[/tex]

By a similar process for the right well I found

[tex]\langle N_R\rangle=\frac{1}{\frac{\hbar\omega_R}{k_BT}e^{\frac{\epsilon-\mu}{k_BT}}-1}[/tex]

This led to the condition for part b that when [itex]T=T^*, \langle N_L\rangle=\langle N_R\rangle[/itex] the relation between the parameters is

[tex]\omega_L=\omega_Re^{\frac{\epsilon}{k_BT^*}}, k_L=k_Re^{\frac{2\epsilon}{k_BT^*}}[/tex]

However, I went back and asked the professor about this as my functions for average number of particles could produce negative values for certain temperature ranges (usually as T approached infinity). I was then told that I couldn't perform the dx integral of my partition function over all x. Obviously there is an intersection point between the two parabolas between 0 and a (say it occurs at x=c, 0<c<a), thus the left well integral covers the interval [itex](-\infty,c][/itex] and the right covers [itex][c,\infty)[/itex].

Trying to do the integrals in this manner proved more difficult as the integrands were Gaussian so there wouldn't be an exact closed form over the new intervals (I did find it could be expressed in terms of error functions, but that seems to make things more complicated). I was also told that this could be solved solely in the canonical ensemble, but I don't see exactly how. The particle number in each well isn't fixed as they are allowed to move between the different wells.