STEM Problem Detail

Problem Statement

Prompt

Provide the step by step solution of part a, b, c, d of question 4.5.

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**Solution :**

**(a) The total density of states is a convolution:**

$$D(E) = \int_{-\infty}^{\infty} dE_1 \int_{-\infty}^{\infty} dE_2 \int_{-\infty}^{\infty} dE_3 D_1(E_1) D_2(E_2) D_3(E_3) \delta(E - E_1 - E_2 - E_3) \text{.}$$

**(b) The joint probability density $$P(E_1, E_2)$$ is given by**

$$P(E_1, E_2) = \frac{D_1(E_1) D_2(E_2) D_3(E - E_1 - E_2)}{D(E)} \text{.}$$

**(c) We set the derivatives $$\partial \ln P / \partial E_{1,2} = 0$$, which gives**

$$\frac{\partial \ln P}{\partial E_1} = \frac{\partial \ln D_1}{\partial E_1} - \frac{\partial \ln D_3}{\partial E_3} = 0 \quad , \quad \frac{\partial \ln P}{\partial E_2} = \frac{\partial \ln D_2}{\partial E_2} - \frac{\partial \ln D_3}{\partial E_3} = 0 \text{,}$$

where $$E_3 = E - E_1 - E_2$$ in the argument of $$D_3(E_3)$$. Thus, we have

$$\frac{\partial \ln D_1}{\partial E_1} = \frac{\partial \ln D_2}{\partial E_2} = \frac{\partial \ln D_3}{\partial E_3} \equiv \frac{1}{T} \text{.}$$

Expanding $$\ln P(E_1^* + \delta E_1, E_2^* + \delta E_2)$$ to second order in the variations $$\delta E_j$$, we find the first order terms cancel, leaving

$$\ln P(E_1^* + \delta E_1, E_2^* + \delta E_2) = \ln P(E_1^*, E_2^*) - \frac{(\delta E_1)^2}{2k_B T^2 C_1} - \frac{(\delta E_2)^2}{2k_B T^2 C_2} - \frac{(\delta E_1 + \delta E_2)^2}{2k_B T^2 C_3} + \dots \text{,}$$

where $$\partial^2 \ln D_j / \partial E_j^2 = -1/k_B T^2 C_j$$, with $$C_j$$ the heat capacity at constant volume and particle number. Thus,

$$P(E_1, E_2) = \frac{\sqrt{\det(\mathcal{C}^{-1})}}{2\pi k_B T^2} \exp \left( -\frac{1}{2k_B T^2} \mathcal{C}_{ij}^{-1} \delta E_i \delta E_j \right) \text{,}$$

where the matrix $$\mathcal{C}^{-1}$$ is defined as

$$\mathcal{C}^{-1} = \begin{pmatrix} C_1^{-1} + C_3^{-1} & C_3^{-1} \\ C_3^{-1} & C_2^{-1} + C_3^{-1} \end{pmatrix} \text{.}$$

One finds

$$\det(\mathcal{C}^{-1}) = C_1^{-1} C_2^{-1} + C_1^{-1} C_3^{-1} + C_2^{-1} C_3^{-1} \text{.}$$

The prefactor in the above expression for $$P(E_1, E_2)$$ has been fixed by the normalization condition $$\int dE_1 \int dE_2 P(E_1, E_2) = 1$$.

**(d) Integrating over $$E_2$$, we obtain $$P(E_1)$$:**

$$P(E_1) = \int_{-\infty}^{\infty} dE_2 P(E_1, E_2) = \frac{1}{\sqrt{2\pi k_B \tilde{C}_1 T^2}} e^{-(\delta E_1)^2 / 2k_B \tilde{C}_1 T^2} \text{,}$$

where

$$\tilde{C}_1 = \frac{C_2^{-1} + C_3^{-1}}{C_1^{-1} C_2^{-1} + C_1^{-1} C_3^{-1} + C_2^{-1} C_3^{-1}} \text{.}$$

Thus,

$$\langle (\delta E_1)^2 \rangle = \int_{-\infty}^{\infty} dE_1 (\delta E_1)^2 P(E_1) = k_B \tilde{C}_1 T^2 \text{.}$$

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Problem #23