STEM Problem Detail

Problem Statement

Prompt

provide the solution for the question 57 present in the image given.

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We are given the Maxwell-Boltzmann probability distribution function:
$$f_0 = e^{-(p_x^2 + p_y^2 + p_z^2)/2mk_BT}$$
where $$f_0 = (2\pi mk_BT)^{-3/2}$$.

We need to calculate the expectation value $$\langle e^{i\vec{p}\cdot\vec{a}/\hbar} \rangle$$.

$$ \langle e^{i\vec{p}\cdot\vec{a}/\hbar} \rangle = \int e^{-(p_x^2 + p_y^2 + p_z^2)/2mk_BT} e^{i(a_x p_x + a_y p_y + a_z p_z)/\hbar} \frac{dp_x dp_y dp_z}{f_0} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_0 e^{-(p_x^2 + p_y^2 + p_z^2)/2mk_BT} e^{i(a_x p_x + a_y p_y + a_z p_z)/\hbar} \frac{dp_x dp_y dp_z}{f_0} = f_0 \left[ \int_{-\infty}^{\infty} e^{-p_x^2/2mk_BT} e^{ia_x p_x/\hbar} dp_x \right] \left[ \int_{-\infty}^{\infty} e^{-p_y^2/2mk_BT} e^{ia_y p_y/\hbar} dp_y \right] \left[ \int_{-\infty}^{\infty} e^{-p_z^2/2mk_BT} e^{ia_z p_z/\hbar} dp_z \right] = f_0 \left[ \sqrt{\frac{\pi}{1/(2mk_BT)}} e^{-(ia_x/\hbar)^2 / (4 \cdot 1/(2mk_BT))} \right] \left[ \sqrt{\frac{\pi}{1/(2mk_BT)}} e^{-(ia_y/\hbar)^2 / (4 \cdot 1/(2mk_BT))} \right] \left[ \sqrt{\frac{\pi}{1/(2mk_BT)}} e^{-(ia_z/\hbar)^2 / (4 \cdot 1/(2mk_BT))} \right] = f_0 \left( \sqrt{2\pi mk_BT} \right)^3 e^{-(a_x^2/\hbar^2)mk_BT/2} e^{-(a_y^2/\hbar^2)mk_BT/2} e^{-(a_z^2/\hbar^2)mk_BT/2} = (2\pi mk_BT)^{-3/2} (2\pi mk_BT)^{3/2} e^{-(a_x^2+a_y^2+a_z^2)mk_BT/2\hbar^2} = e^{-(a_x^2+a_y^2+a_z^2)mk_BT/2\hbar^2} $$

Here, we used the integral formula:
$$ \int_{-\infty}^{\infty} e^{-ax^2+bx} dx = \sqrt{\frac{\pi}{a}} e^{-b^2/4a} $$
In our case, $$a = \frac{1}{2mk_BT}$$ and $$b = \frac{ia}{\hbar}$$.
So, $$b^2 = \frac{-a^2}{\hbar^2}$$.
And $$\frac{b^2}{4a} = \frac{-a^2/\hbar^2}{4/(2mk_BT)} = \frac{-a^2 mk_BT}{2\hbar^2}$$.
Therefore, the integral is $$\sqrt{2\pi mk_BT} e^{-a^2 mk_BT / 2\hbar^2}$$.

Let's re-evaluate the expression with the correct substitution:
$$ \langle e^{i\vec{p}\cdot\vec{a}/\hbar} \rangle = f_0 \left[ \int_{-\infty}^{\infty} e^{-p_x^2/2mk_BT} e^{ia_x p_x/\hbar} dp_x \right] \left[ \int_{-\infty}^{\infty} e^{-p_y^2/2mk_BT} e^{ia_y p_y/\hbar} dp_y \right] \left[ \int_{-\infty}^{\infty} e^{-p_z^2/2mk_BT} e^{ia_z p_z/\hbar} dp_z \right] $$
For each integral, let $$p = x$$, $$a = \frac{1}{2mk_BT}$$, $$b = \frac{ia_x}{\hbar}$$.
$$ \int_{-\infty}^{\infty} e^{-ax^2+bx} dx = \sqrt{\frac{\pi}{a}} e^{-b^2/4a} \sqrt{\frac{\pi}{1/(2mk_BT)}} e^{-(\frac{ia_x}{\hbar})^2 / (4 \cdot 1/(2mk_BT))} = \sqrt{2\pi mk_BT} e^{-(\frac{-a_x^2}{\hbar^2}) / (2/(mk_BT))} = \sqrt{2\pi mk_BT} e^{a_x^2 mk_BT / 2\hbar^2} $$
There seems to be a sign error in the provided solution's final step. Let's re-examine the problem statement and the image carefully.

The image shows:
$$ = f_0 \left( \sqrt{\frac{\pi}{1/(2mk_BT)}} e^{-(ia_x/\hbar)^2 / (4 \cdot 1/(2mk_BT))} \right) \left( \sqrt{\frac{\pi}{1/(2mk_BT)}} e^{-(ia_y/\hbar)^2 / (4 \cdot 1/(2mk_BT))} \right) \left( \sqrt{\frac{\pi}{1/(2mk_BT)}} e^{-(ia_z/\hbar)^2 / (4 \cdot 1/(2mk_BT))} \right) = f_0 \left( \sqrt{2\pi mk_BT} e^{-a_x^2 mk_BT / 2\hbar^2} \right) \left( \sqrt{2\pi mk_BT} e^{-a_y^2 mk_BT / 2\hbar^2} \right) \left( \sqrt{2\pi mk_BT} e^{-a_z^2 mk_BT / 2\hbar^2} \right) = (2\pi mk_BT)^{-3/2} (2\pi mk_BT)^{3/2} e^{-(a_x^2+a_y^2+a_z^2)mk_BT/2\hbar^2} = e^{-(a_x^2+a_y^2+a_z^2)mk_BT/2\hbar^2} $$

The image shows the exponent as $$e^{-(\frac{ia}{\hbar})^2 / (4 \cdot \frac{1}{2mk_BT})}$$.
$$(\frac{ia}{\hbar})^2 = \frac{-a^2}{\hbar^2}$$.
$$4 \cdot \frac{1}{2mk_BT} = \frac{2}{mk_BT}$$.
So, $$\frac{-a^2/\hbar^2}{2/mk_BT} = \frac{-a^2 mk_BT}{2\hbar^2}$$.
This matches the image. The final step in the image is:
$$ = e^{-\frac{1}{2} a^2 m k_B T} $$
This seems to be a typo in the image, as it should be $$e^{-\frac{1}{2\hbar^2} (a_x^2+a_y^2+a_z^2) m k_B T}$$. Assuming the question intended to ask for $$\langle e^{iec{p}{a}} angle$$ instead of $$\langle e^{iec{p}{a}/\hbar} angle$$, then the exponent would be $$e^{-rac{1}{2} (a_x^2+a_y^2+a_z^2) m k_B T}$$. However, the presence of $$\hbar$$ in the exponent of the integral suggests it should be there.

Let's assume the question is as written and the final step in the image has a typo and should be:
$$ = e^{-(a_x^2+a_y^2+a_z^2)mk_BT/2\hbar^2} $$
If we consider the last line of the image as a separate calculation, it states:
$$ = e^{-\frac{1}{2} a^2 m k_B T} $$
This implies that $$a_x=a_y=a_z=a$$ and $$\hbar=1$$. Or it is a simplified notation for the sum of squares. Given the context, it is most likely a typo in the final simplified expression.

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Problem #22