I am asked to find the density of states for a free particle as a function of $|p|$, $\Delta|p|$ and $\Delta V$. I also have the expression for the number of states $\Delta\Omega(p) = g(p)\Delta p = \frac{V}{h^3}\Delta p_x \Delta p_y \Delta p_z$. So how is $\Delta |p|$ different from $\Delta p$, and what do I do with $\Delta V$? $p$ is momentum, $V$ is volume and $h$ is Planck's constant.

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• As a start please define your symbols. – Farcher Aug 30 '17 at 7:46
• @Farcher edited. – user2132672 Aug 30 '17 at 8:12

Momentum $p$ is a vector quantity so: $$p=(p_x, p_y, p_z)$$ $$\Delta p=(\Delta p_x,\Delta p_y,\Delta p_z)$$ However $|p|$ is length of the momentum vector thus a scalar quantity: $$|p|=\sqrt{p_x^2+p_y^2+p_z^2}$$

I am not quite sure what $\Delta V$ is but I think in this case $\Delta V = V$. You already have density of states as a function of $p$. It is the number of states in cube element $\Delta p_x\Delta p_y\Delta p_z$.

If you want a function of $\Delta|p|$ you need to get a number of states in thin spherical shell of radius $|p|$ and thickness $\Delta |p|\to 0$. Simply transform the cube element $\Delta p_x\Delta p_y\Delta p_z$ to radial element $\Delta |p|$.

You should get $$\underline{\Delta p_x\Delta p_y\Delta p_z=4\pi |p|^2 \Delta |p|}$$

To clarify the difference between $\Delta p$ and $\Delta |p|$ just look at this: $$\Delta p=p_2-p_1=(p_{2x}, p_{2y}, p_{2z})-(p_{1x}, p_{1y}, p_{z1})=(p_{2x}-p_{1x}, p_{2y}-p_{1y}, p_{2z}-p_{1z})=(\Delta p_x,\Delta p_y,\Delta p_z)$$

$$\Delta |p|=|p_2|-|p_1|=\sqrt{p_{2x}^2+p_{2y}^2+p_{2z}^2}-\sqrt{p_{1x}^2+p_{1y}^2+p_{1z}^2}$$

• $\Delta V$ is an element of volume. We calculate the density of states in 6 dimensional configuration space. A cube element in this space should be $\Delta p_x \Delta p_y \Delta p_z \Delta V=\Delta p_x \Delta p_y \Delta p_z \Delta x \Delta y \Delta z$. So I think the number of states should have been given as follows: $\Omega(p,V)=g(p,V)=\frac{1}{h^3}\Delta p_x \Delta p_y \Delta p_z \Delta V$ – Puzzled student Aug 30 '17 at 13:03