Two- and Three-Dimensional Electron Traps — Question 10

Section 39-6 · Physics pages 6530–6531

Question 10 Quantum wall forces satisfy a pressure-energy relation under isotropic scaling In a three-dimensional rectangular infinite well, E = C(n2 x/L2 x + n2 y/L2 y + n2 z/L2 z), with positive integer indices and C = π2ℏ2/(2m). Ignore spin. A particle occupies the ground state of a cube of side L. Define the force on the x wall by Fx = −∂E/∂Lx with the other sides fixed, and pressure by Px = Fx/(LyLz). Lx Lz Ly F Schematic; labels, not drawing lengths, specify the data. (a) Find the force and pressure at the cubic geometry. (b) Express pressure using total energy and volume, and verify by differentiating energy during isotropic volume changes.
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Question 10 - Solution Quantum wall forces satisfy a pressure-energy relation under isotropic scaling Lx Lz Ly F (a) Method and calculation E = C(L−2 x + L−2 y + L−2 z ) ⇒Fx = 2C/L3 . P = 2C/L5 , E = 3C/L2, V = L3. (b) Interpretation and further calculation P = 2E/(3V) . E(V) = 3CV−2/3 ⇒−dE/dV = 2CV−5/3 = P . Physical check The partial derivative for one moving wall differs from differentiating all three lengths together. Both give the same pressure when their geometric factors are handled correctly.

Question 10

Quantum wall forces satisfy a pressure-energy relation under isotropic scaling

In a three-dimensional rectangular infinite well, E=C(nx2/Lx2+ny2/Ly2+nz2/Lz2)E=C(n_x^2/L_x^2+n_y^2/L_y^2+n_z^2/L_z^2), with positive integer indices and C=π2ℏ2/(2m)C=\pi^2\hbar^2/(2m). Ignore spin. A particle occupies the ground state of a cube of side L. Define the force on the x wall by Fx=−∂E/∂LxF_x=-\partial E/\partial L_x with the other sides fixed, and pressure by Px=Fx/(LyLz)P_x=F_x/(L_yL_z).

Diagram shown in the original SVG worksheet.

  1. Find the force and pressure at the cubic geometry.

  2. Express pressure using total energy and volume, and verify by differentiating energy during isotropic volume changes.

Question 10 - Solution

Quantum wall forces satisfy a pressure-energy relation under isotropic scaling

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E=C(Lx−2+Ly−2+Lz−2)⇒Fx=2C/L3.E=C(L_x^{-2}+L_y^{-2}+L_z^{-2})\Rightarrow\boxed{F_x=2C/L^3}.P=2C/L5,E=3C/L2,𝒱=L3.\boxed{P=2C/L^5},\quad E=3C/L^2,\quad\mathcal V=L^3.

(b) Interpretation and further calculation

P=2E/(3𝒱).\boxed{P=2E/(3\mathcal V)}.E(𝒱)=3C𝒱−2/3⇒−dE/d𝒱=2C𝒱−5/3=P.E(\mathcal V)=3C\mathcal V^{-2/3}\Rightarrow\boxed{-dE/d\mathcal V=2C\mathcal V^{-5/3}=P}.

Physical check

The partial derivative for one moving wall differs from differentiating all three lengths together. Both give the same pressure when their geometric factors are handled correctly.