Pressure exerted by a single electron in an spherical bubble

This is from APhO 2010, Problem 3, Part (A)

When an electron is planted inside liquid helium, it can repel atoms of liquid helium and form what is called an electron bubble. The bubble contains nothing but the electron itself. We shall be interested mainly in its size and stability.

We use $$\Delta f$$ to denote the uncertainty of a quantity $$f$$. The components of an electron’s position vector $$\vec{q}=(x,y,z)$$ and momentum vector $$\vec{p} = (p_x, p_y, p_z)$$ must obey Heisenberg’s uncertainty relations $$\Delta q_\alpha \Delta p_\alpha \geq \frac{\hbar}{2}$$ where $$\hbar$$ is the Planck constant divided by $$2\pi$$ and $$\alpha = x, y, z$$.

We shall assume the electron bubble to be isotropic and its interface with liquid helium is a sharp spherical surface. The liquid is kept at a constant temperature very close to $$0$$ K with its surface tension $$\sigma$$ given by $$3.75 \times 10^{-4} \text{ N m}^{-1}$$ and its electrostatic responses to the electron bubble may be neglected.

Consider an electron bubble in liquid helium with an equilibrium radius $$R$$. The electron, of mass $$m$$, moves freely inside the bubble with kinetic energy $$E_k$$ and exerts pressure $$P_e$$ on the inner side of the bubble-liquid interface. The pressure exerted by liquid helium on the outer side of the interface is $$P_\text{He}$$.

Find a relation between $$E_k$$ and $$P_e$$

I'm having a bit of trouble with this problem, and with the official solutions.

By assuming that 'the electron moves freely' and 'its interface ... is a sharp spherical surface'; I believe that the potential energy is $$0$$ inside the sphere and $$\infty$$ outside. Hence this is the case of an infinite spherical well (which I do NOT know how to solve, and I believe is out of the scope of this question).

Of course, the wave function must vanish at the surface of the sphere. Hence the electron is not uniformly distributed throughout the sphere. As a result, the usual method to find the pressure as in the kinetic theory of gases (like the derivation given in Blundell and Blundell section 6.1) would be irrelevant.

It seems puzzling then that the result comes to be identical, ie $$PV = \frac{2}{3} E_k$$.

Of course, if I knew the wavefunction, or more specifically $$E_k$$; I could then employ classically $$4\pi R^2 P = \frac{\partial E_k}{\partial R}$$ (or since $$E_k$$ should be quantized, going through the standard calculations through $$Z$$, the partition function).

I have looked at the solutions, and I feel they treat the problem superficially. How would I solve this problem rigorously?

• I found this that may help researchgate.net/publication/… Commented Jan 13, 2022 at 5:08
• pdf also here , citeseerx.ist.psu.edu/viewdoc/… Commented Jan 13, 2022 at 5:22
• @annav, It seems that the article you linked uses the zero-point energy of a particle in a box; but in this case, our potential is spherical. Could you maybe elaborate on why? Commented Jan 13, 2022 at 6:49
• The spherical is the interface , the bubble surface. The potential inside the sphere cannot be zero because the electron has a 1/r one and the surface spheres will also have one, The sphere is not a conductor. The paper used a model to solve your questions, models can only be rigorous within their assumptions. from the statements you are supposed to use general arguments Commented Jan 13, 2022 at 9:11
• @annav So from what I gather, different models would each give $E =\alpha \dfrac{\hbar^2}{m_e R^2}$ with different values of $\alpha$. The actual value does not matter for our problem and gives the same answer. Thank you! Commented Jan 14, 2022 at 10:23