Physics 4301: Homework 10. Problem 1 A particle of ...

Physics 4301: Homework 10. Problem 1 A particle of mass M “lives” in the infinite spherical potential well of radius a. We will be interested in pressure P that this particle exerts on the walls of the well and its dependence on the quantum state of the particle. You may want to recall the considerations of the first law of thermodynamics (there is no “heat” here, of course) and principle of the virtual work in evaluating P from the expectation value of the energy of the system. (We used such considerations earlier for a particle living in the infinite 1D square potential well.) Let us use the standard nomenclature |nr lm⟩ to label the stationary eigenstates, with nr ≥ 1 being a radial quantum number that “numbers” different solutions of the effective radial part of the stationary Schr¨odinger equation: ( ) ~2 d2 u ~2 l(l + 1) − + V (r) + u = Eu. (1) 2m d r2 2m r2 As discussed in class, function u(r) must vanish at r = 0. For our infinite potential well, V (r) = 0 for r < a and V (r) = ∞ for r > a, thereby making the second boundary condition u(a) = 0 requested for our solutions defined within the segment 0 ≤ r ≤ a. Such solutions of (1) for our potential are particularly easy to indicate for angular momentum l = 0: they are just u(r) ∝ sin(kr) with allowed values of k satisfying ka = πnr (using standard E = ~2 k 2 /2m). For arbitrary angular momentum l = 0, 1, 2, . . ., these solutions relate to so-called spherical Bessel functions jl : u(r) ∝ r jl (kr).

(2)

Solutions (2) already satisfy u(0) = 0 and allowed values of k are now found from the condition jl (ka) = 0. The numerical values for a range of these zeroes of Bessel functions (such x that jl (x) = 0) for different l values as they appear sequentially (nr = 1, 2, 3, . . .) are indicated in the following table for your convenience: 0 1 2 3 4 5

3.14 6.28 4.49 7.72 5.76 9.10 6.99 10.42 8.18 11.7 9.36 12.97

9.42 10.9 12.3 13.7 15.04 ...

12.56 14.07 15.51 16.92 ... ...

... ... ... ... ... ...

The first column shows the values of l, which result in their respective rows of consecutive zeroes. You immediately recognize that the first row (for l = 0) is nothing but multiples of π as we already knew. You can relate to other values in this table to see how different |nr lm⟩ states are ordered energy-wise and answer some of the questions below. (a) How does pressure P depend on mass M and radius a? Would this particular functional dependence be affected by value l of the orbital momentum? 1

(b) The product of pressure P and volume V , P V is the quantity that we remember well on the left-hand-side of the equation of state of the ideal gas: “P V = N T ”. What do we have on the right-hand-side in our case? (c) Does the pressure depend on the angular momentum and should it - in your gut feeling? In assessing the effect of the orbital angular momentum, find the numerical ratio of the pressures for the particle in a 1f state (“1” refers to nr and letter “f” to the specific value of l) and for the particle in the ground state. (d) Find pressure P for a particle in state 1 i |ψ⟩ = √ |200⟩ + √ |121⟩. 2 2

2

(3)

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