PHYS 385 Lecture 17 - Particle in a 3D box 17 - 1 Lecture 17 ...

PHYS 385 Lecture 17 - Particle in a 3D box

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Lecture 17 - Particle in a 3D box What's Important: • particle in a box • Fermi energy Text: Gasiorowicz, Chap. 9. In this lecture, we address the situation in which localized interactions are unimportant, so that particle wavefunctions span an entire system, perhaps even as large as a star. We start by considering the motion of a particle between two reflective walls in one dimension, and then generalize the result to three dimensions. Application of our findings to neutron stars are made in the next lecture. Particle in a one-dimensional box Let's consider first the motion of a particle in one dimension between two perfectly reflective walls (from Lec. 9):

p

As there is no potential energy gradient in the region between the walls, the onedimensional (time independent) Schrödinger equation reads h2 d 2u(x ) − = E u(x) (1) 2m dx 2 The solution to this equation has the usual plane-wave form, u(x) = (1/N) sin(nπx / L)

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which satisfies the boundary conditions that u(x) = u(L) = 0, where L is the length of the box. From ∫oπ sin2(n ) d = π/2, the normalization constant N is 1/N 2 = (L/π) ∫oπ sin2(n ) d = (L/π)•(π/2) = L/2. (3) The allowed values of the particle's momentum p obey p = h k = h 2π / (2L /n) = nπ h /L.

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The corresponding kinetic energy is En = n 2π 2 h 2 / 2mL 2.

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n = 1, 2, 3, ...

Particle in a three-dimensional cube Next we place the particle in a three-dimensional cube (not a sphere). Because the directions are orthogonal, the solutions add independently in three dimensions. In three dimensions, the time-independent S. E. reads © 2003 by David Boal, Simon Fraser University. All rights reserved; further resale or copying is strictly prohibited.

PHYS 385 Lecture 17 - Particle in a 3D box

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h2  d 2 d2 d2  + + u (x, y, z) = E uE (x,y, z) 2m  dx 2 dy 2 dz 2  E

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This is another separation of variables problem, in which the wavefunction can be written as a product uE(x, y, z) = ue1(x)•ve2(y)•we2(z), where the component wavefunctions satisfy independent equations like (1) 2 h2 d ue1 (x ) − = e1 ue1 (x ) 2m dx 2 2 h2 d ve2 (y) − = e2 v e2 (y) 2m dy 2

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2 h2 d we3 (z) − = e3 we3 (z) 2m dz 2

The corresponding kinetic energy is then E = e1 + e2 + e3 = (π 2 h 2 / 2mL2)•(nx 2 + ny 2 + nz 2)

ni = 1, 2, 3, ...

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As we have set up this problem, each n runs from 1 to ∞: clearly, n = 0 is not a wave ( = ∞) and n < 0 makes no sense in our context. T=0 Fermi gas So far, we have determined the states allowed for a single particle without interactions and without reference to its spin. Now, let's add fermions to these states. Physically, this applies to systems of electrons, protons, neutrons... individually or as multicomponent systems. In three dimensions, the lowest lying states are ___

states ___ ___

___

___

___

___

nx 2+ny 2+nz 2 9 9 9

number 3

nx 1 2 2

ny 2 1 2

nz 2 2 1

3

1 1 2

1 2 1

2 1 1

6 6 6

1

1

1

1

3

Placing fermions into these levels at T = 0 fills up the levels to some maximum value of the energy Emax • all states with E ≤ Emax are occupied • all states with E ≥ Emax are empty

© 2003 by David Boal, Simon Fraser University. All rights reserved; further resale or copying is strictly prohibited.

PHYS 385 Lecture 17 - Particle in a 3D box

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The value of Emax depends on the number of fermions N, and the value of their spin: each energy state can accommodate 2S + 1 fermions of a given species, where S is the spin of the particle. For protons, neutrons and electrons, S = 1/2, so that each energy state can hold two particles. Our next task is to find Emax . Corresponding to Emax there is a maximal momentum pmax given by Emax = pmax 2 / 2m. (9) According to Eq. (4), the maximal value of n is nmax = Lp max / π h .

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The allowed states of the system correspond to any combination of nx , ny , or nz which satisfies nx 2 + ny 2 + nz 2 ≤ nmax 2 (so E ≤ Emax ), as a consequence of which no individual nx , ny , or nz is greater than nmax . nz nmax

nmax

ny

nmax nx Each value of (nx , ny , nz ) inside the octant corresponds to one unique state. Thus, the number of states N with E ≤ Emax is just the volume of the octant with positive ni , or N(E ≤ E max ) =

1 4π 3 • n 8 3 max

(times 2S + 1)

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where the factor of 1/8 arises from the volume of the octant. We can work backwards from n's to physical quantities as follows 3 1 4π 3 1 4π  Lpmax  N= • n = • 8 3 max 8 3  πh  (times 2S + 1) (12) 3 1  p = L3 2 max  6π  h  The first factor on the RHS is just the volume of the box, L3. It turns out that even if the

© 2003 by David Boal, Simon Fraser University. All rights reserved; further resale or copying is strictly prohibited.

PHYS 385 Lecture 17 - Particle in a 3D box

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shape of the boundary is a sphere instead of a cube, the same expression still applies, with L3 replaced by the volume V enclosed by the sphere. Thus, we can rewrite Eq. (12) as 3 N 1  pmax  = (times 2S + 1) (13) V 6π 2  h  This expression can be inverted to give pmax , which we now call the Fermi momentum pF. And, of course, there is a Fermi energy EF = pF2 / 2m (14) corresponding to the maximum kinetic energy of the occupied states at nmax . The integrated energy over all occupied states is given in Gasiorowicz, p. 163. S = 1/2 gas Our interest is primarily in electrons and protons, so let's explicitly write out the quantities of the T = 0 Fermi gas. Eq. (13) becomes 3 N 1  pmax  = (15) V 3π 2  h  which yields pF = h (3π 2N /V)1/3.

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© 2003 by David Boal, Simon Fraser University. All rights reserved; further resale or copying is strictly prohibited.