NRQM10sol_2

NRQM10sol_2 - Physics 216 Solutions to Problem Set 2 Spring...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 216 Solutions to Problem Set 2 Spring 2010 1. We wish to determine the correct form of the Schrodinger equation for a spin- 1 2 particle in an external electromagnetic field. The wavefunction for a spin- 1 2 particle is a two-component “spinor” wave function, where each component is an independent function of vectorx and t . Thus, the Hamiltonian must be a 2 × 2 matrix, whose elements are also operators on the Hilbert space of square integrable functions. (a) First consider a free spin- 1 2 particle. We demand that H is rotationally invari- ant. Two candidate Hamiltonians are: ( i ) H 1 = vector p 2 2 m I , ( ii ) H 2 = ( vectorσ · vector p ) 2 2 m . where I is the 2 × 2 identity matrix. Show that H 1 and H 2 are in fact identical. The Pauli matrices, σ i satisfy the identity, σ i σ j = δ ij I + iǫ ijk σ k , (1) where there is an implicit sum over the repeated index k . Note that if we multiply eq. (1) by A i B j and sum over i and j , we recover eq. (14.3.39) on p. 382 of Shankar. Moreover, if A = B = p , then the resulting identity is ( vectorσ · vector p ) 2 = vector p 2 I , since ǫ ijk p i p j = 0 follows from the fact that ǫ ijk is antisymmetric under the interchange of i and j whereas p i p j is symmetric under the interchange of i and j . Hence, if follows that the two candidate Hamiltonians H 1 and H 2 are identical. (b) Consider now a spin- 1 2 particle with charge q in an external electromagnetic field. Using the principle of minimal coupling, deduce the Hamiltonians corresponding to (i) and (ii) above, which include the dependence on the scalar potential φ ( vectorx , t ) and the vector potential vector A ( vectorx , t ) [do not choose a specific gauge]. Show that the two resulting Hamiltonians differ. In particular, in case (ii) above, a term in the Hamiltonian arises of the form ge 2 mc vector S · vector B , where g is called the “g”–factor and vector S ≡ 1 2 planckover2pi1 vectorσ is the spin operator. Using the fact that the electron has q = − e , what value of g is predicted by this approach? Applying the principle of minimal substitution to H 1 , we obtain a spin-independent Hamiltonian, H = 1 2 m parenleftBigg vector p − q vector A c parenrightBigg 2 I + qφ I , (2) 1 for a particle in an external electromagnetic field that derives from a scalar potential φ and a vector potential vector A . In contrast, if we apply the principle of minimal substitution to H 2 , we obtain H = 1 2 m vectorσ · parenleftBigg vector p − q vector A c parenrightBigg vectorσ · parenleftBigg vector p − q vector A c parenrightBigg + qφ I . We can simplify the first term above by writing: vectorσ · parenleftBigg vector p − q vector A c parenrightBigg vectorσ · parenleftBigg vector p − q vector A c parenrightBigg = summationdisplay ij σ i σ j parenleftbigg p i − qA i c parenrightbiggparenleftbigg p j − qA j c parenrightbigg = summationdisplay ijk ( δ ij I + iǫ ijk σ k ) parenleftbigg p i −...
View Full Document

This document was uploaded on 10/31/2011.

Page1 / 16

NRQM10sol_2 - Physics 216 Solutions to Problem Set 2 Spring...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online