Physics 443, Solutions to PS 7

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1 Physcs 443, Solutons to PS 7. Grffths 4.50 The snglet confguraton state s χ ) χ + χ χ χ + ) where that second equaton defnes the abbrevated notaton χ + and χ. S a ) S ) b χ â S )ˆb S ) χ In sphercal coordnates the unt vector n the θ, φ drecton s ˆr sn θ cos φî+ sn θ cos φĵ+cos θˆk Let s choose â to be along the z-drecton. Then θ φ 0 and â ˆk. And choose ˆb to be n the x-y plane at an angle θ wth respect to the z-drecton so that φ 0. Then ˆb sn θî + cos θˆk. Now S ) a S ) b χ â S )ˆb S ) χ χ S ) z sn θs ) x + cos θs ) z ) χ χ + χ χ χ + ) S ) z sn θs ) x + cos θs ) z ) χ + χ χ χ + ) sn θ χ + χ χ χ + ) S ) z S ) x χ + χ χ χ + ) + cos θ χ + χ χ χ + ) S ) ) cos θ h h 4 4 cos θ z S ) z χ + χ χ χ + ) We have used the fact that χ ± S z χ ± ± h/ and χ ± S x χ ± χ± h χ 0.. Grffths 4.59 To determne the tme rate of change of expectaton values we use d Q ī Q dt h [H, Q] + a) We have that H m p qa) qφ m p qp A qa p + q A ) + qφ

2 Then d r dt ī [H, r] h [p, r] q[p A, r] q[a p, r] + q [A, r]) + q[φ, r] ) m Let s look at the commutators one at a tme. Sum over pars of ndces) [p, r] [p p, x j ]ˆx j p [p, x j ]ˆx j + [p, x j ]p ˆx j hp δ j ˆx j hp ) [p A, r] [p A, x j ]ˆx j p [A, x j ]ˆx j + [p, x j ]A ˆx j hδ j A ˆx j ha ) [A p, r] [A p, x j ]ˆx j A [p, x j ]ˆx j + [A, x j ]p ˆx j ha δ j ˆx j ha 3) [A, r] A [A, x j ]ˆx j + [A, x j ]A ˆx j 0 4) [φ, r] 0 5) Puttng the peces together we get h [H, r] hp + q ha + q ha) m h p qa) m b) d v dt ī h [H, m p qa)] + m p qa) m h [p, p] q[p A, p] q[a p, p] + q [A, p] + q[p, A] + q [p A, A] + q [A p, A] q 3 [A, A] ) q [φ, p] q[φ, A]) m h Let s collect terms and wrte h [H, p qa)] ) m m h qi + q q I m h I 3

3 I ncludes the terms wth one power of A and powers of p I [p, A] + [p A, p] + [A p, p] I has the terms wth powers of A and one power of p I [A, p] + [p A, A] + [A p, A] And fnally I 3 s the term wth φ and p All of the other terms are zero. Let s see what we can do wth I Now we need I 3 [φ, p] I [p, A] + [p A, p] + [A p, p] Then I becomes [p p, A j ]ˆx j + [p A, p j ]ˆx j + [A p, p j ]ˆx j p [p, A j ] + [p, A j ]p + p [A, p j ] + [A, p j ]p ) ˆx j [p, A j ]ψ p A j A j p )ψ p A j )ψ + A j p ψ A j p ψ p A j )ψ h A j x ψ I h x A j x + A j x h A j A A x x x j x j x x A) + A j ) A j x j h A) + A) ) ˆx j hp B B p)ˆx j In the last step we used the vector dentty A) A + A) ) ˆx j ) ˆx j 3

4 and note that A j ) A ) x ˆx j A) ) ˆx j j Next we evaluate I I [A, p] + [p A, A] + [A p, A] A [A, p j ] [p j, A ]A [A j, p ]A A [p, A j ]) ˆx j A h A A A + A ) j A j + A ˆx j x j x j x x h A A j A ) + A j A ) )A ˆx j x x j x x j h A A) A) A ) ˆx j ha B B A) ha B And I 3 h φ Puttng the peces together we get h [H, m p qa)] m h And fnally d v dt ) qi + q q I m h I 3 q hp B B p) + q ha B) ) + q φ m m h q q p B B p) m m A B) + q φ m q q p B B p) m m A B) + q m φ A ) q q p B B p) m m A B) + q m E c) If E and B are unform over the volume where the wave functon s non zero then, p and B commute and there s a sgn change due to reversng the order of the cross product. That s: p Bψ) h Bψ) h B)ψ B ψ) ) B pψ 4

5 Then m d v dt q q p B) m m A B) + q m E q p qa) B) + E ) m q v B + E) 3. Grffths 4.6 a) The effect of the gauge transformaton on the electrc feld s E φ A E φ A φ Λ ) A Λ φ A E The effect of the gauge transformaton on the magnetc feld s B A B A A + Λ) A B where we use the fact that Λ) 0. b) We am to show that f h Ψ ) h m qa + qφ Ψ 6) then h Ψ ) h m qa + qφ Ψ 5

6 where Ψ e qλ/ h Ψ and φ φ λ, A A + Λ Substtuton nto Schodnger s equaton gves LHS m ) h qa + Λ) + qφ Λ ) Ψe qλ/ h 7) Takng the dervatve on the left hand sde of Equaton we get h ΨeqΛ/ h e qλ/ h q Λ + h ) Ψ 8) On the rght hand sde we get RHS m h m e qλ/ h [ m e qλ/ h [ m ) ) h qa + Λ) e qλ/ h h + q Λ qa + Λ) Ψ + qφ Λ )ΨeqΛ/ h ) ) qa + Λ) e h qλ/ h + qa Ψ + qφ Λ )ΨeqΛ/ h ) ) h h + q Λ qa + Λ) + qa Ψ + qφ Λ ] )Ψ ) ) h h qa + qa Ψ + qφ Λ ] )Ψ Settng the LHS equal to the RHS we recover Equaton Grffths 5. a) If r r r and 9) R m r + m r m + m 0) 6

7 then solvng for r n terms of R and r gves r R + µ/m))r and solvng for r gves r R µ/m )r. If î + ĵ + x y ˆk z r î x + ĵ y + ˆk z R î X + ĵ Y + ˆk Z then x X x From Equaton 9 we get that x x so that X + x x x and from 0 that X x m m +m x m m + m Rx + rx µ m Rx + rx and smlarly for the y and z components. And so µ m R + r Meanwhle Equaton 9 gves x x that, and 0 that X x m m +m so x m m + m Rx rx µ m Rx rx µ m R r b) The two partcle Schrodnger equaton s ] [ h h + V r r ) ψ Eψ m m 7

8 [ h µ R + r ) h µ ] R r ) + V r) ψ Eψ m m m m ) [ h µ ) ] R + m m r) h µ R + m m r) + V r) ψ Eψ [ h µ + ) R h + ) ] r + V r) ψ Eψ m m m m m m [ h ] m + m ) R h µ r + V r) ψ Eψ c) Separate the varables so that ψr, r) ψ R R)ψ r r, substtute nto the Schrodnger equaton and dvde by ψ and we get h m + m ) Rψ R ψ R h rψ r µ ψ r + V r) E The frst term depends only on R and the second and thrd only on r so h m + m ) Rψ R E R ψ R h µ rψ r + V r)ψ r E r ψ r and E E R + E r 5. Entangled States a) GG ψ 0 0 GG RG α + GG GR β + GG GG γ 0 G R G G α + G G G R β + G G G G γ 0 G R β + G G γ 8

9 where we use the fact that G G, and G R 0. Smlarly GG ψ 0 0 GG RG α + GG GR β + GG GG γ 0 G R G G α + G G G R β + G G G G γ 0 G R α + G G γ For future reference, we have that G G α β γ G R ) b) p GG ψ G R G G α + G G G R β + G G G G γ γ G G γ G G + γ G G γ G G where we have used Equaton. Fnally, snce α + β + γ we have that γ G G + G R And γ Snce we are gven) G R G G + G R p γ G G ) G R G G ) G G + G R ) G q G + q R R q R q G 9

10 we get that G G q 3) R G q 4) G R q 5) R R q 6) Then substtuton of Equatons 3-6 nto gves p q)q q + q q) q q c) To maxmze p wth respect to q set 0 d q) q ) dq q q q) q q) + q q) ) q) q q ) 0 q) q + q + q 0 q) + q) + q q q 7) q ± 5) But q > 0 so we choose the + sgn and the q that mnmzes p s q z 5 ). For future reference note that γ G R G G + G R q q + q q + q α β γ G G G R q q + q q Then when p s maxmzed so that z q and z z and + z z/ z) whch follows from Equaton 7) GG ψ α G R G G + β G G G R + γ G G G G 0

11 γ z + z z3 GR ψ α G R R G + β G G R R + γ G G R G β z + z z RG ψ α R R G G + β R G G R + γ R G G G α z RR ψ α R R R G + β R G R R + γ R G R G 0 GG ψ α G R G G + β G G G R + γ G G G G etc. β z + γ z z γ z + γ z 0 z