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1 Physics 443, Solutions to PS. Griffiths.9 For Φ(x, t A exp[ a( mx + it], we need that + h Φ(x, t dx. Using the known result of a Gaussian intergral + exp[ ax ]dx /a, we find that: am A h. ( The Schrödinger Equation is given by HΦ i h Φ, with H t ( h /m V (x. Plugging our Wavefunction into this Equation, we find: h m ( am + ( amx h h + V a h, a h a mx + V a h, V (x ma x. ( x + Being odd functions, x and p are zero. And using + x exp[ ax ]dx 0.5 /a 3, we have:. Griffiths.6 x h 4am, p ham, σ x x x h am, σ p p p am h, σ x σ p h. (3 d ψ dt ψ dx ( ψ t ψ + ψ ψ dx (4 t Courtesy Shaffique Adam

2 We use Schrodinger s equation and its complex conjugate to replace the time derivatives with space derivatives and the potential V (x. V (x is assumed real and independent of time. d ψ dt ψ dx ī h [ ( i h ψ h m x + V ψ ψ ī ( h h ψ m ( h ( ψ m x ψ ψ ψ dx x Integrate by parts and use the fact that normalizable wave functions are zero at infinity to drop the total derivative. Then ( d i h ( ψ ψ dt ψ dx ψ m x x ψ ψ dx x x 0 3. Griffiths.5 (a We have that Ψ(x, 0 A[ψ (x+ψ (x]. We know that the states ψ n (x sin k L nx, with k n n/l are normalized, orthogonal and real. Then L 0 L ( Ψ (x, 0Ψ(x, 0dx A ψ + ψ + ψ ψ dx 0 A ( A ] ψ x + V ψ dx (b Ψ(x, t ( ψ (xe iω t + ψ (xe iω t (5 where ω n hk n m. Then Ψ(x, t L [ ψ (x + ψ (x + ψ (xψ (x cos( ωt ] [ sin (k x + sin (k x + sin(k x sin(k x cos ωt ] where ω ω ω

3 (c (d x L 0 L x Ψ(x, t dx [ x sin (k x + x sin (k x + x sin(k x sin(k x cos ωt ] dx L 0 [ L + L + ( ] L (/L + cos ωt (3/L [ L 8 ] 9 cos ωt We got the first two terms in the integration by inspection since sin kx is symmetric about the midpoint of the well. We find the integral of the third term in a table. It is appended at the end of this solution set. The amplitude of the oscillation is p m d dt 8L 9 8L x ω sin ωt 9 (e An energy measurement would yield either E or E. The probability of measuring E is equivalent to the probablity that Ψ is in the state ψ, namely P ψ (xψ(x, tdx 4. Griffiths. H (E + E (a To normalize the wave function set Ψ(x, 0 dx where Ψ(x, 0 Ae a x 3

4 Then Ψ(x, 0 dx, ( A e ax + 0 ( A a A a 0 e ax dx (b The momentum distribution (c φ(k φ(k Ψ(x, 0e ikx dx a ( e a x e ikx dx a ( 0 e ax ikx + e ax ikx dx 0 ( a e ax ikx a ik 0 + eax ikx a ik 0 ( a a + ik + a ik a a a + k Ψ(x, t φ(ke i(kx ωkt dk a 3 a + k ei(kx ω kt dk and ω k hk m. (d In the limit where a, Ψ(x, t a e i(kx ω kt dk which looks like a δ(x at t 0 and will spread out in time due to the k dependence of ω. In the limit a 0, Ψ(x, t a3 4 k ei(kx ω kt dk

5 Small k, (long wavelenghts will dominate the distribution at t Griffiths. We have a Ψ(x, 0 exp( ax. The way to solve this problem is take the Fourier transform φ(k, since we know how to time evolve φ(k for a free particle. In particular, we have the following relations Ψ(x, t φ(k + dk φ(k exp( ikx i hk t m + dx Ψ(x, 0 exp( ikx. Putting these two equations together, one can solve for Ψ(x, t as Ψ(x, t ( a ( a 4 exp( ax exp( ikx exp(ikx exp( i hk t m 4 dydk exp( ay iky + ikx i hk t/m. This looks like a mess, but it is really just two integrals of the form given in the hint. Performing the integral over dy first, and then dk, we have Ψ(x, t ( a 4 dk ( a exp k 4a + ikx i hk t, m ( ( a 4 4a /4a + i ht/m exp ( ax exp + i hat/m + i hat/m ( a 4 x 4(/4a + i ht/m, dx dk, 5

6 Using the definition of ω answer as Ψ(x, t a/[ + ( hat/m ], we can rewrite our x ωe ω. For large t, we have that ω is inversely proportional to time, so the wavefunction is of the same form but with its rms spread proportional to t. We can also notice that when written in this form, that x, p 0. For x, we integrate directly and find /4ω. p involves integration by parts as follows p h Ψ Ψ, + h Ψ Ψ, ( ( h ω ax ax exp( ω x, i hat/m + i hat/m h 4 a ω3 x exp( ω x, h a. (6 We see that σ x σ p σ x σ p h/. h a/(ω, which when t 0, ω a, and 6. Current Vector Find the current density carried by a plane wave Ae ikx in one dimension, showing that it is in fact what one would expect from the formula ρv, and verify that it satisfies the equation of continuity. [For ψ A exp(ikx, We know that the current density is j i h ( ψ ψ ψ ψ, m x x i h m ( ika ika. (7 Therefore, j ( hk/m(a (v(ρ. The continuity equation t ρ.j is trivially satisfied.] 6

7 7. Commutators Prove the following: where a is a constant number. [Â, ˆB] [ ˆB, Â] [Â + ˆB, Ĉ] [Â, Ĉ] + [ ˆB, Ĉ] [a, Â] 0 [aâ, ˆB] a[â, ˆB] [Â ˆB, Ĉ] Â[ ˆB, Ĉ] + [Â, Ĉ] ˆB [Using the definition that [A, B] AB BA, it is straight forward to show these relations. We just do one of them here. [AB, C] ABC CAB ABC + ( ACB + ACB + ( ACB + ACB CAB 8. Sum Rule A[B, C] + (ACB ACB + [A, C]B A[B, C] + [A, C]B (8 (a Consider the commutator [H, x] [H, x] [ p + V (x, x] m m [p, x] Then (p[p, x] + [p, x]p (where we have used the result of problem 7. m i h m p ψ n [H, x] ψ n ψ n [Hx xh] ψ n i h m ψ n p ψ n ψ n (E n x xe n ψ n (E n E n ψ n x ψ n ψ n p ψ n i m h (E n E n ψ n x ψ n 7

8 (b ψn p m ψ n ψn [H, x] ψ n h m h m h m h m h m h ψn [H, x] ψ n (9 n ψ n [H, x] ψ n ψ n [H, x] ψ n (0 n (E n E n ψ n x ψ n (E n E n ψ n x ψ n n (E n E n ψ n x ψ n ψ n x ψ n n (E n E n ψ n x ψ n In going from Equation 9 to 0 we use the fact the ψ n is a complete set. Integrals x sin(ax sin(bxdx ( cos(a bx cos(a bx x sin(a bx + (a b (a + b a b x sin(a + bx a + b 8

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