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Unformatted text preview: PHYS851 Quantum Mechanics I, Fall 2008 HOMEWORK ASSIGNMENT 6: Motion in 1D, Probability Current: 1. [10pts] Let P (ta, b) be the probability that a particle is in the range a < x < b. Show that d P (ta, b) = j (a, t) − j (b, t) dt where j (x, t) is the probability current at position x at time t. Answer:
b P (ta, b) = d P (ta, b) = dt Using ψ = xψ , gives
d ψ dt a b a dx ψ (x, t)2 ˙ ˙ dx ψ ∗ ψ + ψ ∗ ψ = − i H ψ , and H = T + V (X ), where T is the kinetic energy operator, i
a b d P (ta, b) = dt = = dx [ ψ H x ψ − ψ ∗ xH ψ ]
b i
a dx [ ψ T x ψ + ψ V (X )x ψ − ψ ∗ xT ψ − ψ ∗ xV (x)ψ ]
b 2 2 i
a dx −
b a 2m (ψ ∗ )′′ ψ + V (x)ψ ∗ ψ − 2m ψ ∗ ψ ′′ − V (x)ψ ∗ ψ i =− 2m integration by parts gives dx [(ψ ∗ )′′ ψ − ψ ∗ ψ ′′ ] b b d i ∗′ dx (ψ ∗ )′ ψ ′ − ψ ∗ ψ ′ (ψ ) ψ − P (ta, b) = − a dt 2m a b i =− [(ψ ∗ )′ ψ − ψ ∗ ψ ] a 2m b a b +
a dx (ψ ∗ )′ ψ ′ with j=− this gives i [ψ ∗ ψ ′ − (ψ ∗ )′ ψ ] 2m P (ta, b) = j (a, t) − j (b, t) 2. [10pts] For the inﬁnite squarewell potential, calculate the momentum uncertainty ∆P for each energy eigenstate. You should write the eigenstate wavefunctions as φn (x) = where u(x) is the unit step function. Answer: P = =
−∞ 2 nπx u(x)u(L − x) sin L L φn  P  φn
∞ dx φ∗ (x) xP φn n
∞ = −i = −i =− =0
2 0 222 dx φn (x)
−∞ φn (∞) φn (−∞) d φn (x) dx dφn φn i φ2 (∞) − φ2 (−∞) n 2n L P 2 =− = dx φn (x)φ′′ (x) n nπx L 0 L 2nπx d 1 − cos L 0 dx sin2
L 2 nπ L3 222 nπ = L3 222 nπ = L2 ∆P = so P2 − P
2 = nπ L 3. Finite square well Bound State:Consider a particle of mass m which feels the potential x < −a 0, −V0 , −a < x < a V (x) = 0, x>a (a) [10pts] Start by writing a general form (ansatz) for the solutions in the regime −V0 < E < 0. By applying the required boundary conditions, ﬁnd an equation which determines the energies of the bound states (those states with energy satisfying −V0 < E < 0). Answer: The general solution has the form: beγx x < −a c cos(kx) −a < x < a ψeven (x) = be−γx a<x x < −a −beγx c sin(kx) −a < x < a ψodd (x) = be−γx a<x 2m(E + V0 )/ 2 . where γ = −2mE/ 2 and k = For even parity, continuity of psi at x = a gives c cos(ka) = be−γa and continuity of the derivative at x = a gives −kc sin(ka) = −γbe−γa due to symmetry, the equations for x = −a are redundant. Solving these equations for b and c gives e−γa − cos(ka) b =0 γe−γa −k sin(ka) c ke−γa sin(ka) + γe−γa cos(ka) = 0 which gives tan(ka) = γ k A solution required det = 0, which gives For odd states we have e−γa − sin(ka) γe−γa k cos(ka) cot(ka) = − γ k b c =0 which leads to (z0 /z )2 − 1 where z = (b) [10pts] Show that this equation can be written as tan z = √ a 2m(E + V0 )/ and z0 = a 2mV0 / . This equation holds for solutions which are even functions of x. What about solutions which are odd functions? Answer: let z = a 2m(E + V0 )/
2 2 = ka and z0 = a 2m
2 2mV0 / k2 + γ 2 = (E + V 0 − E ) = z0 a
2 z0 a 2 γ2 = γa = So the even solutions require tan(z ) = For the odd solutions this gives − k2
2 z0 − z 2 2 z0 − (ka)2 = z0 z 2 −1 cot(z ) = − z0 z 2 −1 (c) [10pts] Make a plot of tan z versus z and on the same plot show (z0 /z )2 − 1 versus z . Indicate on the plot where the curves intersect. Plot must be computer generated. Answer: Here is a plot for z0 = 100, showing 4 bound state solutions. (d) [10pts] Write Ei as a function of zi , where zi would be the z value of the ith intersection point and Ei would be the corresponding energy eigenvalue. You do not need to ﬁnd expressions or values for the zi ’s. In principle, you could use a computer to ﬁnd the zi ’s and that would then give you the energy eigenvalues. Assuming a solution zi were obtained numerically, we would need a formula to get Ei , the boundstate energy: k2 = ka =
2 zi = 22 2m
2 (E + V 0 )
2 2ma2 (E + V 0 ) 2ma2
2 (Ei + V0 )
22 zi 2ma2 so ﬁnally Ei = −V0 + ...
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This note was uploaded on 10/25/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Fall '08 term at Michigan State University.
 Fall '08
 MichaelMoore
 mechanics, Current, Work

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