is even and ψ 1 ( ξ 2 1)e ξ 2 dξ = 2 p 2 dx =! 2 2α 2! 2 π ξ 4 e ξ 2 dξ =
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1 Physics 0 Homework #6 Spring 06 Due Friay 5/3/6. Griffith s. a. The expectation value of the position (an momentum by virtue of Ehrenfest s theorem) will be zero. Since ψ 0 is even an ψ is o, ψ will be even. So, x x = x ψ. This of course means that p = m, by Ehrenfest s theorem. However, the expectation value of the position an momentum square will not be zero. For ψ 0 :! x = α x e ξ = α mω 3 ξ e ξ ξ = p = ψ o i! mω ψ o =! α! For ψ : = m!ω π m!ω ( ξ )e ξ ξ = π! x = α x e ξ = α mω 3 p = ψ i! mω ψ =! α! = m!ω π! π mω e ξ ξ e ξ π π = m!ω ξ e ξ ξ = ξe ξ π =! mω ξ! π mω 3 π = 3! mω ξ ξe ξ ξ m!ω 3 π ( ξ 3ξ )e ξ ξ = π 3 π = 3m!ω c. The expectation values of the kinetic an potential, as well as the total energies are base on part a. We have
2 T = p m = V = mω x = H = T + V =!ω forψ 0 3!ω forψ!ω forψ 0 3!ω forψ!ω forψ 0 3!ω forψ. Griffith s.5 The groun state wave function is ψ 0 = region is given by E 0 = mω x 0 x 0 = mω π! e mω! x an the classically allowe E 0. To calculate the probability we mω use P = ψ * mω 0 ψ 0 ψ 0 = x 0 π! e mω! x, where the factor of two is x 0 from integrating from x 0 to infinity an from minus infinity to x 0 outsie of the classically allowe region. Evaluating the integral on Mathematica (or looking it up in a table of integrals) we fin: P = mω π! mω π! π!ωe 0 mω E 0 Erf [ E 0!ω ] = [ Erf []] using the fact that the groun state energy is E 0 =!ω. Evaluating the error function on Mathematica we fin that the probability is given as P = Erf [] = , or a 5.7%chance of being foun outsie of the classically forbien region! The mathematica coe is given below.
3 3. Griffith s 3.7 Note: In part B, o not construct linear combinations that are orthogonal. Instea etermine the eigenvalue for each of the two wave functions given. The eigenvalue shoul be the same for both of the wavefunctions. What is it? a. For the two functions given we have ˆQf (x) = qf (x)an ˆQg(x) = qg(x). Now to form a linear combination let h(x) = af (x) + bg(x), where a an b are constants. Thus we have operating with ˆQ, ˆQh(x) = ˆQ af (x) + bg(x) ( ) = a ˆQf (x) + b ˆQg(x) = q(af (x) + bg(x)) = qh(x). This is exactly what was one in homework problem #3.8. b. To etermine the eigenvalues for the two given functions take the two erivatives. We have: e x = ( ) ex = ( ) ex = ex an e x = e x ( ) = ( ex ) = e x. Thus these two functions are eigenfunctions of the. Griffith s 3.0 operator an both have the same eigenvalue, +. The groun state wave function of the infinite square well is given as ψ = a sin π a x. Apply the momentum operator an we have ˆpψ = i! a sin π a x = i! π a a cos π a x. Therefore since we o not get the wave function back multiplie by a constant, the groun state wave function of the infinite square well is not an eigenstate of the momentum operator. 5. Griffith s 3.3 [AB,C] = ABC CAB = ABC CAB + (ACB ACB) a. [AB,C] = A[B,C] [C, A]B = A[B,C]+ [A,C]B b. c. f [x n, p] [x n, p] f = i!x n ( i! f (xn f )) = i!x n + i! f nxn f + x n. [x n, p] f = i!nx n f [x n, p] = i!nx n [ f, p] [ f, p]q = fpq pfq = i!f q ( i! ( fq)) = i!f q + i!(f q + f q ) [ f, p]q = i! f f q [ f, p] = i!
4 6. Griffith s. - Note: Only o parts A an B. Skip C. a. The commutation relations: x, y [ ] = xy yx [ y,z] = yz zy [ z, x] = zx xz r i,r j p x, p y f = p x p y f p y p x f =! p y, p z f = p y p z f p z p y f =! f y f y f y z f z y p x, p z f = p p f p p f = f x z z x! z f z p i, p j The position-momentum commutation relations were one in class. See class notes. c. For this problem, we ll use eqn. 3.7 equation Q = ψ Qψ = ψ * Qψ [ H,Q] = ψ H,Q Q = ψ Q ψ Q = i! [ H,Q ] + Q [ ]ψ = ψ * [ H,Q]ψ = ψ * ( HQ QH )ψ = ψ Q * ψ. In this
5 To show the relation between position an momentum, x = i ( ψ * ( Hx xh )ψ! ) + ψ * ψ [ H, x] = p m + V, x = p m, x + [ V, x] [V, x] = Vx xv m p, x = [ m pp, x] = m [H, x] = i! m p x = i! " r = ( ) = i! m p = i!p + pi! m " p m ψ * [ x, pp ] = ([ m x, p ] p + p[x, p] ) i! m p ψ + ψ * ψ = m ψ * ( p)ψ = p m
6 Ehrenfest s Theorem p = i p ( ψ * ( Hp ph )ψ! ) + ψ * ψ [ H, p] = p m + V, p = p m, p + [ V, p] [V, p] f = ( Vp pv ) f = i!v f = i!v f [V, p] = i! V m f i!v p, p = [ m pp, p] == [H, p] = i! V p = i! ψ * i! Vf i!f V = i! V f ([ m p, p ] p + p[ p, p] ) i! V ψ + ψ p * ψ V = ψ * ψ i! ψ * ψ p = V 7. Griffith s. a. To etermine the stationary states of the infinite cubical well, assume that the solution can be written as ψ (x, y,z) = X(x)Y (y)z(z). Inserting this into the Schroinger wave equation gives us! m ψ + Vψ = Eψ m + y + z! YZ X + XZ Y y XYZ = EXYZ + XY Z z = me! XYZ X X + Y Y y + Z Z z = me = k = (k! x + k y + k z ) Here we ve separate the Schroinger equation into three pieces, each of which equals a constant. We can solve each of these:
7 X X Y Y y = k x X = k X X(x) = A x sin(k x x) + B x cos(k x x) = k y Y y = k Y Y (y) = A y sin(k y y) + B y cos(k y y) Z Z z = k z Z z = k Z Z(z) = A z sin(k z x) + B z cos(k z x) Next we apply the bounary conition in each irection that ψ (x = y = z ) =ψ (x = y = z = a). This gives: ψ x=y=z=0 0 = A x sin(0) + B x cos(0) B x 0 = A y sin(0) + B y cos(0) B y 0 = A z sin(0) + B z cos(0) B z ψ x=y=z=a 0 = A x sin(k x a) k x a = n x π k x = n xπ a 0 = A y sin(k y a) k y a = n y π k y = n yπ a 0 = A z sin(k z a) k z a = n z π k z = n zπ a This leas us to the energies of the stationary states: k = (k x + k y + k z ) = n x + n ( y + n z ) π a = me!! π E n = n ma =! π ma n x + n ( y + n z ) The stationary states are given as: ψ (x, y,z) = A x A y A z sin(k x x)sin(k y y)sin(k z z) 3 ψ (x, y,z) = a sin(k x x)sin(k y y)sin(k z z) where we normalize each of the terms accoring to a = A x sin a (k x x) = A x A x = A y = A z = a. 0
8 b. The energies are given in the table below. E n n = n x + n y + n z E n π! ma # egenerate states E 3 E E 3 E E 5 E E E 8 E E 0 E E E E
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