PHY 215A, Quantum Mechanics: Solutions of Homework Set #4
1. 10 points. Time Evolution. The time evolution of two states from time t=0 to time t is given by |ψ(t) >= exp[−iHt/h¯]|ψ(0) >; |φ(t) >= exp[−iHt/h¯]|ψ(0) > . (1) Then <φ(t)|ψ(t) > = <φ(0)|exp[+iHt/h¯]exp[−iHt/h¯]|φ(0) > (2) = <φ(0)| I |φ(0) > |ψ(0) >=<φ(0)|φ(0) > . Thus with both states evolving in time under the same Hamiltonian H, the inner product is unchanged. This is the case even though each state may be changing substantially; recall the Gaussian wave packet for a free particle: the width of the wave packet increases, finally increasing linearly in time, and the center of mass of the wavepacket (expectation value of X) move linearly in time. This invariance of the inner product between the states – their relative direction in Hilbert space, you might say – is unchanged. Mathematically, this follows because the time evolution is a unitary operator, a fact we used above to get the result.
2. 10 points. More time evolution. Problem: Using the energy representation {|E >}, start with the time-dependent Schr¨odinger equation and derive the simple equation for the time evolution operator: U(t) = exp(−iHt/h¯). This was done in class; the object is to work through it clearly and understand it, so present it carefully. Note: The argument is given in the text, and was done in class. The objective was to have you go through it carefully yourself and follow it through. 3. 5 points. Momentum of Gaussian Wave Packet. For a free particle of momentum p =hk ¯ modulated by a real Gaussian wave packet, i.e. ψ(x)= exp(ikx)G(x) where G(x) is the real Gaussian centered at zero (for simplicity), calculate the effect of the Gaussian on the mean momentum
. Solution. d
=<ψ|p|ψ > = dxe−ikxG(x)[−ih¯ [eikxG(x)] Z dx = dxe−ikxG(x)[¯hkeikxG(x)+ eikxG′(x)] Z = dx[¯hkG2(x)+ G(x)G′(x)]=¯hk + zero. Z The integral is over all space. Thehk ¯ results because the Gaussian is normalized, the 2nd term vanishes because G(x) is an even function, G′(x) is odd and their product is odd so it integrates to zero. Thus the Gaussian envelope does not affect
. It will of course evolve in time differently from a planewave. 2 4. 20 points. Scattering from a repulsive potential in 1D. Consider a particle (mass m etc.) impinging on a “square potential” from the left. The potential is nonzero only for 0 2 (c) At E = V the wavevector k′ = 2m(E − V )/h¯ becomes zero, and the q two planewave solutions reduce to only one, a constant. Going back to the Schr¨odinger equation, one finds the general solution is of the form U + W x with constants U, W determined by boundary conditions as usual. This is what should have been figured out. Anyone who did, and worked through to the new solution, will get up to 5 points extra credit. 5. 15 points. Uncertainties in the Potential Well. The eigenfunctions for a ‘particle in the box’ well are given in Shankar Eq. 5.2.17a and 5.2.17b (and discussed in class). For the even n solutions only, (a) give the values of as a function of n. Interpret the result (which can be obtained easily). Both expectation values are zero. There are physical and mathematical ways to arrive at this condition. Just doing the integrals, which are simple, works. Physically: suppose . Also, without doing the integrals, , Pψ has the opposite parity from ψ so that integral vanishes. (b) calculate the uncertainty ∆P as a function of n. Interpret both the dependence on n and the dependence on L. Since vanishes, we just need to calculate < P 2 > and take the square root. Again, thinking saves pencil lead. 2 2 2 2 P 2 π h¯ < P >= 2m< >= 2m