Skip to article frontmatterSkip to article content

Problem Sheet: Week 8

School of Physics, University of Bristol

Wavepackets

  1. In this question we will consider a wavepacket of a free particle. Assume the wavepacket has the form (4.14) from the notes, namely

    Ψ(x,t)=12πc(k)eik2t/2Meikxdk,\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\infint c(k) e^{-i\hbar k^2 t/2M} e^{ikx} dk,

    where c(k)c(k) is given by

    c(k)={aif k1,0otherwise. c(k) = \begin{cases} a& \text{if } |k| \leq 1, \\ 0 &\text{otherwise. } \end{cases}

    and where aa is a positive constant.

    1. Sketch the function c(k)c(k).
    2. Find aa such that c(k)c(k) is normalised.
    3. By evaluating the above integral expression, show that the wavefunction at t=0t = 0 is
      Ψ(x,0)=1πsinxx.\Psi(x,0) = \frac{1}{\sqrt{\pi}} \frac{\sin x}{x}.
    4. Sketch the wavefunction Ψ(x,0)\Psi(x,0) and the probability density P(x,0)P(x,0).
  2. In this question we will derive the result (4.23) from Example 4.1 in the notes (Gaussian wavepackets). Assume again a wavepacket of the form (4.14) (given above), with c(k)c(k) given by the Gaussian function

    c(k)=(2a2π)1/4ea2(kk0)2,c(k) = \left(\frac{2 a^2}{\pi}\right)^{1/4}e^{-a^2(k-k_0)^2},

    where aa and k0k_0 are arbitrary real constants.

    1. By evaluating the above integral expression for Ψ(x,t)\Psi(x,t) when t=0t = 0, show that
    Ψ(x,0)=(1a2π)1/2ex2/4a2eik0x.\Psi(x,0) = \left(\frac{1}{a\sqrt{2\pi}}\right)^{1/2} e^{-x^2/4a^2}e^{ik_0 x}.
    1. Write down the probability density P(x,0)P(x,0) for this wavefunction.
    2. Make a sketch of the probability density and use it (without calculation) to find the expectation value of the position of the particle x\langle x \rangle.
    3. Calculate the standard deviation of the probability density P(x,0)P(x,0). That is, calculate Δx=x2x2\Delta x = \sqrt{\langle x^2\rangle - \langle{x}\rangle^2}, where
      x2=x2P(x,0)dx. \langle x^2 \rangle = \int_{-\infty}^{\infty} x^2 P(x,0) dx.

Momentum

  1. Consider a particle with the same wavefunction at time t0t_0 as Problem 7.1 of Problem Sheet: Week 7,

    Ψ(x,t0)={154(1x2)if x1,0otherwise. \Psi(x,t_0) = \begin{cases} \frac{\sqrt{15}}{4}(1-x^2)& \text{if } |x| \leq 1, \\ 0 &\text{otherwise. } \end{cases}
    1. Show that the momentum wavefunction of the particle at time t0t_0 is
      Ψ~(p,t0)=1532π(sin(p/)p3cos(p/)p2).\tilde{\Psi}(p,t_0) = \sqrt{\frac{15 \hbar^3}{2\pi}}\left(\frac{\hbar \sin (p/\hbar)}{p^3} - \frac{\cos(p/\hbar)}{p^2}\right).
      Hint

      You will need to integrate by parts twice in order to obtain this result.

    2. Sketch the wavefunction Ψ~(p,t0)\tilde{\Psi}(p,t_0) and the probability density P(p,t0)P(p,t_0).
    3. What is the probability amplitude and probability density for the particle to have momentum p=πp = \pi \hbar?
  2. Consider two particles, the first of which has wavefunction Ψ(x,t0)\Psi(x,t_0), and the second of which has wavefunction Ψ(x,t0)\Psi'(x,t_0), related to Ψ(x,t0)\Psi(x,t_0) via

    Ψ(x,t0)=Ψ(x,t0)eik0x,\Psi'(x,t_0) = \Psi(x,t_0)e^{ik_0x},

    where k0k_0 is a real constant. This is the same situation considered in Problem 7.5.

    The momentum wavefunction of the first particle is

    Ψ~(p,t0)=12πΨ(x,t0)eipx/dx.\tilde{\Psi}(p,t_0) = \frac{1}{\sqrt{2 \pi \hbar}}\infint \Psi(x,t_0) e^{-ipx/\hbar} dx.
    1. Write down the momentum wavefunction of the second particle, i.e.~the momentum wavefunction associated to Ψ(x,t0)\Psi'(x,t_0).
    2. Show that the momentum wavefunctions of the two particles are related via
      Ψ~(p,t0)=Ψ~(pk0,t0).\tilde{\Psi}'(p,t_0) = \tilde{\Psi}(p-\hbar k_0,t_0).
      Hint

      It may be useful to introduce the new variable p=pk0p' = p-\hbar k_0 in your answer to part 1.

    3. In a single plot, make representative sketches of Ψ~(p,t0)2|\tilde\Psi(p,t_0)|^2 and Ψ~(p,t0)2|\tilde\Psi'(p,t_0)|^2.
    4. Use your answers to part 2 and 3 to explain how the state of a particle changes when we multiply the spatial wavefunction by eik0xe^{ik_0x}. How does this relate to your answer to Problem 7.5.3.
  3. Consider a particle with the following momentum wavefunction at time t0t_0,

    Ψ~(p,t0)={1pbpaif pappb,0otherwise. \tilde{\Psi}(p,t_0) = \begin{cases} \frac{1}{\sqrt{p_b - p_a}} & \text{if } p_a \leq p \leq p_b, \\ 0 &\text{otherwise. } \end{cases}

    where pa<pbp_a < p_b. This wavefunction has the form of a ``box’', of width Δ=pbpa\Delta = p_b - p_a and centre at pc=(pa+pb)/2p_c = (p_a + p_b)/2.

    1. Show that the spatial wavefunction Ψ(x,t0)\Psi(x,t_0) of the particle at time t0t_0 is
    Ψ(x,t0)=2π(pbpa)(eipbx/eipax/)ix.\Psi(x,t_0) = \sqrt{\frac{\hbar}{2\pi (p_b - p_a)}} \frac{(e^{ip_bx/\hbar} - e^{ip_ax/\hbar})}{ix}.
    1. By expressing pap_a and pbp_b in terms of the centre pcp_c and width Δ, show that the wavefunction can alternatively be written as
    Ψ(x,t0)=Δ2πeipcx/sinc(xΔ/2),\Psi(x,t_0) = \sqrt{\frac{\Delta}{2\pi\hbar}} e^{ip_c x/\hbar}\mathrm{sinc}(x\Delta/2\hbar),

    where sinc(y)=sin(y)/y\mathrm{sinc}(y) = \sin(y)/y.

    1. (Tricky) Consider now a particle with a spatial wavefunction
    Ψ(x,t0)=Δ2πsinc(xΔ/2).\Psi'(x,t_0) = \sqrt{\frac{\Delta}{2\pi\hbar}}\mathrm{sinc}(x\Delta /2\hbar).

    Use Problem 8.4 to write down (i.e. without calculating explicitly) the momentum wavefunction Ψ~(p,t0)\tilde{\Psi}'(p,t_0) of the particle at t0t_0. What is the centre and width of this wavefunction?