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A problem based on Quantum Physics.

  1. Dec 22, 2013 #1
    1. The problem statement, all variables and given/known data

    A system has 2 possible energy values E0 and 2E0. At a certain instant it is in a state in which the expectation value of the energy is 3/2 E0.Calculate the wave function in this state given that ψ0 , ψ1;are the wavefuntion corresponding to two possible stationary states.What is the wave function after a time t has elapsed?

    2. Relevant equations






    3. The attempt at a solution
    <E>=∫ ψ*ψdτ=∫(c0ψ0 + c1ψ1)*Ê(c0ψ0 + c1ψ1)dτ
    <E>=c02E0 + 2c12E0=3/2E0
    Co^2 + 2C1^2=3/2.
    Co=√2/3 and C1 =√4/3.so ψ =√2/3ψο +√4/3 ψ1.
    But in the text book given answer is 1/√2 (Ψο + ψ1). Where did I go wrong?also I can't figure out wavefuntion after t time elapse analytically.please help me out.
     
    Last edited: Dec 22, 2013
  2. jcsd
  3. Dec 22, 2013 #2
    Can you explain how you got those specific values of C0 and C1 out of the equation you've derived? I'm not sure I see how that works.

    Also, that equation is just one equation in two unknowns, so there are an infinite number of C0/C1 pairs that satisfy it. Remember that your wavefunction also needs to be normalized, so in addition to your equation, you also need to have C0^2 + C1^2 = 1. With those two equations together, you should be able to derive the answer that they got.
     
  4. Dec 22, 2013 #3
    Thanks for the hint .I figured it out.what is not clear to me is time evolution in stationary state concept.could you give me some hint about the 2nd part of the question.
     
  5. Dec 22, 2013 #4
    Well, remember that time evolution is governed by the Schrodinger Equation: [itex]i\hbar\frac{d}{dt}\psi(t) = \hat{H}\psi(t)[/itex]. That means that the wavefunction obeys the equation [itex]\psi(t) = e^{-i\hat{H}t/\hbar}\psi(0)[/itex]. So now you just need to determine what the [itex]\hat{H}[/itex] operator does to energy eigenstates, and you should be able to work out what happens to the wavefunction as it moves through time.
     
  6. Dec 22, 2013 #5

    ShayanJ

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    I think the integral should not be with respect to time but position!
    Because we have [itex] \langle A \rangle=\langle \psi |A|\psi\rangle=\langle \psi|A\psi\rangle=\langle A\psi|\psi\rangle [/itex] which means just the inner product of two kets and that is an integral with respect to position when you're working in the position space!
     
  7. Dec 23, 2013 #6
    My knowledge of quantum physics is very elementary and I still haven't digested dirac notation well enough.could you please explain your point using old notation.
     
  8. Dec 23, 2013 #7
    But we aren't working in position space here. There isn't any description of what the system's position-space dependence is, only that it has a Hamiltonian with two energy eigenstates, and that the system is in a superposition of them. So there's no way to write out a position wavefunction for this system, meaning that expressing the wavefunction in a position basis isn't going to be useful.

    sudipmaity, do you know how to write down the time dependence of a system that is in a single energy eigenstate? If so, it's not too much more work to write down the time dependence of a system that is in a superposition of eigenstates.
     
  9. Dec 23, 2013 #8

    vanhees71

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    Since this is a two-level system, it doesn't make much sense to me to work in position space. You have given all information on the system, provided the energy eigenvalues are not degenerate. Thus the Hilbert space is two-dimensional and spanned by two energy eigenkets [itex]|u_1 \rangle[/itex] and [itex]|u_2 \rangle[/itex]. These are orthonormal vectors, because they are the eigenvectors of the self-adjoint Hamilton operator of the system. The eigenvalues are also given to be [itex]E_0[/itex] and [itex]2 E_0[/itex].

    A general pure state is represented by a unit vector, i.e., by
    [tex]|\psi \rangle = \psi_1 |u_1 \rangle + \psi_2 |u_2 \rangle.[/tex]
    The components [itex]\psi_1[/itex] and [itex]\psi_2[/itex] must fulfill the normalization condition
    [tex]|\psi_1|^2 + |\psi_2|^2=1.[/tex]

    The other condition given is that the energy expectation value is [itex]3/2 E_0[/itex], i.e.,
    [tex]\langle \psi |H| \psi \rangle = E_0 |C_1|^2+2 E_0 |C_2|^2=\frac{3}{2} E_0,[/tex]
    which you should prove yourself from the formalism!

    Now you have a linear system of equations for [itex]|C_1|^2[/itex] and [itex]|C_2|^2[/itex]. Find the solution first. Then you'll see that your texbook is not wrong, because it gives you indeed a state with the desired expectation value of the energy.

    Think, however, a bit further! Is the state really determined completely by just giving the one expectation value of the energy? Note that [itex]C_1[/itex] and [itex]C_2[/itex] both can be complex and that any unit vector that differs from [itex]|\psi \rangle[/itex] simply multiplied by a phase factor describes the same pure state, i.e., any ket
    [tex]|\psi' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle[/tex]
    is equivalent with [itex]|\psi \rangle[/itex].

    Further you can think about the time evolution of the state in the Schrödinger picture, as also seems to be discussed somewhat in this thread.

    For sure, integration of a wave function over time in the proposed in the original posting doesn't make much sense at all. Where does this idea come from? Maybe you mix this up from classical electrodynamics, where you consider time averaged values of observables like field energy? That's not how quantum theory works! Here the expectation values are of different nature, and you must not consider the wave function in quantum theory as a kind of field describing a particle as a "smeared something". This was the original idea of Schrödinger's when he invented his "wave mechanics". It has been rapidly abandoned by Born to contradict the experimental findings about particles like electrons and substituted by the probabilistic interpretation of the wave function, which still is valid today. It is very important to get used to that ideas, which however takes some time and is the real challenge of learning quantum mechanics!

    For that purpose it's also very goot to get used to the bra-ket (or more generally the abstract Hilberst-space) formalism of quantum mechanics as soon as you can. Wave functions are nice but not the most versatile way to express quantum theory. They are simply the kets in position representation. As in this example, there is not always a position representation available or it's not necessary to be used.
     
  10. Dec 23, 2013 #9

    ShayanJ

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    Well,you may choose not to work in position space,but I know of no situation where the inner product of two wavefunctions becomes an integral with respect to time!!!
     
  11. Dec 23, 2013 #10
    Right, but vanhees71's point is that in a two-level system, you don't need an integral at all. It's just a sum of discrete terms for each energy level.
     
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