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Homework Help: Time independant wave-function can always be taken to be real?

  1. Oct 6, 2007 #1
    1. The problem statement, all variables and given/known data

    I need to prove that. I know conceptually how that is. If it's complex, I can split it up into a bunch of real solutions that have complex constants or what not, so Psi(x) is still always real. That much makes sense.

    I also see why it has to be used as real, because then |Psi(x,t)|^2 is real. Otherwise it would be complex and the Terrorists win.

    2. Relevant equations

    The book (Griffith's Intro to QM) tells me to use

    [tex]\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2} + V(x)\Psi = E\Psi[/tex]

    "If Psi(x) works for the equation, then so does its complex conjugate and any linear combination of them." Great.

    3. The attempt at a solution

    I don't think simple plugging in Psi(x)* for Psi(x) will give me anything useful.

    Although, I think I see why it plugging in the complex conjugate would work if Psi(x) works. It means Psi(x) is real and so is the CC, right? If it wasn't real, then you'd get a complex energy.

    Or something...

    I just don't understand how I can prove this mathematically.

    Only thing I can think of right now is saying that if Psi(x) is complex, I can just make a complex constant in front of a real Psi(x) (more if I have to), then show that even though I have a bunch of complex constants, it will all cancel out once I multipy Psi(x)* by Psi(x).

    Although that would mean that either all the constants are complex or all of them are real, or else I would still end up with complex stuff at the end.

    Then again, the only time you would even write Psi(x) in the form of an infinite sum with constants is if you wanted the constants to be complex, right? There's no point in writing it out if it's already real...

    This is the point of QM, right? To make your head explode?
  2. jcsd
  3. Oct 7, 2007 #2


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    |Psi(x,t)|^2 can be real even though psi(x) is real, modulus and modulus square is always real.

    Do an Ansatz:

    [tex] \Psi (x) = \psi (x) + i \phi (x) [/tex]
    [tex] \psi , \phi [/tex] are real functions.

    Now if [tex] \Psi(x) [/tex] is a solution, then also [tex] \Psi (x)^{*} [/tex] is; i.e
    [tex] \Psi (x)^* = \psi (x) - i \phi (x) [/tex]

    Show that this is not possible.

    And perhaps also have a pure imaginary function:
    [tex] \Psi _v(x) = i \phi _v(x) [/tex]
    [tex] \Psi _v(x)^* = - i \phi _v(x) [/tex]
    Last edited: Oct 7, 2007
  4. Oct 7, 2007 #3


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    I don't see why not!?

    I think that the point is that if [itex] \Psi [/itex] is a solution, [\Psi^*[/itex] is also a solution with the same energy (a key point!), as long as there is no time dependence.
    Therefore, the sum [itex] \Psi + \Psi^* [/itex] is also a solution with the same energy (divide by root 2 if you want it normalized). And this combination is real.

    Therefore, given any time independent complex wavefunction, just build [itex] (\Psi + \Psi^*)/\sqrt{2}[/itex] and this will be real and have the same energy as the initial wavefunction and obey the same Schrodinger equation. Basically, it contains the same physics as the initial function but is real.
  5. Oct 7, 2007 #4
    The book says that the complex conjugate and any linear combination (with the same energy) will work. But I still don't see how to prove that mathematically.

    By "same energy", does it mean that the entire solution have the same energy, and it's just split between the terms, or that each term has the same energy?
  6. Oct 7, 2007 #5


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    Not sure I understand what exactly the question is asking, but my best guess of its intention suggests it's wrong. For instance, it's easy to show that if you want to describe a state with a non-zero expectation value for momentum, then you need a wavefunction with a non-zero imaginary part. All real-valued wavefunctions have zero average momentum.

    Poop-loops: Which book is this, and where is it in the book?
  7. Oct 7, 2007 #6


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    How can you prove your assertions ?
  8. Oct 7, 2007 #7
    It's Griffith's "Introduction to Quantum Mechanics" second edition. Problem 2.1b

  9. Oct 7, 2007 #8


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    For simplicity, in 1D:
    [tex]\langle p \rangle = -i \hbar \int_{- \infty}^{\infty} dx \psi ^*(x) \psi '(x) = -i \hbar \int_{- \infty}^{\infty} dx \psi (x) \psi '(x) [/tex]
    for all real-valued [itex]\psi(x)[/itex].

    Now, integrating by parts, you get:
    [tex]\langle p \rangle = -i \hbar [\psi ^2 (x) ]_{- \infty}^{\infty} + i \hbar \int_{- \infty}^{\infty} dx \psi '(x) \psi (x) [/tex]

    For a well-behaved wavefunction, the first term above should vanish, leaving you with

    [tex]\langle p \rangle = -\langle p \rangle \implies \langle p \rangle = 0 [/tex]
    Last edited: Oct 8, 2007
  10. Oct 8, 2007 #9

    Doc Al

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    Take the complex conjugate of the time-independent S.E. That will show you that both [itex]\psi^* (x)[/itex] and [itex]\psi (x)[/itex] are solutions with the same energy.
  11. Oct 8, 2007 #10
    Since the S.E. is a real function, taking the CC of it will give me the real function back, right?

    EDIT: Okay, so what I did was to show that the complex conjugate of the Hamiltonian x Psi(x) = Hamiltonian x Psi(x) (which is the time independent SE, right?), and the complex conjugate of a complex conjugate is just that thing, so (H x Psi(x)*) = H x Psi(x), right? So it's still all the same?
    Last edited: Oct 8, 2007
  12. Oct 8, 2007 #11

    Doc Al

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    The time-independent S.E. is the equation you gave in post #1. By taking its complex conjugate, you can show that if [itex]\psi (x)[/itex] is a solution for some value of E, then so is [itex]\psi^* (x)[/itex]. Then you prove that those linear combinations given in the hint (which are real) are also solutions.
  13. Oct 8, 2007 #12


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    The SE is an equation, not a function.

    First of all, you do not want to be multiplying the Hamiltonian with the wavefunction. The Hamiltonian is an operator; it operates on the wavefunction.

    Next, you can take any equation, and if you replace both sides with their complex conjugates, the two sides that result will still equal each other. Doing this with the LHS of the (TI)SE, you see that [itex](\cal{H} \psi )^* = \cal{H}^{\dagger} \psi ^* [/itex], but since the Hamiltonian is hermitian, [itex]\cal{H}^{\dagger} = \cal{H} [/itex]. The RHS is just the product of a real number (E) and [itex] \psi[/itex]. Writing its complex conjugate is even easier. Comparing the new LHS and the new RHS tells you that if [itex]\psi[/itex] is an eigenfunction of [itex]\cal{H} [/itex] with eigenvalue E, then so is [itex]\psi ^*[/itex].
  14. Oct 8, 2007 #13
    That's kind of what I did. Except that I treated H as a function... so it was real and then H* would also be real.

    I think I read in the book about the Hamiltonian being Hermitian, but I had no idea how it could be a matrix (only Hermitian thing I've learned about so far...) so I guess it just went in one eye and out the other.

    Thanks a lot, though. I asked my professor about this today and... I still didn't get it really.
  15. Oct 8, 2007 #14
    Actually, no, the book says nothing about the Hamiltonian operator being Hermetian.

    It just states what it is, and goes over the expectation value of it.

    I don't know if I'm only supposed to use what's given in the book, but I don't care.
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