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The Schrodinger equation

  1. Jan 4, 2009 #1

    tpg

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    Hi everyone, I'm having an issue trying to make the abstract form of the schrodinger equation:

    [tex]i\hbar\frac{\partial}{\partial t}\left|\psi\right\rangle = H\left|\psi\right\rangle[/tex]

    be consistent with the form that operates on wavefunctions in the position representation:

    [tex]i\hbar\frac{\partial}{\partial t}\psi(x) = H\psi(x)[/tex]

    If I try to do this by plugging in [tex]\left|\psi\right\rangle =\int dx\,\psi(x)\left|x\right\rangle[/tex] to the abstract form, I end up with a contradiction. Starting with the RHS:

    [tex]i\hbar\frac{\partial}{\partial t}\left|\psi\right\rangle & = & i\hbar\frac{\partial}{\partial t}\left(\int dx\,\psi(x)\left|x\right\rangle \right)[/tex]
    [tex] & = & i\hbar\int dx\,\left(\frac{\partial\psi}{\partial t}\left|x\right\rangle +\psi\frac{\partial\left|x\right\rangle }{\partial t}\right)[/tex]

    If we then use the Schrodinger equation again on [tex]\frac{\partial\left|x\right\rangle }{\partial t}[/tex]

    [tex]\frac{\partial\left|x\right\rangle }{\partial t} & = & -\frac{i}{\hbar}H\left|x\right\rangle =\frac{i\hbar}{2m}\int dx'\,\left|x'\right\rangle \frac{\partial^{2}}{\partial x'^{2}}\left(\left\langle x'|x\right\rangle \right)[/tex]
    [tex] & = & \frac{i\hbar}{2m}\int dx'\,\left|x'\right\rangle \delta''(x-x')[/tex]
    [tex] & = & \frac{i\hbar}{2m}\frac{\partial^{2}\left|x\right\rangle }{\partial x^{2}}[/tex]

    Where above we use the identity [tex]\int dx\,\phi(x)\delta''(x-a)=\phi''(a)[/tex]. I think we can now use parts a couple of times, together with the fact that [tex]\psi (x)[/tex] and [tex]\psi' (x)[/tex] go to zero at infinity, to say that

    [tex]\int dx\,\psi\frac{\partial\left|x\right\rangle }{\partial t} & = & \frac{i\hbar}{2m}\int dx\,\psi\frac{\partial^{2}\left|x\right\rangle }{\partial x^{2}}[/tex]
    [tex] & = & \frac{i\hbar}{2m}\int dx\,\frac{\partial^{2}\psi}{\partial x^{2}}\left|x\right\rangle[/tex]

    Now, the RHS of the TDSE will go to [tex]-\frac{\hbar^{2}}{2m}\int dx\,\frac{\partial^{2}\psi}{\partial x^{2}}\left|x\right\rangle [/tex].
    If we multiply by [tex]\left\langle x\right|[/tex] from the left on both sides,
    we end up with

    [tex]i\hbar\left(\frac{\partial\psi}{\partial t}+\frac{i\hbar}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}\right) & = & -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}[/tex]
    [tex]\Rightarrow\quad\frac{\partial\psi}{\partial t} & = & 0[/tex]

    Clearly this is not right, as from the second form of the Schrodinger equation written down above, we get

    [tex]\frac{\partial\psi}{\partial t} & = & \frac{i\hbar}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}[/tex]

    Can somebody help me find whereabouts I am going wrong here? Thanks.
     
  2. jcsd
  3. Jan 4, 2009 #2

    Hurkyl

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    The Schrödinger equation is not an identity. In particular, there's absolutely no reason to think

    [tex]\frac{\partial\left|x\right\rangle }{\partial t} & = & -\frac{i}{\hbar}H\left|x\right\rangle[/tex]

    In general, for a (possibly time-varying) ket [itex]| \psi \rangle[/itex], there absolutely no reason to think

    [tex]
    i\hbar\frac{\partial}{\partial t}\left|\psi\right\rangle = H\left|\psi\right\rangle
    [/tex]

    Kets that do satisfy this equation are very special; they are the ones that describe a system governed by quantum dynamics.
     
  4. Jan 4, 2009 #3

    tpg

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    So is the application of the Schödinger equation to a theoretical state of well-defined position invalid? I'm working on a theoretical basis here, and under that assumption surely the two forms of the Schödinger equation ought to be compatible?
     
  5. Jan 4, 2009 #4

    Hurkyl

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    It is a very rare thing for a (possibly) time-varying ket to satisfy the Schrödinger equation; therefore, it is almost never valid to apply the Schrödinger equation to such a ket. The only times you can do so are when the ket has been assumed to satisfy the equation (or has been proven to do so, or has been specifically constructed to satisfy the equation, or other similar circumstances)

    Quantum mechanics is primarily concerned with studying those (possibly) time-varying kets that do satisfy the Schrödinger equation, so you will see and use a lot of them despite their rarity.


    We can look at a comparable situation in classical mechanics; consider a massive particle with position function [itex]\vec{r}(t)[/itex] travelling through some uniform force field. The corresponding physical equation is Newton's law: [itex]F = m d^2 r(t) / dt^2[/itex]. However, very, very few functions [itex]\vec{r}(t)[/itex] actually satisfy Newton's law!

    Apply Schrödinger's equation to an arbitrary ket is directly analogous to applying Newton's law to an arbitrary function. Schrödinger's equation is not a property of general kets -- it is only a property of those kets describing the time evolution of a quantum system! (And possibly a few other kets that satisfy it by chance or by construction)
     
  6. Jan 4, 2009 #5

    jambaugh

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    At first glance I would say the step:
    is not correct. You shouldn't assume [itex]|x\rangle[/itex] satisfies the Schrodinger equation.


    Also [itex]|x\rangle[/itex] is time independent so its time derivative should be zero.

    This with your derivation of the r.h.s. of TDSE should give you the correct wave-function equation.
     
  7. Jan 4, 2009 #6

    tpg

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    Ah - I see. Of course, a state of well defined position will not remain that way, hence the Schrödinger equation does not apply. Thank you both very much for clearing that up.

    In that case, does anyone know how to derive one form of the Schrödinger equation from the other?
     
  8. Jan 5, 2009 #7
    Hi tpg

    It is usual to write psi(x)=<x|psi>, where bra <x| is the dual vector of ket |x>. In this expression, |x> means the "freezed" state of the traveling particle |psi> when it is at position x. So |x> is a fixed vector without time-dependence. Therefore you can retrieve one form from the other by factorizing <x|.

    Btw, if we introduce an integral in expression: [tex]\left|\psi\right\rangle =\int dx\,\psi(x)\left|x\right\rangle[/tex], we should do it on both sides. I would say that the correct form should be: [tex]\int dx\,\left|\psi\right\rangle =\int dx\,\psi(x)\left|x\right\rangle[/tex]
     
  9. Jan 5, 2009 #8

    jambaugh

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    Think that through.
     
  10. Jan 6, 2009 #9
    Hello jambaugh,

    You're right, what I wrote in that part was nonsense. [tex]\left|\psi\right\rangle =\int dx\,\psi(x)\left|x\right\rangle[/tex] is a good expression. Or otherwise stated: [tex]\left|\psi\right\rangle =\int dx\,\left|x\right\rangle\langle x\left|\psi\rangle[/tex]

    Best regards,
    Arjen
     
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