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How do we know a state function will stay normalized?

  1. Oct 1, 2006 #1
    According to Shrodinger Equation, The state function must change with time, soI wonder how do we know that after we normalize this state function, it will stay normalized ?

    Thanks
     
  2. jcsd
  3. Oct 1, 2006 #2
    I don't remember the actual proof, but the notion that the wave function must stay normalized is fairly logical.

    Think of it this way... in a non relativistic process matter must be conserved... The fact that the wave function is normalized means that there is a probability to observe the particle somewhere in the universe. Given that it's here one minute (ie in our universe) we would expect it to remain within our observable universe in the next.

    All it's saying is that particles don't suddenly disappear.

    I believe griffith's intro to QM has the actual proof.
     
  4. Oct 1, 2006 #3
    Schrodinger's equation can be expressed as an unitary transform from a state a t=0 to a state t=t' and unitary transforms leave the inner product of wavefunctions unchanged. So if the initial wave function was normalized then evolution under the Schrodinger equation will lead to another wavefunction which is also normalized.

    This is not true for the process of measurement since it cannot be represented by a unitary transform.
     
  5. Oct 1, 2006 #4
    Consider [tex]\frac{d}{dt}\left(\psi\psi^{\ast}\right)[/tex] using Schrodinger's Equation and if memory serves it drops out fairly easily.
     
  6. Oct 1, 2006 #5

    quasar987

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    This is just a copy/paste of post I made in another thread..

    When solving the SE by the method of separation of variables, we find that the time dependant part of the solution is [itex]\exp{iEt/\hbar}[/itex], and the position dependant part satisfies the time-independant SE. Denote [itex]\psi(x)[/itex] the solution to the time independant SE for a given potential. Then the general solution to the SE is [itex]\Psi(x,t)=\psi(x)e^{iEt/\hbar}[/itex], and according to the Born interpretation, [itex]\Psi \Psi^*[/itex] is a probability density function for the position of the particle. But [itex]\Psi \Psi^* = \psi\psi^*[/itex]. I.e. the probability density is is time independant!

    From there, showing that the normalisation constant is time-independant is just one step away.
     
  7. Oct 2, 2006 #6

    dextercioby

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    [tex] \frac{d}{dt}\left(\langle \psi(t),\psi(t)\rangle\right) =0 \Leftrightarrow \frac{d\psi(t)}{dt}=\frac{1}{i\hbar}\hat{H}\psi(t) \ ,\ \forall \psi(t)\in D(\hat{H}) [/tex]

    Daniel.
     
  8. Oct 2, 2006 #7
    The normalisation remains because of the unitarity of the evolution operator.
    The operator U = exp(H t /i hbar) preserves the normalisation because H is hermitian.
    This is easy to check by developping
    <phi(t)|phi(t)> = <phi(0)|Ut U|phi(o)> ​
    for small time steps.
    Therefore is applies also for time-dependent hamiltonians.

    Michel
     
    Last edited: Oct 2, 2006
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