Normalisation of Schrodinger Eq.

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Homework Help Overview

The discussion revolves around the normalization of a wave function in the context of the Schrödinger equation. The original poster seeks to understand how normalization is preserved over time after initially normalizing the wave function at t = 0.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the Schrödinger equation on the time evolution of wave functions and question the methods for demonstrating the preservation of normalization. There are discussions about using derivatives and bra-ket notation, as well as the potential use of specific wave function forms.

Discussion Status

Several participants have provided insights into the mathematical relationships involved, including the use of derivatives and the properties of the Hamiltonian operator. There is a recognition that while some progress has been made, further work is needed to generalize the proof beyond stationary states.

Contextual Notes

Participants note the importance of distinguishing between stationary states and more general solutions to the Schrödinger equation. There is also mention of the properties of the Hamiltonian operator and its implications for the proof of normalization preservation.

Shredface
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Homework Statement



Suppose you assume that you have normalised a wave function at t = 0. How do you know that it will stay normalised as time goes on? Show explicitly that the Schrödinger equation has the property that it preserves normalistion over time.

Homework Equations



From my notes I have deduced:

[tex]\frac{d}{dt} \int_{-\infty}^{+\infty} \left| \Psi\left(x,t\right)\right|^2 dx = \int_{-\infty}^{+\infty} \frac{\partial}{\partial t} \left| \Psi\left(x,t\right)\right|^2 dx = 0[/tex]

However I have failed to come up with a solution.

EDIT: Should I rearrange the Schrödinger equation for d/dt of psi then solve that way? I'm confused.
 
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Shredface said:
[tex]\frac{\partial}{\partial t} \left| \Psi\left(x,t\right)\right|^2[/tex]
You know how to compute derivatives, right?

(bra-ket notation might be easier, though... no sense in using integrals when you can just do algebra!)
 
The schrödinger equation

[tex]i\hbar \frac{\partial}{\partial t} \Psi (x,t) = E\Psi (x,t)[/tex]

suggest that you can write the time-evolution of a state according to:

[tex]\Psi(x,t) = \psi(x) e^{-i{E t\, \hbar}}[/tex]

Now, what will this give you?

(this is the way I was taught how to show that [itex] \int_{-\infty}^{+\infty} \left| \Psi\left(x,t\right)\right|^2 dx =[/itex] is constant over time. )
 
So I get: [tex]\frac{\partial}{\partial t} \left|\Psi\right|^2 = \Psi \frac{\partial \Psi^*}{\partial t} + \Psi^* \frac{\partial \Psi}{\partial t}[/tex]
What should I use for my wave function? [tex]\Psi\left(x,t\right) = Ae^{i(kx-wt)/\hbar}[/tex] ?
 
Shredface said:
What should I use for my wave function?
What's wrong with just calling it [itex]\Psi[/itex], and working with it symbolically?

Although... you could prove the theorem first for stationary states, as malawi_glenn is implicitly suggesting...


Shredface said:
[tex]\frac{\partial \Psi}{\partial t}[/tex]
Don't you know something about that? ...
 
[tex] \frac{\partial \Psi}{\partial t}[/tex]

hint: The Schrödinger equation.

And for [tex] \frac{\partial \Psi^*}{\partial t}[/tex]

The complex conjugate of the schrödinger equation.
 
How's this...

[tex]E \Psi(x,t) = i\hbar \frac{\partial}{\partial t} \Psi(x,t), \Rightarrow E \Psi^*(x,t) = -i\hbar \frac{\partial}{\partial t} \Psi^*(x,t)[/tex]

Giving:

[tex]\frac{\partial \Psi}{\partial t} = -\frac{iE}{\hbar} \Psi(x,t)[/tex] and [tex]\frac{\partial \Psi^*}{\partial t} = \frac{iE}{\hbar} \Psi^*(x,t)[/tex]

So:

[tex]\frac{\partial}{\partial t} |\Psi|^2 = |\Psi|^2 (\frac{iE}{\hbar} - \frac{iE}{\hbar}) = 0[/tex]

Is this making sense?
 
yes, very good! Recall that this is possible since E is real.
 
  • #10
So the integral of zero is zero thus proving the initial statement. Yay!

Cheers for the help guys, all seems so easy now!
 
  • #11
Shredface said:
So the integral of zero is zero thus proving the initial statement. Yay!
But only in the case of stationary states! You still have to do some more work for an arbitrary solution to the Schrödinger equation.

Fortunately, it's not much work, since the general case features the unitary operator H rather than the real number E. Fortunately, unitary operators share enough of the properties of real numbers that the proof works out pretty much unchanged.



Incidentally, I think the main lesson here is when you don't know what you're supposed to do, instead try what you know how to do. You know how to take derivatives, you know to use the Schrödinger equation to compute time derivatives of quantum states -- you just had to try and string them along together. A great many problems can be solved this way: just do what you can and eventually you get to a point where you can see how to connect what you can do to what you're trying to do, and then you've done it!

(Special cases can help too -- e.g. trying stationary states first because it's simpler)
 
  • #12
Is H unitary? I thought one use the fact that since [tex]H^{\dagger}=H[/tex], i.e hermitian, the energy eigenvaules E are real.

But Hurkyl has correct that using that E is real still only proves the stationary case.

What the real proof is about is that [tex]H^{\dagger}=H[/tex], the Schrödinger equation is
[tex] H\Psi(x,t) = i\hbar \frac{\partial}{\partial t} \Psi(x,t)[/tex]

And the complex conjugated will use [tex]H^{\dagger}[/tex]

I agree on Hurkyl's advice, don't think TOO much, use what you know and see if it works.
 
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  • #13
malawi_glenn said:
Is H unitary? I thought one use the fact that since [tex]H^{\dagger}=H[/tex], i.e hermitian, the energy eigenvaules E are real.
Bleh, that's what I meant. Thanks for the correction!
 

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