Wavefunction Normalization at Different Times

[user3]

In Introduction to Quantum Mechanics by Griffith, when he is normalizing a wave function that's dependent on both x and t, he let's t=0 , and solves for the constant (A). But if the integration of ψ^2 at any time t is 1, then is it correct to let t = 2, for instance, instead of 0 and solve for A? If yes, that would mean that A could be an infinite number of different A's, and that would be wrong because we would get different values for expectation values of position for every constant A. And if No, explain why please.

 

[stevendaryl]

In Introduction to Quantum Mechanics by Griffith, when he is normalizing a wave function that's dependent on both x and t, he let's t=0 , and solves for the constant (A). But if the integration of ψ^2 at any time t is 1, then is it correct to let t = 2, for instance, instead of 0 and solve for A? If yes, that would mean that A could be an infinite number of different A's, and that would be wrong because we would get different values for expectation values of position for every constant A. And if No, explain why please.

Since the normalization is a constant over time, it doesn't matter which moment you choose to normalize it -- you'll get the same normalization constant.

 

[user3]

If you normalize this wave function at t=0 and at t=2, you don't get the same constant:
ψ=Ae^[ -a[(mx^2)/hbar + i t] ] where m, a ,and A are positive real constants.


at t = 0 , you get A=(2am/hbar pi) ^1/4 (solution)

at t=2 , I got A = e^4ai * (2am/hbar pi) ^1/4

if i hadn't done a stupid mistake in the calculations.

 

[stevendaryl]

If you normalize this wave function at t=0 and at t=2, you don't get the same constant:
ψ=Ae^[ -a[(mx^2)/hbar + i t] ] where m, a ,and A are positive real constants.


at t = 0 , you get A=(2am/hbar pi) ^1/4 (solution)

at t=2 , I got A = e^4ai * (2am/hbar pi) ^1/4

if i hadn't done a stupid mistake in the calculations.

The only thing that matters for normalization constant [itex]A[/itex] is the absolute value, [itex]|A|[/itex]. The phase can be chosen arbitrarily.

 

[jtbell]

at t = 0 , you get A=(2am/hbar pi) ^1/4 (solution)

This is correct.

at t=2 , I got A = e^4ai * (2am/hbar pi) ^1/4

This is incorrect. Hint: write your wave function in the form
$$\Psi(x,t) = Ae^{-amx^2/\hbar}e^{-iat}$$
which might make things slightly more obvious.

When you calculate ##|\Psi|^2 = \Psi^*\Psi##, what happens to the factor that contains t? Remember that ##\Psi^*## means complex conjugate of ##\Psi##.

 

[user3]

but what happens if the "i" was not there: ψ=Ae^[ -a[(mx^2)/hbar + t] ] ?

 

[Dr Transport]

but what happens if the "i" was not there: ψ=Ae^[ -a[(mx^2)/hbar + t] ] ?

That is not an eigen-function of the Schrodinger equation...

 

[Vanadium 50]

Then it's a mistake. (It's not a phase)

 

[atyy]

To choose the normalization, fix the time (any time will do). Even at any fixed time, there are an infinite number of possible normalizations, just different by ##\exp(i\theta)##, where ##\theta## is a constant. These different normalizations are physically equivalent, just like in electrostatics the zero of potential is arbitrary. So just pick one.

Schroedinger's equation will take care that the wave function at other times remains correct.