Normalizing a Wavefunction of a harmonic oscillator

Bready
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1. At a certain time the wavefunction of a one-dimensional harmonic oscillator is

[tex]\psi[/tex](x) = 3[tex]\phi[/tex]0(x) + 4[tex]\phi[/tex]1(x)

where [tex]\phi[/tex]0(x) and [tex]\phi[/tex]1(x) are normalized energy eigenfunctions of the ground and first excited states respectively. Normalize the wavefunction and determine the probability of finding the oscillator in the ground state.




3. I'm not really sure if I'm normalizing the wavefunction correctly, I get the normalizing constant as 1/7. However, when I calculate the probability of the ground state and first state combined they don't equal one. Aren't they supposed to and have I normalized correctly?
 
Bready said:
I'm not really sure if I'm normalizing the wavefunction correctly, I get the normalizing constant as 1/7.
Redo this. What's the definition of normalization? (What must equal 1?)
 
I've got the equation for normalising a wavefunction; one over the square root of the wavefunction squared.


= ([tex]\int[/tex][3[tex]\phi[/tex]0(x) + 4[tex]\phi[/tex]1(x)]2)1/2 =

I still get [tex]\sqrt{1/49}[/tex] when I should be getting [tex]\sqrt{1/25}[/tex]; I think I'm squaring the function wrong. The coefficients I get upon squaring are (9 + 12 + 12 + 16), I'm guessing the 12s cancel each other out to leave 9+16=25 but I don't see how. Do I need to know the energy eigenfunctions to calculate this? I think I'm getting myself confused, can you point me in the right direction of squaring the wavefunction? It's just all previous examples I have give the wavefunction experssed as a single function rather than an addition of two. Thanks very much.
 
show your intermediate steps.

I think you are forgetting that phi_0 and phi_2 are ortogonal to each other..
 
Bready said:
I'm guessing the 12s cancel each other out to leave 9+16=25 but I don't see how. Do I need to know the energy eigenfunctions to calculate this?
All you need to know is:
malawi_glenn said:
I think you are forgetting that phi_0 and phi_2 are ortogonal to each other..
That's the secret. :wink:

Show your steps, keeping the eigenfunctions in your expansion.
 
psi(x) = 3phi_0(x) + 4phi_1(x)

=> [psi(x)]^2 = [3phi_0(x) + 4phi_1(x)]^2

=> [9(phi_0(x))^2 + 12phi*_0(x)phi_1(x) + 12phi_0(x)phi*_1(x) + 16(phi_1(x))^2

Is this right?

When I integrate with limits +-infinity does it give

9+0+0+16?
 
Good! Now you've got it.
 
Thanks very much! :D
 

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