Solving a tricky integral to normalize a wave function

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Homework Statement


A particle of mass m is moving in one dimension in a potential V(x,t). The wave
function for the particle is: ψ = Axe^([-sqrt(km)/2h_bar]*x^2)e^([-isqrt(k/m)]*3t/2). For -infinitity < x < infinity, where k and A are constants. Normalize this wave function.


Homework Equations


Normalization of wave function in one dimension:
∫ψ*ψdx = 1

The Attempt at a Solution


So I said that ψ* would be the same as ψ except i would be negative, so ψ*ψ would be A^2*x^2*e^([-sqrt(km)/h_bar]*x^2). To normalize, ∫ψ*ψdx = A^2*∫x^2*e^([-sqrt(km)/h_bar]*x^2)dx = 1, with the integration being from negative infinity. I am mostly just having trouble getting through this integral. I tried integration by parts:

u = x^2, du = 2xdx
v = [-h_bar/sqrt(km)] * [1/2x] * e^([-sqrt(km)/h_bar]*x^2), dv = e^-([sqrt(km/h_bar]*x^2) dx.

So ∫ψ*ψdx = A^2 [uv-∫vdu] = A^2 [[x^2*[-h_bar/sqrt(km)] * [1/2x] * e^([-sqrt(km)/h_bar]*x^2)] (evaluated from negative infinity to positive infinity) - ∫[-h_bar/sqrt(km)] * e^([-sqrt(km)/h_bar]*x^2) dx, again from negative infinity to positive infinity.

So for the first term, I don't know how to evaluate that from negative infinity to infinity because there is a 2x in the denominator and an e^(x^2) in the numerator, and the only way I really know how to evaluate is by just plugging values in. I also don't see an obvious way to evaluate that integral. So is there a way to evaluate this or is integration by parts just not going to work for me? Once I find this integral I can just take the square root of its reciprocal to get the value of A.

This is my first post so I have read the rules and am hoping I have posted in the right forum, etc. Any help on this problem would be very much appreciated.
 

Answers and Replies

  • #2
Redbelly98
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Welcome to PF!

Homework Statement


A particle of mass m is moving in one dimension in a potential V(x,t). The wave
function for the particle is: ψ = Axe^([-sqrt(km)/2h_bar]*x^2)e^([-isqrt(k/m)]*3t/2). For -infinitity < x < infinity, where k and A are constants. Normalize this wave function.


Homework Equations


Normalization of wave function in one dimension:
∫ψ*ψdx = 1

The Attempt at a Solution


So I said that ψ* would be the same as ψ except i would be negative, so ψ*ψ would be A^2*x^2*e^([-sqrt(km)/h_bar]*x^2). To normalize, ∫ψ*ψdx = A^2*∫x^2*e^([-sqrt(km)/h_bar]*x^2)dx = 1, with the integration being from negative infinity. I am mostly just having trouble getting through this integral. I tried integration by parts:

u = x^2, du = 2xdx
v = [-h_bar/sqrt(km)] * [1/2x] * e^([-sqrt(km)/h_bar]*x^2), dv = e^-([sqrt(km/h_bar]*x^2) dx.

So ∫ψ*ψdx = A^2 [uv-∫vdu] = A^2 [[x^2*[-h_bar/sqrt(km)] * [1/2x] * e^([-sqrt(km)/h_bar]*x^2)] (evaluated from negative infinity to positive infinity) - ∫[-h_bar/sqrt(km)] * e^([-sqrt(km)/h_bar]*x^2) dx, again from negative infinity to positive infinity.

So for the first term, I don't know how to evaluate that from negative infinity to infinity because there is a 2x in the denominator and an e^(x^2) in the numerator, and the only way I really know how to evaluate is by just plugging values in. I also don't see an obvious way to evaluate that integral. So is there a way to evaluate this or is integration by parts just not going to work for me? Once I find this integral I can just take the square root of its reciprocal to get the value of A.

This is my first post so I have read the rules and am hoping I have posted in the right forum, etc. Any help on this problem would be very much appreciated.
You won't be able to solve that integral analytically. But since this is physics (not math), surely you are allowed to look it up? Check in a table of integrals, but be sure to look in the definite integrals section. It will probably look something like
0 x2 e-x2 = ___?​
or
0 x2 e-ax2 = ___?​
 
  • #3
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Thanks for your response! Darn, that is something we have little experience with and my teacher made it sound like we should know how to evaluate this. So since this integral runs from -infinity to infinity, would I double the integral from 0 to infinity?

The fact that this can't be solved analytically actually makes me suspicious, are you sure that I am trying to evaluate the right integral based on the problem?
 
  • #4
Redbelly98
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Thanks for your response! Darn, that is something we have little experience with and my teacher made it sound like we should know how to evaluate this. So since this integral runs from -infinity to infinity, would I double the integral from 0 to infinity?
Yes. When the integrand is an even function, as it is here, you can do that.
The fact that this can't be solved analytically actually makes me suspicious, are you sure that I am trying to evaluate the right integral based on the problem?
Welcome to advanced physics! :biggrin:

Looks right to me. Just to be sure, we have
[tex]\Psi = A \ x \ e^{-\sqrt{km} \ \cdot \ x^2 / (2 \hbar)} \cdot e^{-i \sqrt{k/m} \ \cdot \ 3 t/2},[/tex]
correct?
 
  • #5
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Yes. When the integrand is an even function, as it is here, you can do that.

Welcome to advanced physics! :biggrin:

Looks right to me. Just to be sure, we have
[tex]\Psi = A \ x \ e^{-\sqrt{km} \ \cdot \ x^2 / (2 \hbar)} \cdot e^{-i \sqrt{k/m} \ \cdot \ 3 t/2},[/tex]
correct?

Yep that's right, and I finally got a response from my prof, evidently this is called the gamma function and you do have to look it up. Thanks so much for your help though!
 
  • #7
What was/is the answer to this problem? I have the exact same problem and just can't make it work!
 
  • #8
Ray Vickson
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Homework Statement


A particle of mass m is moving in one dimension in a potential V(x,t). The wave
function for the particle is: ψ = Axe^([-sqrt(km)/2h_bar]*x^2)e^([-isqrt(k/m)]*3t/2). For -infinitity < x < infinity, where k and A are constants. Normalize this wave function.


Homework Equations


Normalization of wave function in one dimension:
∫ψ*ψdx = 1

The Attempt at a Solution


So I said that ψ* would be the same as ψ except i would be negative, so ψ*ψ would be A^2*x^2*e^([-sqrt(km)/h_bar]*x^2). To normalize, ∫ψ*ψdx = A^2*∫x^2*e^([-sqrt(km)/h_bar]*x^2)dx = 1, with the integration being from negative infinity. I am mostly just having trouble getting through this integral. I tried integration by parts:

u = x^2, du = 2xdx
v = [-h_bar/sqrt(km)] * [1/2x] * e^([-sqrt(km)/h_bar]*x^2), dv = e^-([sqrt(km/h_bar]*x^2) dx.

So ∫ψ*ψdx = A^2 [uv-∫vdu] = A^2 [[x^2*[-h_bar/sqrt(km)] * [1/2x] * e^([-sqrt(km)/h_bar]*x^2)] (evaluated from negative infinity to positive infinity) - ∫[-h_bar/sqrt(km)] * e^([-sqrt(km)/h_bar]*x^2) dx, again from negative infinity to positive infinity.

So for the first term, I don't know how to evaluate that from negative infinity to infinity because there is a 2x in the denominator and an e^(x^2) in the numerator, and the only way I really know how to evaluate is by just plugging values in. I also don't see an obvious way to evaluate that integral. So is there a way to evaluate this or is integration by parts just not going to work for me? Once I find this integral I can just take the square root of its reciprocal to get the value of A.

This is my first post so I have read the rules and am hoping I have posted in the right forum, etc. Any help on this problem would be very much appreciated.

You need to evaluate [itex] I = \int_{-\infty}^{\infty} x^2 e^{-a x^2} dx, [/itex] where a > 0. You can integrate by parts, using [itex]u = x, \: dv = x e^{-ax^2} dx = (1/2a) d(ax^2) e^{-ax^2}.[/itex] You will end up having to do an integral of the form [itex] \int_{-\infty}^{\infty} e^{-ax^2} dx, [/itex] which you are certainly supposed to have seen before.

RGV
 
  • #9
Redbelly98
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What was/is the answer to this problem? I have the exact same problem and just can't make it work!
Please show us what you have tried towards solving this. Also, you should familiarize yourself with our forum policy on getting help with homework problems by reading the section titled Homework Help at this link:

https://www.physicsforums.com/showthread.php?t=414380

You need to . . .
Note, the OP seems to be finished with the problem.
 
  • #10
My humble apologies! I will return with explanation at a later point in time.

For now, thank you for your help.
 

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