Quantum Wave Function to Momentum Amplitude

In summary, the student is trying to solve for the momentum between a small momentum range and is stuck on a nasty integral. He plans on using an equation from his textbook to find the value of the uncertainty for the momentum.
  • #1
deedsy
81
0

Homework Statement


I was hoping someone could verify that I've set up the integral correctly for this problem..

Suppose that at t=0, the wavefunction of a free particle is
[tex]ψ(x,0) = \sqrt{b} e^{-|x|b + ip_0 x/\hbar} [/tex]

a) what is the momentum amplitude for this wave function?

Homework Equations


*see below

The Attempt at a Solution



So, I think I know how to complete this, and just want to make sure I'm setting up the integral correctly before I dive into solving it.

a) I will use [tex]\phi(p,0)= \frac{1}{\sqrt{2\pi\hbar}}\int_∞^∞ ψ(x,0) e^{-ipx/\hbar}\,dx [/tex] ---> that's minus ∞ to ∞

so, for my wavefunction...
[tex]\phi(p,0)= \frac{1}{\sqrt{2\pi\hbar}}\int_\infty^\infty \sqrt{b} e^{-|x|b + ip_0 x/\hbar} e^{-ipx/\hbar}\,dx [/tex] ---> again, -∞ to ∞

Here, it looks like i can't cancel [itex]e^{ip_0 x/\hbar}[/itex] and [itex]e^{-ipx/\hbar} [/itex], and I'm stuck with what looks like a very nasty integral...

[tex]\phi(p,0)= \frac{1}{\sqrt{2\pi\hbar}}\int_\infty^\infty \sqrt{b} e^{-|x|b} e^{ix/\hbar (p_0-p)}\,dx [/tex] ---> again, -∞ to ∞
 
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  • #2
Often the thing to do when you have an absolute value is to break whatever it is into piece-wise segments. Will that work here?
 
  • #3
MisterX said:
Often the thing to do when you have an absolute value is to break whatever it is into piece-wise segments. Will that work here?

Yeah, that's what I was planning on doing - dividing it into two integrals, one from 0 to infinity (|x|=x) and one from -infinity to 0 (|x|=-x).
 
  • #4
Nasty? It's essentially integrating ##e^{-kx}## where ##k## is a constant.
 
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  • #5
ok, so for the integral I got..

[tex]\phi(p,0)= \frac{\sqrt{b}}{\sqrt{2\pi\hbar}} \frac{2b}{b^2 + (p_0-p)^2/\hbar^2}[/tex]

Now, since my original wavefunction was normalized, is this [itex]\phi(p,0) [/itex] already normalized?

The next part had me find the probability for finding the momentum between p and p+dp, so I just found what [itex]|\phi(p,0)|^2[/itex] equaled.
After some simplification, this equaled [itex]\frac{2 b^3 \hbar^3}{\pi (b^2 h^2+p_0^2-2 p_0 p +p^2)^2}[/itex] ----> so this is the probability that I will find a momentum between a small momentum range.

If all looks good up until this stage, I'll get started on the next parts having me find the values of the uncertainties Δx and Δp. Right now, I'm trying to find a way to do it with the things I've already solved for
 
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  • #6
ok, so I plan on using this equation to find Δp at t=0

[tex]<(p-p_0)^2> = \int_∞^∞ (p-p_0)^2 |\phi(p,0)|^2 dp [/tex]

Then, [itex] Δp = \sqrt{<(p-p_0)^2>} [/itex]
 
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  • #7
Then for Δx at t=0, I could do

[tex]<(x-x_0)^2> = \int_∞^∞ (x-x_0)^2 |\psi(x,0)|^2 dx [/tex]

Then, [itex] Δx = \sqrt{<(x-x_0)^2>} [/itex]

I don't see that particular equation in my book, but it's just the root mean square, so it should work for finding Δx?

Now, when I multiply these together, I should get [itex]\Delta x \Delta p >= \frac{\hbar}{2}[/itex]
 
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  • #8
deedsy said:
ok, so for the integral I got..

[tex]\phi(p,0)= \frac{\sqrt{b}}{\sqrt{2\pi\hbar}} \frac{2b}{b^2 + (p_0-p)^2/\hbar^2}[/tex]

Now, since my original wavefunction was normalized, is this [itex]\phi(p,0) [/itex] already normalized?
Yes.

The next part had me find the probability for finding the momentum between p and p+dp, so I just found what [itex]|\phi(p,0)|^2[/itex] equaled.
After some simplification, this equaled [itex]\frac{2 b^3 \hbar^3}{\pi (b^2 h^2+p_0^2-2 p_0 p +p^2)^2}[/itex] ----> so this is the probability that I will find a momentum between a small momentum range.
Actually, the probability to find the particle with momentum between p and p+dp is ##\lvert \phi(p,0) \rvert^2 dp##. The amplitude squared alone isn't the probability.

deedsy said:
ok, so I plan on using this equation to find Δp at t=0

[tex]<(p-p_0)^2> = \int_∞^∞ (p-p_0)^2 |\phi(p,0)|^2 dp [/tex]

Then, [itex] Δp = \sqrt{<(p-p_0)^2>} [/itex]

deedsy said:
Then for Δx at t=0, I could do

[tex]<(x-x_0)^2> = \int_∞^∞ (x-x_0)^2 |\psi(x,0)|^2 dp [/tex]

Then, [itex] Δx = \sqrt{<(x-x_0)^2>} [/itex]
You might find it a little less work to use ##\Delta p^2## = ##\langle p^2 \rangle - \langle p \rangle^2## and ##\Delta x^2## = ##\langle x^2 \rangle - \langle x \rangle^2##.
 

What is a quantum wave function?

A quantum wave function is a mathematical representation of the state of a quantum system. It describes the probability of finding a particle in a certain position or with a certain momentum.

What is momentum amplitude?

Momentum amplitude is a measure of the amplitude of a quantum wave function at a certain momentum value. It is related to the probability of finding a particle with a certain momentum in a particular region.

How is quantum wave function related to momentum amplitude?

The quantum wave function and momentum amplitude are related through the mathematical process of Fourier transform. Momentum amplitude is the Fourier transform of the quantum wave function, allowing for the representation of a particle's momentum in terms of its position.

What is the significance of quantum wave function to momentum amplitude?

The relationship between quantum wave function and momentum amplitude is significant because it allows for the prediction and understanding of a particle's behavior in terms of both its position and momentum. It also forms the basis of many quantum mechanical calculations and theories.

How is the quantum wave function to momentum amplitude used in experiments?

In experiments, the quantum wave function to momentum amplitude relationship is used to determine the probability of a particle being detected at a certain momentum value. This information is then compared to the actual results of the experiment to validate the predictions of quantum mechanics.

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