Quantum Mechanics (position uncertainty)

In summary: It might look messy but it should work out.Here is a sample of a summary:In summary, the conversation discusses a problem involving finding the expectation value of x for a particle in state Ψ. The equation for this expectation value is given, and the wave function in ground state is also provided. The person attempting the problem uses a substitution and evaluates the integral, but encounters an issue with the limits resulting in an answer of infinity. Another integral for <x2> is also discussed, but the person is unsure if they are on the correct path. The expert reassures that they are and suggests evaluating the integral to get the correct answer.
  • #1
roam
1,271
12

Homework Statement



I need some help with the following problem:

http://img827.imageshack.us/img827/4061/prob1y.jpg

Homework Equations



For some particle in state Ψ the expectation value of x is given by:

[itex]\left\langle x \right\rangle = \int^{+\infty}_{-\infty} x |\Psi(x, t)|^2 \ dx[/itex]

The wave function in ground state:

[itex]\psi_0 (x)= A e^{\frac{-m \omega}{\hbar}x^2}[/itex]

The Attempt at a Solution



Let α = mω/2ħ, so

[itex]\left\langle x \right\rangle = \int^{+\infty}_{-\infty} x A^2 (e^{-\alpha x^2})^2 \ dx[/itex]

[itex]= A^2 \int^{+\infty}_{-\infty} x (e^{-2 \alpha x^2}) \ dx[/itex]

[itex]=A^2 \left[ \frac{-e^{-2 \alpha x^2}}{4 \alpha} \right]^{+\infty}_{-\infty} \ dx[/itex]

But since e=∞ and e-∞=0 (i.e. the limits as x approaches ±∞) we get:

<x>=A2 ∞ = ∞

So, what have I done wrong here? :confused:

Also for <x2> we have the following

[itex]\left\langle x^2 \right\rangle = \int^{+\infty}_{-\infty} x^2 A^2 (e^{-\alpha x^2})^2 \ dx[/itex]

But similarly here I will encounter the same problem. So how can I evaluate the <x> and <x2> without getting infinity?

Clearly they can't equal to infinity since we must obtain:

[itex]\Delta x = \sqrt{\left\langle x^2 \right\rangle -\left\langle x \right\rangle^2} = \sqrt{\frac{\hbar}{2m \omega}}[/itex]

Any help is greatly appreciated.
 
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  • #2
Check the evaluation of your upper and lower limits on your integrals again. ;)
 
  • #3
klawlor419 said:
Check the evaluation of your upper and lower limits on your integrals again. ;)

The upper limit is +∞ and the lowe limit is -∞. Here's what I did:

[itex]A^2 \left( \frac{e^{-2 \alpha \infty^2}}{4 \alpha} - \frac{e^{2 \alpha \infty^2}}{4 \alpha} \right)[/itex]

[itex]= A^2 \left( \frac{e^{-\infty^2}}{4 \alpha} - \frac{e^{\infty^2}}{4 \alpha} \right)[/itex]

[itex]= A^2 (0- 0) = 0[/itex]

Edit: this time I get 0. So <x>=0?
 
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  • #4
Yes! Not infinity as you were getting before. It makes sense to, a simple ground state harmonic oscillator spends most of the time close to origin.
 
  • #5
Okay, but what about <x2>? The integral [itex]\int^{+\infty}_{-\infty} x^2 A^2 (e^{-\alpha x^2})^2 \ dx[/itex] turns out to be extremely messy. I'm just wondering if I am on the correct path? :confused:

If I integrate that I get this:

[itex]\left[ \frac{\sqrt{\pi/2} \ erf(\sqrt{2} \sqrt{\alpha} x)}{8 \alpha^{3/2}} - \frac{x e^{-2\alpha x^2}}{4 \alpha} \right]^{+\infty}_{-\infty}[/itex]

But I must get <x2>=ħ/2mω so that the position uncertainty would be the square root of that.
 
  • #6
You are the right path, if you evaluate the integrals you should get the right value.
 

FAQ: Quantum Mechanics (position uncertainty)

What is the uncertainty principle in quantum mechanics?

The uncertainty principle in quantum mechanics is a fundamental principle that states that it is impossible to know the exact position and momentum of a particle at the same time. This means that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa.

How does the uncertainty principle relate to position uncertainty in quantum mechanics?

The uncertainty principle is closely related to position uncertainty in quantum mechanics. This is because the position of a particle is one of the properties that cannot be known with absolute certainty due to the uncertainty principle. This means that the more precisely we know the position of a particle, the less certain we are about its momentum and vice versa.

How is position uncertainty calculated in quantum mechanics?

In quantum mechanics, position uncertainty is calculated using the Heisenberg uncertainty principle. This principle states that the product of the uncertainties in position and momentum must be greater than or equal to a constant value (h/4π). The exact calculation of position uncertainty involves mathematical operations on the wave function of a particle.

What are the implications of position uncertainty in quantum mechanics?

The implications of position uncertainty in quantum mechanics are significant. This principle highlights the inherently probabilistic nature of quantum mechanics, where we can only predict the likelihood of finding a particle in a certain position and not its exact location. It also has practical applications, such as in the design of electronic devices and understanding the behavior of subatomic particles.

Can position uncertainty be overcome in quantum mechanics?

No, position uncertainty cannot be overcome in quantum mechanics. It is a fundamental principle that arises from the nature of particles at the quantum level. However, scientists have developed techniques such as quantum entanglement and quantum teleportation to overcome some of the limitations posed by the uncertainty principle in certain scenarios.

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