Energy-Time Uncertainty of Gaussian

In summary, the homework statement says that the energy-time uncertainty principle states that the energy or momentum of a particle cannot be known with absolute certainty at any given time.
  • #1
sspitz
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0

Homework Statement


Check the energy-time uncertainty principle for:
[tex]\Psi(x,0)=Ae^{-ax^2}e^{ilx}[/tex]
using the observable x.

Homework Equations


[tex]
\Delta{E}\Delta{t}\geq\hbar/2\\
\Delta{E}=\sigma_H=\sqrt{<H^2>-<H>^2}\\
\Delta{t}=\frac{\sigma_x}{\frac{d<x>}{dt}}
[/tex]


The Attempt at a Solution


First I find the wavefunction with time dependence
[tex]
\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int{\phi(k)e^{ikx-i\frac{\hbar k^{2}t}{2m}}dk}\\
\phi(k)=\frac{1}{\sqrt{2\pi}}\int{\Psi(x,0)e^{-ikx}dx}
[/tex]
This gives me a fairly complicated expression
[tex]
\Psi(x,t)=\frac{Aexp(\frac{-ax^{2}+lix-\frac{tl^{2}i\hbar}{2m}}{w^{2}})}{w}\\
A^{2}=\sqrt{\frac{2a}{\pi}},w^{2}=1+\frac{\hbar2ita}{m}
[/tex]
Just finding <H> (not to mention <H^2> is giving me a lot of trouble. Is grinding out the derivatives and integrals the only way to do this problem?

The only other options I can think of:

1: <H> and <H^2> should not change in time (?), so I can just use the wavefunction at t=0 to find them? (But, still have to use the wavefunction with time to get the std dev of x)

2: Considering [tex] \frac{e^{ikx}}{\sqrt{2\pi}} [/tex] are like orthonormal eigenfunctions for H, could I say:

[tex]
<H>=\int{\phi(k)^{2}E(k)dk}
[/tex]
 
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  • #2
remember that

[itex] \Psi (x, t) = e^{(-i/ \hbar) H t} \Psi (x, 0) [/itex]

H commutes past the time evolution operator

x does not, use a series expansion to determine the correct commutation relation

Furthermore complete the square in the gaussian
 
  • #3
You are definitely on the right track. Your expression is going to be ugly. Now, set
[tex] z\equiv 2 \hbar at/m[/tex]
and use
[tex] \lvert \psi \lvert^2= \sqrt{ \frac{2a}{ \pi}} \frac{1}{ \sqrt{1+ z^2}} e^{-l^2/2a}e^{a\big\{ \frac{(ix+l/2a)^2}{(l+iz)}+ \frac{(-ix +l/2a)^2}{(1-iz)}\big\}}[/tex]
Expand the term in the brackets {} of the right-hand exponent, and plug it back into
[tex] \lvert \psi(x,t) \lvert^2[/tex]
where your psi should be
[tex] \psi(x,t)= \bigg( \frac{2a}{ \pi}\bigg)^{1/4} \frac{1}{ \sqrt{1+2i \hbar at/m}} e^{-l^2/4a}e^{a(ix+l/2a)^2/(1+2ia \hbar t/m)} [/tex]
 
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  • #4
Thanks for the help. I think I got it.

As a check on comprehension, I'd be interested in knowing if #2 in the original post is right or wrong and why. It seems to me that phi(k) above is for energy eigenfunctions what the momentum space wavefunction is for momentum eigenfunctions. That is, roughly, both show the amount of each eigenfunction in psi.
 
  • #5
Code:
Sorry, I misunderstood your last question, so I'm editing my post. I'm going to have to come back to your #2 point later, since it's getting late where I am. The following was written when I misunderstood your question, so I'm just going to post it rather than waste it. If we're being honest, this problem is designed just to make you do some math. It doesn't really help you with any physical intuition. By convention, the time evolution operator is
[tex] \hat U (t,t_0) = e^{-i(t-t_0) \, \hat H/ \hbar}[/tex]
or in application,
[tex] \big\lvert \, \psi(t) \big>= e^{-i(t-t_0) \, \hat H/ \hbar} \big\lvert \, \psi(t_0) \big> [/tex]
So, for example, if you've got an infinite square well, at t_0 =0, in its ground state,
[tex] \big\lvert \, \psi_{n=1} (t=0) \big>= \big\lvert \, \psi_1 (0) \big>= \sqrt{ \frac{2}{a}} \sin\frac{ \pi x}{a} [/tex]
You know the energy is
[tex]E_1= \frac{ \hbar^2 \pi^2}{2ma^2} [/tex]
So at a future time t,
[tex] \big\lvert \, \psi_1 (t) \big>= e^{-it \, \hat H/ \hbar} \bigg(\sqrt{ \frac{2}{a}} \sin\frac{ \pi x}{a}\bigg) = \bigg(\sqrt{ \frac{2}{a}} \sin\frac{ \pi x}{a}\bigg)e^{-it \, E_1/ \hbar} =\sqrt{ \frac{2}{a}} \sin\frac{ \pi x}{a}e^{-it \, \hbar \pi^2/2m a^2}[/tex]
Time evolution is really pretty simple. It gets convoluted with Green's functions and stuff like that, but really for all intents and purposes, you should just think of it as in the example above.
 
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FAQ: Energy-Time Uncertainty of Gaussian

1. What is the Energy-Time Uncertainty of Gaussian?

The Energy-Time Uncertainty of Gaussian is a mathematical concept that describes the relationship between the uncertainty in a particle's energy and the uncertainty in the time at which the particle is observed.

2. How is the Energy-Time Uncertainty of Gaussian calculated?

The Energy-Time Uncertainty of Gaussian is calculated using the Gaussian wave function, which is a probability distribution function that describes the position and momentum of a particle. The product of the standard deviations of the position and momentum in the Gaussian wave function gives the Energy-Time Uncertainty of Gaussian.

3. What is the significance of the Energy-Time Uncertainty of Gaussian?

The Energy-Time Uncertainty of Gaussian is significant because it is a fundamental property of quantum mechanics that limits our ability to know both the energy and time of a particle with absolute certainty. It also plays a crucial role in understanding the behavior of quantum systems.

4. How does the Energy-Time Uncertainty of Gaussian affect measurements in experiments?

The Energy-Time Uncertainty of Gaussian means that there is always some inherent uncertainty in the measurements of a particle's energy and time. This can lead to errors and limitations in experimental measurements, particularly in systems where the energy and time of particles are closely linked.

5. Can the Energy-Time Uncertainty of Gaussian be reduced?

No, the Energy-Time Uncertainty of Gaussian is a fundamental property of quantum mechanics and cannot be reduced. However, it can be minimized by reducing the uncertainties in either the energy or time measurement, or by using specialized techniques such as quantum entanglement.

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