Energy-Time Uncertainty Problem.

  • Thread starter alexgmcm
  • Start date
  • Tags
    Uncertainty
In summary: Delta f = f \cdot \Delta L + L \cdot \Delta f Since \Delta L = - \Delta f, we can rearrange the equation to get:\Delta v = L \cdot \Delta f - f \cdot \Delta f \Delta v = (L-f) \cdot \Delta f But L-f is just the difference between the two values, which in this case is 0 since the speed of the wave is constant. Therefore:\Delta v = 0 \cdot \Delta f \Delta v = 0 This means that the uncertainty in the velocity is 0, which is what we expect since the speed of the wave is constant. Therefore,
  • #1
alexgmcm
77
0

Homework Statement


Following a question that asks for the Energy-Time uncertainty principle, [tex]\Delta E \Delta T \geq \frac{\hbar}{2}[/tex]

Show that for any wave, the fractional uncertainty in wavelength, [tex]\frac{\Delta \lambda}{\lambda}[/tex] is the same in magnitude as the fractional uncertainty in frequency [tex]\frac{\Delta f}{f}[/tex]

Homework Equations


Energy-Time uncertainty principle as stated above:
[tex]\Delta E \Delta T \geq \frac{\hbar}{2}[/tex]
The relation between Energy and Frequency of a wave:
[tex] E = hf [/tex]

The Attempt at a Solution



Using the above equations we can deduce that:
[tex]\Delta E = h \Delta f [/tex]
but from E = hf we can say that [tex] h = \frac{E}{f}[/tex]
[tex] \therefore \Delta E = \frac{E}{f} \Delta f [/tex]
[tex] \therefore \frac{\Delta E}{E} = \frac{\Delta f}{f} [/tex]

Now I thought I might be able to do a similar thing for wavelength using [tex]E = \frac{hv}{\lambda}[/tex] but this runs into problems stemming from the fact that lambda is the denominator of that equation not the numerator. I have tried several rearrangements and even considered using Position-Momentum Uncertainty with De Broglie, but I still can't get see how I can equate the fractional uncertainties. I'm not even sure if my rearrangement for the frequency is on the right track or not.

Any help would be greatly appreciated.
 
Physics news on Phys.org
  • #2
Did you recall that [itex] c = f \lambda [/itex] and [itex] f = \frac{1}{T} [/itex] ? These might be of use.
 
Last edited:
  • #3
[tex] c= f \lambda [/tex] is only for photons though, whereas it says show that for ANY wave, so I guess that wouldn't be relevant.

I could try using the frequency-period relation but I'm not sure exactly how.
 
  • #4
[itex]v = f\lambda[/itex] applies for any wave, where [itex]v[/itex] is the speed of the wave. The formula for photons is just a special case.
 
  • #5
diazona said:
[itex]v = f\lambda[/itex] applies for any wave, where [itex]v[/itex] is the speed of the wave. The formula for photons is just a special case.

Yeah I mentioned trying to use [tex]E=\frac{hv}{\lambda}[/tex] In my original post, I have been trying it for several hours now and still can't seem to make any progress, my main issue is that whilst something like [tex] \Delta E = h \Delta f [/tex] is true, the similar statement, [tex] \Delta t = \frac{1}{\Delta f} [/tex] is not true. Hmm... will try some more.
 
  • #6
Hmm... I've made a little more progress this time by starting with:
[tex]\Delta t = \frac{\Delta \lambda}{v} = \frac{\Delta \lambda}{f \lambda} = t \frac{\Delta \lambda}{\lambda}[/tex]

Therefore [tex]\frac{\Delta t}{t} = \frac{\Delta \lambda}{\lambda}[/tex]

Which is very similar to my expression [tex]\frac{\Delta E}{E} = \frac{\Delta f}{f}[/tex]

But I can't see how to equate these two in anyway, or any other way of getting the two expressions equated, despite the fact that energy and time are obviously related via the uncertainty principle.

The statement that the proportional uncertainty is the same for both wavelength and frequency seems obvious, but proving it is turning out to be very difficult :(
 
  • #7
Wait, surely one of derivations is wrong, and I have a feeling it is the one in terms of wavelength. The reason is because if I assume my working to be true, then we know that somehow we can equate the proportional uncertainty in frequency and wavelength. But from my workings thus far that would also mean that the proportional uncertainty in energy is equivalent to the proportional uncertainty in time. Which I don't think is necessarily the case, as they are linked only by the uncertainty principle which I don't think implies such a relation.

I should note that although this is posted in the homework section, it is not really homework but a revision problem for exams, and therefore I will gain no marks from its solution nor am I ever likely to be given the solution by my lecturers. So any help on the problem would be greatly appreciated. I chose to post it in the homework section as it is a university question and not a general discussion, but I worry that this was perhaps the wrong thing to do if people are reticent to help me for fear of helping someone to cheat.
 
  • #8
I do not see why is the problem related to the uncertainty principle.

Any kind of wave is a periodic traveling disturbance. Consider plane waves traveling in the x direction. The functional form is f(t/T-x/L) where T is the time period and L is the period in space, the wavelength. The speed of wave is that of a "wave-front" the place where the wave has a specified property, for example a maximum. For a wavefront, t/T-x/L = const. ---> x=L/T*t +constant, the wavefront travels with the speed v= L/T. As T is the reciprocal of frequency, v=L*f. The speed of wave is determined by the kind of wave and the medium it travels, its uncertainty is zero:

[tex]\Delta v = f*\Delta L+L*\Delta f}=0 \rightarrow \frac{\Delta L}{L}=-\frac{\Delta f}{f}[/tex]



ehild
 
  • #9
ehild, I think your method is correct, I didn't realize you could express the uncertainty in the velocity as a sum like that, but that's awesome!

Thank you for your help.
 
  • #10
Hmm, now I realize if I use the partial derivative method for the treatment of errors I arrive at the same answer.
[tex] \Delta v = \sqrt{ ( f \Delta \lambda )^2 + ( \Delta f \lambda)^2} = 0 [/tex]
[tex] \therefore f^2 \Delta \lambda ^2 = - \Delta f ^2 \lambda ^2 [/tex]
[tex] \therefore f \Delta \lambda = - \Delta f \lambda [/tex]
[tex] \therefore \frac{\Delta f}{f} = - \frac{ \Delta \lambda}{\lambda}[/tex]

Hmm... the minus sign seems to mess it up as wouldn't that go to i? But the partial derivative method seems to get very close to the solution you gave and I can see why you would express the uncertainty as the sum of the two products via such a method.
 
  • #11
Any small change of a function f(x,y) can be approximated with the partial derivatives of f, multiplied by the change of x and y, respectively, and summed.


[tex]\Delta f = \partial f/\partial x \cdot \Delta x +\partial f/\partial y \cdot \Delta y [/tex]

You have the function v(L,f) = L*f.

[tex]\Delta v = \partial v/\partial L \cdot \Delta L +\partial v/\partial f \cdot \Delta f=f \Delta L + L \Delta f [/tex]


But v has no uncertainty, it is specified for the kind of wave. [tex]\Delta v =0 \rightarrow f \Delta L = -L \Delta f[/tex]

The same method can be applied for the derivation of an implicit function.

For example, you have a circle x2+y2=R2=25, and you want to find the slope of its tangent at the point (3,4)
Instead of expressing y with x, and taking the derivative with respect to x, you can consider R2 as function of x and y and find the differential of R2 which is obviously zero.
[tex] d(R^2)=2x\cdot dx +2y\cdot dy=0[/tex]
From here, the differential quotient is dy/dx = -x/y, the slope of the tangent is -3/4. It looks very formal, but is is correct and works!

In error calculus, x and y are independent variables and their variances add up to give the variance of a function f(x,y).
In our case, L and f are not independent.

ehild
 

Related to Energy-Time Uncertainty Problem.

What is the energy-time uncertainty problem?

The energy-time uncertainty problem is a principle in quantum mechanics that states that it is impossible to precisely measure the energy and time of a particle at the same time. This is due to the inherent uncertainty in the properties of quantum particles.

Why is the energy-time uncertainty problem important?

The energy-time uncertainty problem is important because it affects the way we understand and measure quantum systems. It also has implications for technologies such as quantum computing and communication.

What is the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle is a fundamental principle in quantum mechanics that states that it is impossible to know both the position and momentum of a particle with absolute certainty. The energy-time uncertainty problem is a specific application of this principle.

How does the energy-time uncertainty problem impact experiments?

The energy-time uncertainty problem can impact experiments by limiting the precision with which certain measurements can be made. This can lead to uncertainties in the results of experiments and may require multiple measurements to be taken to obtain a more accurate understanding of the system being studied.

Are there any ways to overcome the energy-time uncertainty problem?

There are some techniques that can be used to decrease the impact of the energy-time uncertainty problem in experiments. These include using shorter pulses of energy and increasing the precision of measurement devices. However, the uncertainty principle is a fundamental aspect of quantum mechanics and cannot be completely overcome.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
292
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
417
  • Introductory Physics Homework Help
Replies
18
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top