Uncertainty Principle Equations

[erok81]

I have a few questions regarding how/when to use which uncertainty principle formulas. I am not sure of the difference between them. I've checked my book and wiki and neither really helped.

Here are the formulas in question.

[tex]\Delta p \Delta x \approx \hbar[/tex]

[tex]\Delta p \Delta x \geq \frac{\hbar}{2}[/tex]

One of our homework assignments we had to find the velocity of an electron with a given x. So we used the first one and solved for v. I understand you can use, at the first one, for things like this when dealing with quantum level stuff. Is that correct?

I also saw the second one used by replacing the ≥ with an equal sign to find the minimum value of whatever variable we were working with. Which makes sense.

The only thing I don't understand is when is one preferred over the other? And can these be used to solve things like I mentioned above?

 

[maverick_starstrider]

The absolute limit on accuracy allowed by nature is [itex]\Delta x \Delta p = \frac{h-bar}{2} [/itex]. Of course one can always make measurements that are LESS accurate than this (which, in reality is always the case). So we include the greater than. The h-bar was just Heisenberg's initial, out-dated guess. It is now understood that the exact value is h-bar/2

 

[Matterwave]

Actually, as far as I know, the hbar/2 value comes from our definition of standard deviations. If we defined the standard deviation of some set of measurements in a different way, we would get a different value in that inequality.

Really, this principle gives a fundamental limit on how narrow the line-widths of the wavefunction can be in momentum or position space, if we defined our line-widths to be different, the value we would get would be different.

The second equation is correct based on the conventional definition of standard deviation. Whereas the first equation is an approximation which is useful when we want to get order of magnitude estimates.

 

[maverick_starstrider]

Actually, as far as I know, the hbar/2 value comes from our definition of standard deviations. If we defined the standard deviation of some set of measurements in a different way, we would get a different value in that inequality.

Really, this principle gives a fundamental limit on how narrow the line-widths of the wavefunction can be in momentum or position space, if we defined our line-widths to be different, the value we would get would be different.

The second equation is correct based on the conventional definition of standard deviation. Whereas the first equation is an approximation which is useful when we want to get order of magnitude estimates.

Yes this is all correct. Despite all the hoop-lah during its initial discovery the Heisenberg Uncertainty principle is rather obvious now (if we accept that particles are modeled by a wave-like wavefunction) and is simply just a property of Fourier transforms which is

[itex]\Delta x \Delta k = \pi [/itex]

 

[erok81]

Perfect. That makes sense. Thanks for the direction both of you.

I'm going to give a few problems a try that involve this principle and see if I have any more questions.

Another example we did was with a baseball. Using the first question and calculated it's velocity (some really small value or course). What this answer showed was the velocity it could have but necessarily. Or is it the discrepancy in the velocity that we can't measure?

I hope that question makes sense. I am at work posting between tasks.

 

[Matterwave]

I think you would have to go into more detail on how you got this "velocity" to see what you actually calculated.

 

[Dickfore]

I don't know if this would be useful to the OP or not, but here is an alternative approach to try and find the absolute upper bound on the Heisenberg uncertainty principle for a position vector component and its conjugate momentum:

The definition of a variance of an observable is:

[tex]
\sigma^{2}[f] \equiv \langle(f - \langle f \rangle)^{2}\rangle
[/tex]

By expanding the binomial and using the linear property of the operation of taking the average value, this definition can be brought to a form that is more convenient for calculations:

[tex]
\sigma^{2}[f] = \langle f^{2} \rangle - \langle f \rangle^{2}
[/tex]

Finally, given a wave function [itex]\psi(x)[/itex], the average of an observable can be calculated as:

[tex]
\langle f \rangle = \int_{-\infty}^{\infty}{\psi^{\ast}(x) \hat{f} \, \psi(x) \, dx}
[/tex]

assuming the wave function is normalized so that:

[tex]
\langle 1 \rangle = \int_{-\infty}^{\infty}{\psi^{\ast}(x) \, \psi(x) \, dx} = 1
[/tex]

The position operator (like any operator in the basis of its eigenfunctions) is simply a multiplication by a coordinate:

[tex]
\hat{x} \equiv x
[/tex]

whereas the momentum operator can be shown to be a differential operator:

[tex]
p = \frac{\hbar}{i} \, \frac{d}{d x}
[/tex]

Combining everything together, we propose the following variational problem:

[tex]
F[\psi(x), \psi^{\ast}(x)] = \sigma^{2}[x] \cdot \sigma^{2}[p] \rightarrow \mathrm{min.}
[/tex]

under the subsidary normalization condition given above. Varying with respect to [itex]\psi^{\ast}(x)[/itex], gives the following differential equation:

[tex]
\left(x^{2} \, \psi(x) - 2 \, x \, \psi(x) \, \langle x \rangle \right) \, \sigma^{2}[p] + \sigma^{2}[x] \, \left(-\hbar^{2} \frac{d^{2}\psi(x)}{d x^{2}} - 2 \, \frac{\hbar}{i} \, \frac{d \psi(x)}{d x} \, \langle p \rangle
\right) - \lambda \, \psi(x) = 0
[/tex]

which has 5 parameters ([itex]\langle x \rangle, \sigma^{2}[x], \langle p \rangle, \sigma^{2}[p], \lambda[/itex]) that need to be determined self-consistently.

I will post the solution later.

 

[erok81]

I think you would have to go into more detail on how you got this "velocity" to see what you actually calculated.

Let's say we did this. Assume > 10^-15 type of values. Or does that really matter?

[tex]\Delta p \Delta x = \frac{\hbar}{2}[/tex]

Solving for velocity...

[tex]v= \frac{\hbar}{2 \Delta x m}[/tex]

This is actually a good example because it's where I am confused the most.

What exactly is v? Is it an actual velocity the particle can have? Or the uncertainty of the velocity measurement having nothing to do with how fast the particle is actually moving?

I don't know if this would be useful to the OP or not, but here is an alternative approach to try and find the absolute upper bound on the Heisenberg uncertainty principle for a position vector component and its conjugate momentum:

/snip

Thanks for posting this. I'll go over with a bit more detail later tonight.

 

[Matterwave]

So that would be some uncertainty in the velocity (and as such, you need to keep the delta sign).

It wouldn't really give you the actual velocity of the ball though, the statement is just saying that you can't measure your velocity to be any more accurate than within delta-v of the actual velocity without "messing up" your x-measurement.

This argument is "semi-classical" and not very rigorous, but the act of trying to apply the HUP to macroscopic objects is not very rigorous.

 

[erok81]

So that would be some uncertainty in the velocity (and as such, you need to keep the delta sign).

It wouldn't really give you the actual velocity of the ball though, the statement is just saying that you can't measure your velocity to be any more accurate than within delta-v of the actual velocity without "messing up" your x-measurement.

This argument is "semi-classical" and not very rigorous, but the act of trying to apply the HUP to macroscopic objects is not very rigorous.

And if it is applied to non-classical situations, it's still the same thing though? I have no idea why this is giving me so much trouble. I get the concept, just not when it's broken down into its individual components.

 

[maverick_starstrider]

And if it is applied to non-classical situations, it's still the same thing though? I have no idea why this is giving me so much trouble. I get the concept, just not when it's broken down into its individual components.

No it's really not but you probably don't have to worry about it. The notion that Heisenberg's Uncertainty is "due to some thought experiment where to determine the position of some particle you need to give it some energy which changes its momentum etc." is actually false. This is really the observer effect or what is sometimes called Heisenberg's Microscope. It was an idea put for by Heisenberg himself to explain this strange uncertainty in the early days of quantum but then HE HIMSELF rescinded it when he realized that wasn't actually what was happening. And yet it's an idea that lingers.

As was previously mentioned Heisenberg's uncertainty is really just an observation about waves. If one considers Fourier Transforms one realizes that any function can be made by simply adding together sin and cos waves of different wavelengths and amplitudes. It is a property of this math that the more localized the function I'm constructing is (i.e. imagine trying to make a single localized bell curve in space) the wider range of sin and cos wavelengths I need to construct this function. wavelength is [itex] \lambda = \frac{2 \pi}{k} [/itex] is related to k, our wavenumber. So essentially when doing Fourier transforms the more localized our function is in space the more values of k I need to use to make it. Given that momentum is [itex] h k /2 \pi [/itex] I see that the more local my wavefunction the larger the range of k's that must be used to construct. In the math of Fourier series the trade-off between k and x is

[itex]\Delta x \Delta k = \pi[/itex]

which in terms of momentum becomes our famous Heisenberg relation. Nothing to do with thought experiments involving microscopes and such.

 

[Dickfore]

[tex]
\left(x^{2} \, \psi(x) - 2 \, x \, \psi(x) \, \langle x \rangle \right) \, \sigma^{2}[p] + \sigma^{2}[x] \, \left(-\hbar^{2} \frac{d^{2}\psi(x)}{d x^{2}} - 2 \, \frac{\hbar}{i} \, \frac{d \psi(x)}{d x} \, \langle p \rangle
\right) - \lambda \, \psi(x) = 0
[/tex]

which has 5 parameters ([itex]\langle x \rangle, \sigma^{2}[x], \langle p \rangle, \sigma^{2}[p], \lambda[/itex]) that need to be determined self-consistently.

I will post the solution later.

The equation can be simplified to:

[tex]
\frac{d^{2} \psi(x)}{d x^{2}} - 2 \frac{i \langle p \rangle}{\hbar} \, \frac{d \psi(x)}{d x} + \left[ \frac{\lambda + \langle x \rangle^{2} \sigma^{2}[p]}{\hbar^{2} \, \sigma^{2}[x]} - \frac{\sigma^{2}[p]}{\hbar^{2} \, \sigma^{2}[x]} \, (x - \langle x \rangle)^{2} \right] \, \psi(x) = 0
[/tex]

We can cancel the first derivative by introducing the new wave function:

[tex]
\psi(x) = \exp\left(\frac{i \, \langle p \rangle \, x}{\hbar}\right) \, \phi(x)
[/tex]

after which the equation becomes:

[tex]
\frac{d^{2} \phi(x)}{d x^{2}} + \left[ \frac{\lambda + \langle x \rangle^{2} \sigma^{2}[p] + \sigma^{2}[x] \, \langle p \rangle^{2}}{\hbar^{2} \, \sigma^{2}[x]} - \frac{\sigma^{2}[p]}{\hbar^{2} \, \sigma^{2}[x]} \, (x - \langle x \rangle)^{2} \right] = 0
[/tex]

Next, we introduce a dimensionless coordinate:

[tex]
\zeta = \frac{x - \langle x \rangle}{a}
[/tex]

and obtain the following equation:

[tex]
\frac{d^{2} \phi}{d \zeta^{2}} + (\epsilon - \zeta^{2}) \, \phi = 0
[/tex]

where:

[tex]
a = \left(\frac{\hbar \, \sigma[x]}{\sigma[p]} \right)^{2}
[/tex]

[tex]
\epsilon = \frac{\lambda + \langle x \rangle^{2} \sigma^{2}[p] + \sigma^{2}[x] \, \langle p \rangle^{2}}{\hbar \, \sigma[x] \, \sigma[p]}
[/tex]

This equation is the same as the equation we obtain for a time-independent Schroedinger equation for a linear harmonic oscillator. The ground state is of the form of a Gaussian packet:

[tex]
\phi(\zeta) = A \, \exp\left(-\frac{\zeta^{2}}{2}\right)
[/tex]

The coordinate probability distribution is normal:

[tex]
f(x) = |\phi(\zeta)|^{2} = |A|^{2} \, \exp\left[-\frac{(x - \langle x \rangle)^{2}}{a^{2}} \right]
[/tex]

The variation in the coordinate is:

[tex]
2 \, \sigma^{2}[x] = a^{2}
[/tex]

Using the above equation for a, we get:

[tex]
\sigma[x] \, \sigma[p] = \frac{\hbar}{2}
[/tex]

This is the self-consistent condition. At the same time, it is the absolute minimum of the product of the two uncertainties. For any other wave function, we would have:

[tex]
\sigma[x] \, \sigma[p] \ge \frac{\hbar}{2}
[/tex]

This is the Heisenberg uncertainty relation.