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Uncertainty values for non-Gaussian functions

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PhyPsy
#1
Apr6-12, 03:58 PM
P: 39
1. The problem statement, all variables and given/known data
[itex]\phi(k)= \left\{ \begin{array}{cc} \sqrt{\frac{3}{2a^3}}(a-|k|), & |k| \leq a \\ 0, & |k|>a \end{array} \right.[/itex]
[itex]\psi(x)= \frac{4}{x^2}sin^2 (\frac{ax}{2})[/itex]
Calculate the uncertainties [itex]\Delta x[/itex] and [itex]\Delta p[/itex] and check whether they satisfy the uncertainty principle.

2. Relevant equations
[itex]\Delta x\Delta p \geq h/2[/itex]

3. The attempt at a solution
The solution is worked out in the book, which is [itex]\Delta k=a[/itex] and [itex]\Delta x=\pi /a[/itex]. I understand that for a Gaussian distribution, you can use the standard deviation as [itex]\Delta x[/itex] and [itex]\Delta k[/itex], and this leads to the lowest limit of the uncertainty relation, [itex]h/2[/itex]. I don't see how I'm supposed to come up with [itex]\Delta x[/itex] and [itex]\Delta k[/itex] for non-Gaussian functions, though. The book seems to just pick [itex]\Delta k=a[/itex] and [itex]\Delta x=\pi /a[/itex] somewhat arbitrarily without explaining why these values were chosen. Could someone tell me if there is a method for figuring out what the uncertainties should be for non-Gaussian functions like this one?
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fzero
#2
Apr6-12, 05:04 PM
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P: 2,603
If we are measuring the eigenvalue [itex]a[/itex] of an observable [itex]A[/itex], then we may characterize the uncertainty [itex]\Delta a[/itex] in [itex]a[/itex] using the statistical variance:

[tex](\Delta a)^2 = \langle A^2 \rangle - \langle A\rangle^2.[/tex]

This definition works for any state we use to compute the expectation values. In fact, this definition is the one used in the derivation of the generalized uncertainty principle.
PhyPsy
#3
Apr7-12, 04:13 PM
P: 39
[itex]\langle A\rangle=0[/itex] for both functions, so [itex](\Delta a)^2= \int ^\infty _{-\infty}a^2 f(a) da[/itex]

[itex](\Delta k)^2= \int ^\infty _{-\infty} k^2 |\phi(k)|^2 dk = \frac{3}{2a^3} [\int ^0 _{-a} k^2 (a+k)^2 dk + \int ^a _0 k^2 (a-k)^2 dk][/itex]
[itex]=\frac{3}{2a^3} [(\frac{1}{3} a^2 k^3 + \frac{1}{2} a k^4 + \frac{1}{5} k^5)| ^0 _{-a} + (\frac{1}{3} a^2 k^3 - \frac{1}{2} a k^4 + \frac{1}{5} k^5)| ^a _0[/itex]
[itex]=\frac{3}{2a^3} \frac{2}{30} a^5 = \frac{a^2}{10}[/itex]
[itex]\Delta k = \frac{a}{\sqrt{10}}[/itex]
[itex]\Delta p = \frac{a \hbar}{\sqrt{10}}[/itex]

[itex](\Delta x)^2= \int ^\infty _{-\infty} x^2 |\psi(x)|^2 dx = 16 \int ^\infty _{-\infty} \sin ^4 (\frac{ax}{2}) dx / x^2[/itex]
[itex]=16[\frac{1}{2} a Si(ax) - \frac{1}{4} a Si(2ax) - \sin ^4 (\frac{ax}{2}) / x]|^\infty _{-\infty} = 16(\frac{a \pi}{2} - \frac{a \pi}{4}) = 4a \pi[/itex]
[itex]\Delta x = 2 \sqrt{a \pi}[/itex]

[itex]\Delta p \Delta x = \frac{2a \hbar \sqrt{a \pi}}{\sqrt{10}}[/itex]
So whether or not this is greater than [itex]\hbar / 2[/itex] depends on what a is. Did I do something wrong here?

fzero
#4
Apr7-12, 11:23 PM
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P: 2,603
Uncertainty values for non-Gaussian functions

Quote Quote by PhyPsy View Post
[itex]\langle A\rangle=0[/itex] for both functions, so [itex](\Delta a)^2= \int ^\infty _{-\infty}a^2 f(a) da[/itex]

[itex](\Delta k)^2= \int ^\infty _{-\infty} k^2 |\phi(k)|^2 dk = \frac{3}{2a^3} [\int ^0 _{-a} k^2 (a+k)^2 dk + \int ^a _0 k^2 (a-k)^2 dk][/itex]
[itex]=\frac{3}{2a^3} [(\frac{1}{3} a^2 k^3 + \frac{1}{2} a k^4 + \frac{1}{5} k^5)| ^0 _{-a} + (\frac{1}{3} a^2 k^3 - \frac{1}{2} a k^4 + \frac{1}{5} k^5)| ^a _0[/itex]
[itex]=\frac{3}{2a^3} \frac{2}{30} a^5 = \frac{a^2}{10}[/itex]
[itex]\Delta k = \frac{a}{\sqrt{10}}[/itex]
[itex]\Delta p = \frac{a \hbar}{\sqrt{10}}[/itex]
I agree with this, but we'd also want to calculate [itex]\Delta p[/itex] using [itex] \hat{x} = -i\hbar d/dp[/itex] in the momentum representation. Note that [itex]\psi(x)[/itex] is not the Fourier transform of [itex]\phi(k)[/itex], so we really have two problems to solve here.

[itex](\Delta x)^2= \int ^\infty _{-\infty} x^2 |\psi(x)|^2 dx = 16 \int ^\infty _{-\infty} \sin ^4 (\frac{ax}{2}) dx / x^2[/itex]
[itex]=16[\frac{1}{2} a Si(ax) - \frac{1}{4} a Si(2ax) - \sin ^4 (\frac{ax}{2}) / x]|^\infty _{-\infty} = 16(\frac{a \pi}{2} - \frac{a \pi}{4}) = 4a \pi[/itex]
[itex]\Delta x = 2 \sqrt{a \pi}[/itex]

[itex]\Delta p \Delta x = \frac{2a \hbar \sqrt{a \pi}}{\sqrt{10}}[/itex]
So whether or not this is greater than [itex]\hbar / 2[/itex] depends on what a is. Did I do something wrong here?
Again, this is a completely different wavefunction, so we can't use the uncertainty in momentum that we computed from the other one. Use [itex]\hat{p} = -i\hbar d/dx[/itex] to compute it.
PhyPsy
#5
Apr8-12, 01:14 PM
P: 39
Actually, [itex]\psi (x)[/itex] is the Fourier transform of [itex]\phi (k)[/itex], or at least it is supposed to be...

[itex]\psi (x)= \frac{1}{\sqrt{2 \pi}} \int ^\infty _{-\infty} \phi(k) e^{ikx} dk[/itex]
[itex]=\frac{\sqrt{3}}{2 \sqrt{\pi a^3}} [\int ^0 _{-a} (a+k)e^{ikx} dk + \int ^a _0 (a-k)e^{ikx} dk][/itex]
[itex]=\frac{\sqrt{3}}{2 \sqrt{\pi a^3}} [\int ^0 _{-a} ke^{ikx} dk - \int ^a _0 ke^{ikx} dk + a \int ^a _{-a} e^{ikx} dk][/itex]
[itex]=\frac{\sqrt{3}}{2 \sqrt{\pi a^3}} [\frac{a}{ix} e^{-iax} + (1 - e^{-iax} )/ x^2 - \frac{a}{ix} e^{iax} - (e^{iax} - 1)/ x^2 + 2a \sin (ax) /x][/itex]

At this point, the book just says that after some calculations, the answer is [itex]\psi (x) = 4 \sin ^2 (\frac{ax}{2}) / x^2[/itex]. I don't know how they came up with that; the answer I got is [itex]\frac{\sqrt{3}}{x^2 \sqrt{\pi a^3}} [1 - \frac{ax}{2} \sin (ax) - \cos (ax)][/itex].


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