Uncertainties For Wave Packets

  • Thread starter Thread starter flyusx
  • Start date Start date
  • Tags Tags
    Standard Wave
AI Thread Summary
The discussion centers on the normalization and Fourier transformation of wave packets, specifically addressing the calculations for the position wave function ##\psi(x)## and its associated probability density functions. The normalization of the momentum wave function ##\tilde{\phi}(p)## is successfully achieved, leading to a normalized position wave function. However, difficulties arise when calculating the standard deviation for ##\psi(x)##, as the integral diverges, suggesting infinite uncertainty. Despite this, the relationship defined by Heisenberg's uncertainty principle remains satisfied, indicating that the calculations may still be valid. The participant expresses concern over the diverging integral but acknowledges that it aligns with expected theoretical outcomes.
flyusx
Messages
63
Reaction score
9
Homework Statement
Zettili Quantum, 3rd Edition, Exercise 1.39 (page 89)

Part (a): Find the Fourier transform ##\psi(x)## of $$\tilde{\phi}(p)=\begin{cases}0,&\left|p\right|\geq p_{0}\\A&\left|p\right|<p_{0}\end{cases}$$
Part (b): Find A so that ##\psi(x)## is normalised.
Part (c): Estimate the uncertainties ##\Delta p## and ##\Delta x## and then verify that ##\Delta x\Delta p## satisfies Heisenberg's uncertainty relation.
Relevant Equations
$$\psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde{\phi}(p)\exp\left(\frac{ixp}{\hbar}\right)\text{d}p$$
$$\sin(\theta)=\frac{\exp(i\theta)-\exp(-i\theta)}{2i}$$
$$\Delta x=\sqrt{\int_{-\infty}^{\infty}(x-\mu)^{2}f(x)\text{d}x}$$
$$\mu=\int_{-\infty}^{\infty}xf(x)\text{d}x$$
I believe I have solved parts (a) and (b) correctly. I'll first normalise ##\tilde{\phi}\left(p\right)## since I have found that doing so and then Fourier-transforming it to return ##\psi\left(x\right)## keeps the resultant (position) wave packet normalised.
$$\int_{-\infty}^{\infty}\left|\tilde{\phi}\left(p\right)\right|^{2}\text{d}p=\int_{-p_{0}}^{p_{0}}\left|A\right|^{2}\text{d}p=2p_{0}\left|A\right|^{2}=1$$
Which returns
$$A=\frac{1}{\sqrt{2p_{0}}}$$
Now performing the Fourier transform for a time-independent position-basis wave function, I use
$$\psi\left(x\right)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde{\phi}\left(p\right)\exp\left(\frac{ixp}{\hbar}\right)\text{d}p=\frac{1}{\sqrt{2\pi\hbar}}\int_{-p_{0}}^{p_{0}}\frac{1}{\sqrt{2p_{0}}}\exp\left(\frac{ixp}{\hbar}\right)\text{d}p$$

Where ##\tilde{\phi}\left(p\right)## is effectively a constant so the integral is that of an exponential.
$$\psi\left(x\right)=\frac{1}{\sqrt{4\pi\hbar p_{0}}}\int_{-p_{0}}^{p_{0}}\exp\left(\frac{ixp}{\hbar}\right)\text{d}p=\frac{1}{\sqrt{4\pi\hbar p_{0}}}\frac{\hbar}{ix}\left[\exp\left(\frac{ixp}{\hbar}\right)\right]_{p=-p_{0}}^{p=p_{0}}$$
The exponential term evaluated can be converted into an expression with sine.
$$\exp\left(\frac{ixp_{0}}{\hbar}\right)-\exp\left(\frac{-ixp_{0}}{\hbar}\right)=2i\sin\left(\frac{xp_{0}}{\hbar}\right)$$
So the Fourier transform is
$$\psi(x)=\frac{1}{\sqrt{4\pi\hbar p_{0}}}\frac{\hbar}{ix}2i\sin\left(\frac{xp_{0}}{\hbar}\right)=\boxed{\sqrt{\frac{\hbar}{xp_{0}}}\frac{\sin\left(\frac{xp_{0}}{\hbar}\right)}{x}}$$
When I compute a squared-amplitude integration over all space, I do get one so ##\psi(x)## is normalised.

It is this third part that I have trouble with. I obtain the standard deviation expressions for a function from the Wikipedia page on standard deviations. There, ##\mu## is the expected value of the density function ##f(x)##. From what I understand, the squared-amplitude ##\left|\psi(x)\right|^{2}## and ##\left|\tilde{\phi}(p)\right|^{2}## are probability density functions, eg if I have a random wave function ##\left|\chi(x)\right|^{2}\text{d}x##, it represents the probability of finding the associated particle of ##\chi(x)## between ##x## and ##x+\text{d}x##. The ##\mu## values for both ##x\left|\tilde{\phi}(p)\right|^{2}## and ##x\left|\psi(x)\right|^{2}##, I compute to be zero since the integrands are odd functions. Setting ##\mu=0##, the standard deviation for ##\tilde{\phi}(p)## is
$$\Delta p=\sqrt{\int_{-p_{0}}^{p_{0}}\frac{1}{2p_{0}}p^{2}\text{d}p}=\sqrt{\frac{1}{2p_{0}\left[\frac{p^{3}}{3}\right]^{p=p_{0}}_{p=-p_{0}}}}=\frac{p_{0}}{\sqrt{3}}$$
However, when I go to do the same calculation for ##\psi(x)##, I obtain the integrand
$$x^{2}\left|\psi(x)\right|^{2}=x^{2}\frac{\hbar}{\pi p_{0}}\frac{\sin^{2}\left(\frac{xp_{0}}{\hbar}\right)}{x^{2}}=\frac{\hbar}{\pi p_{0}}\sin^{2}\left(\frac{xp_{0}}{\hbar}\right)$$
Which when integrated from ##(-\infty,\infty)## is non-converging, hence I cannot obtain ##\Delta x\Delta p## unless I was to say that the uncertainty is infinite?

I will say that on page 54, Zettili uses the example wave packets ##\psi(x)=\left(\frac{2}{\pi a^{2}}\right)^{\frac{1}{4}}\exp\left(-\frac{x^{2}}{a^{2}}\right)\exp\left(ik_{0}x\right)## and ##\phi(k)=\left(\frac{a^{2}}{2\pi}\right)^{\frac{1}{4}}\exp\left(-\frac{a^{2}\left(k-k_{0}\right)^{2}}{4}\right)## for a computation (or rather estimate) with result ##\Delta x\Delta k=\frac{1}{2}## (where the ##k##-basis wave packet is centered at ##k_{0}## and ##a## is a constant). When I take these wave packets, compute their square amplitudes and integrate from ##(-\infty,\infty)## using the appropriate standard deviation formulas, I indeed get one-half (the uncertainty/standard deviation is independent of ##a## or ##k_{0}##). This gives me a bit of confidence that the equations I'm using might be right (or in the least might not be fully wrong).
 
Last edited:
Physics news on Phys.org
1736822582682.png
 
anuttarasammyak said:
Yes, I do, thanks for the correction.
 
  • Like
Likes anuttarasammyak
flyusx said:
Which when integrated from (−∞,∞) is non-converging, hence I cannot obtain ΔxΔp unless I was to say that the uncertainty is infinite?
Even if the product is infinite, the relation
\triangle p \triangle x \ge \frac{\hbar}{2}
is satisfied. I am not surprized to find infinite uncertainty in such a square shaped momentum space wave function.
 
anuttarasammyak said:
Even if the product is infinite, the relation
\triangle p \triangle x \ge \frac{\hbar}{2}
is satisfied. I am not surprized to find infinite uncertainty in such a square shaped momentum space wave function.
This is the only time for which going through my process of normalising, Fourier transforming and then evaluating the standard deviations according to the formulas has given me a divergent integral, albeit it has never given me a result that didn't satisfy Heisenberg's uncertainty principle. I suppose I got a bit scared upon seeing a diverging integral and thought my method and understanding was flawed.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top