flyusx
- 69
- 12
- 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).
$$\int_{-\infty}^{\infty}\left|\tilde{\phi}\left(p\right)\right|^{2}\text{d}p=\int_{-p_{0}}^{p_{0}}\left|A\right|^{2}\text{d}p=2p_{0}\left|A\right|^{2}=1$$
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).
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