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We know that all wavefunctions can be written as an sum of plane waves infinite in extent. This often leads to conceptual troubles when thinking about physics. It turns out this isn't necessary because one can also have finite in extent "plane wave" decomposition.

Consider the free particle hamiltonian $H = \frac{p^2}{2m}$. Consider $\psi(x) = \int A(k) e^{i k x} dk$. It evolves like $\psi(x,t) = \int A(k) e^{i k x - i\omega t} dk$. This is all standard and correct.

Now assume that the wavefunction is finite in some region $A$ and zero outside region $B$. Everything else is defined as region $C$ Then we know there exists completely smooth bump functions $B(x) = 1$ for $x \in A$ and $B(x) = 0$ for $x \in C$. Therefore we know that

$$ \psi(x) = B(x)\psi(x) = \int A(k) \underbrace{B(x)e^{i k x}}_\text{Finite "plane waves"} dk $$

This gives two ways to write the initial conditions as a superposition of finite plane waves or infinite plane waves. The infinite plane waves can obviously reconstruct everything because they are everywhere. However these are of finite size. What paradox does this give?

Paradox

We know, from linearity and Ehrenfest, that these finite plane waves will move at the classical velocity $\hbar k /m$. If $\psi$ was stationary and just diffused, we know it's size increases like $\sqrt{\hbar t/m}$.

  • So it seems like the plane waves would all run away unable to reconstruct via interference in the same region as $\psi(x,t)$.
  • The finite plane waves no longer being infinite in extent will no longer be able to interfere to vanish where they should. So the wavefunction under this initial condition should grow in size like $\approx (\hbar \sqrt{\langle k^2 \rangle}/m) t$ in contradiction to the $\sqrt{\hbar t/m}$ result for a gaussian wavepacket.

Where's the flaw in reasoning?

Picture of the First Paradox Wavepacket flying apart

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  • $\begingroup$ shouldn't there be a $\psi(k)$ inside the integral? $\endgroup$ – JEB Jul 5 '18 at 16:56
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    $\begingroup$ @EnriqueMendez You seem to be making the assumption that $B(x)$ is independent of time. I.e. although it is true that $\psi(x)=B(x)\psi(x)$ it is not true that $\psi(x,t)=B(x) \psi(x,t)$. $\endgroup$ – Quantum spaghettification Jul 5 '18 at 19:12
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    $\begingroup$ I agree with @Quantumspaghettification. Moreover, the second illustration the you have shown is wrong in my opinion. It's not true that "the finite plane waves are wavepackets that fly away from the center". To do things in a correct way, you should project each such finite plane wave on a orthonormal basis, e.g. the set of infinite plane waves. You will see that each finite plane wave is a linear combination of infinite plane waves and, therefore, the time evolution of each finite plane wave is much more complex than simply "flying away". $\endgroup$ – AndreaPaco Jul 5 '18 at 20:26
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    $\begingroup$ What you've forgotten is diffraction / wavepacket spreading: your "finite plane waves" spread faster than they move away from each other, so they are always able to interfere; the assumption that fails is the assumption that you can ignore the (huge) effect that adding B(x) will have on the future evolution of the "plane wave". The answer by Quantum Spaghettification has the details. $\endgroup$ – Emilio Pisanty Jul 6 '18 at 8:17
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    $\begingroup$ Related: What is the spreading for rectangular wave packets? $\endgroup$ – Emilio Pisanty Jul 6 '18 at 8:19
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Let us actually look at how the wavefunction: $$\phi_p(x)=B(x)e^{ipx}$$ evolves with time. Where for simplicity I we will take: $$B(x)= \Theta(1-x)\Theta(1+x)$$ i.e. $1$ in the region $[-1,1]$ and $0$ elsewhere. Decomposing $\phi_p(x)$ into it's fourier components (using mathematica) we get: $$\phi_p(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\sqrt{\frac{2}{\pi}}\frac{\sin(p-k)}{p-k} e^{kxi}dk$$ Setting then $\hbar^2/(2m)=1$ we get: $$\phi_p(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\sqrt{\frac{2}{\pi}}\frac{\sin(p-k)}{p-k} e^{(kx-k^2 t)i}dk$$ If we then use Mathematica to plot the modulus squared of this (discretization things to make it faster) we get the following, for $p=1$:

enter image description here

(this starts from $t=3$ as lower $t$ lead to problems in numerical calculation). If we also look at the value of $|\phi_p(0,t)|^2$ as a function of time we get the following:

enter image description here

As we can see from both of this figures, although $\phi(x,t)$ does move away, as shown in the diagram above, it also spreads out giving an appreciable value at $x=0$ even for large $t$. This spreading out is which solves your paradoxes.

Edit (to address comments): Version in terms of Gaussians

Ok the analysis given above did although inline with the question has a lot steps which aren't immediately apparent. To make things clearer let us change to: $$\phi_p(x)=B(x) e^{ipx}\quad \text{where}\quad B(x)=e^{-\frac{x^2}{2\sigma^2}}$$

the same principle holds but using a Gaussian instead of a Heaviside step just means things are easier to calculate. I.e. We have that: $$ \phi_p(x)=\sqrt{\frac{\sigma^2}{2\pi}} \int^\infty_{-\infty} dk e^{-\frac{1}{2}(p-k)^2 \sigma^2+i k x}$$ Thus $$ \phi_p(x,t)=\sqrt{\frac{\sigma^2}{2\pi}} \int^\infty_{-\infty} dk e^{-\frac{1}{2}(p-k)^2 \sigma^2+i k x-i k^2 t}$$ $$=\frac{\sqrt{2 \pi } \exp \left(\frac{2 p \sigma ^2 (x-p t)+i x^2}{4 t-2 i \sigma ^2}\right)}{\sqrt{\sigma ^2+2 i t}}$$ Then: $$|\phi_p(x,t)|^2=\frac{2 \pi e^{-\frac{\sigma ^2 (x-2 p t)^2}{\sigma ^4+4 t^2}}}{\sqrt{\sigma ^4+4 t^2}}$$

as $t\rightarrow \infty$ this expression becomes independent of $x$ for all $p$. The same principle holds when $B(x)$ is the similar (although more annoying) combination of Heaviside step functions.

Appendix: Mathematica Code

Here is my Mathematica code in case anyone is interested:

FourierTransform[HeavisideTheta[1 - x]*HeavisideTheta[1 + x] E^(I p x), x, k]
inF[p_, x_, t_] := 
 NIntegrate[(Sqrt[2/\[Pi]] Sin[-k + p])/(-k + p)
    Cos[x k - k^2 t], {k, -10, 10}]
inG[p_, x_, t_] := 
 NIntegrate[(Sqrt[2/\[Pi]] Sin[-k + p])/(-k + p)
    Sin[x k - k^2 t], {k, -10, 10}]
plot[t_] := 
 DiscretePlot[
  inF[1, x, t]^2 + inG[1, x, t]^2, {x, -10 + 0.1`, 10 + 0.1`, 1}, 
  Joined -> True, PlotRange -> {0, 1}]
frames = Table[plot[t], {t, 3, 10, 0.1}];
Export["wave.gif", frames,"AnimationRepetitions" -> \[Infinity]]
Export["graph.png", 
 DiscretePlot[inF[1, 0, t]^2 + inG[1, 0, t]^2, {t, 1, 100, 1}]]
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  • $\begingroup$ The problem with this is the $\hbar/2m=1$. That's the fully quantum limit. If you take $\hbar/2m=0.001$ or some other small value or increase $k$. Is this still always the case? $\endgroup$ – Enrique Mendez Jul 6 '18 at 14:50
  • $\begingroup$ @EnriqueMendez Concerning $\hbar/2m=1$ if we let $\hbar/2m=0.001$ instead we would simply get $0.001t$ everywhere I have $t$. Thus we can define $t'=0.001t$ and would get the same graphs a function of $t'$ instead of $t$. I.e. they would spread out - just slower (but this would also effect the timescale of the original wavefunction $\psi(x,t)$ in the same manor) so makes no difference Concerning the momentum this is a subtler issue. I can pretty much guarantee you will see the same behavior and I will give a little thought of weather this can be easily explained. $\endgroup$ – Quantum spaghettification Jul 6 '18 at 15:37
  • $\begingroup$ The momentum is definitely the correct criticism then. If you can prove that you always get this behaviour then you've violated the correspondence principle. Particles should be able to be localized in some limit. $\endgroup$ – Enrique Mendez Jul 6 '18 at 18:01
  • $\begingroup$ @EnriqueMendez I have proved the independence of $p$ in the question for $B(x)$ Gaussian (the use of Heaviside step functions just makes it to hard to do analytically). I am not sure I fully understand your point about the correspondence principle and where it fits in here - please could you expand. $\endgroup$ – Quantum spaghettification Jul 6 '18 at 19:19
  • $\begingroup$ We know that particles must be able to move at constant velocity if the mass is large enough. This is the classical limit and schrodinger's equation must reproduce them. If you've shown that the wavefunction always must diffuse faster than it's classical velocity (so that it stays at the origin), you've violated this condition. The bump function essentially give us a nice definition of localization and therefore particle ideas. They must go to classical ideas in the limit m goes to infinity and the velocity $\hbar k /m$ is constant. $\endgroup$ – Enrique Mendez Jul 7 '18 at 13:39
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The thing is that those $B(x)e^{ikx}$ are not plane waves. The plane waves are just the $e^{ikx}$ part. And as such, the velocity of those superposed terms is not $\hbar k/m$. In fact it is not even defined.

To see this, consider a function $B(x)e^{ikx}$. To find its velocity, we need to use the momentum operator acting on it:

$$p\psi(x)=-i\hbar\frac{d\psi}{dx}.$$

Applying this identity, we get

$$pB(x)e^{ikx}=-i\hbar(B'(x)+ikB(x))e^{ikx}$$

$$p=\hbar\left(k-i\frac{B'}{B}\right).$$

As such, unless $B$ is another plane wave (which it clearly is not, from your definition), this function is not an eigenstate of momentum, and therefore does not have a well defined velocity. You can't just say the "wave packets" "diffuse" with the velocity $\hbar k/m$. It simply is not mathematically well defined.

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  • $\begingroup$ Yes the velocity is well defined. Consider just one $B(x)e^{i k x}$. It’s mean velocity is $hk/m = \langle p \rangle$. Therefore it’s mean position moves at the velocity $hk/m$ I don’t see how that could be any better defined. I never said they were plane waves. I said they were ‘finite plane waves’. $\endgroup$ – Enrique Mendez Jul 5 '18 at 23:42
  • $\begingroup$ Can you properly define $B$ for me? You said it was smooth but if it's a step function then it definitely isn't. $\endgroup$ – Gabriel Golfetti Jul 5 '18 at 23:54
  • $\begingroup$ Yeah a bump function is defined to be non-zero on a set, and zero in some surrounding area. en.wikipedia.org/wiki/Bump_function $\endgroup$ – Enrique Mendez Jul 6 '18 at 18:05

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