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Apologies if this is a little vague. It might not have a good answer.

Given the interpretation of $|\psi(x)|^2$ as a probability distribution it's unsurprising that a wave function that is concentrated around a point $x$ should behave at least a little like a classical particle at the point $x$.

Is there a similarly intuitive explanation for why a plane wave function $\exp(ik\cdot x)$ behaves somewhat like a classical particle with momentum $\hbar k$? I'm not looking for the standard explanation in terms of eigenstates of the momentum operator, but something that can be used pedagogically for people whose linear algebra isn't sophisticated enough for that.

For example it's not hard to see that a plane wave in n-dimensions has a direction associated with it but it's not intuitively obvious to me that a higher frequency wave should have a higher momentum (unless I reason via the Schrödinger equation which I don't want to do). It's also not surprising that a plane EM wave carries momentum, after all it can interact with charged matter via the Lorentz force and transfer momentum to it, but wave functions don't have such a straightforwardly interpretable interaction.

So how can we make it unsurprising that a plane wave function has a definite momentum?

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Does the concept of group velocity of a wavepacket help? –  Michael Brown Apr 11 '13 at 15:20
    
@MichaelBrown Having just yesterday worked through the basics of anomalous dispersion in media, seeing examples of group velocities that do all kinds of weird things, I think the answer has to be no :-) –  Dan Piponi Apr 11 '13 at 15:48

1 Answer 1

The statement "the plane-wave wavefunction $\psi(x)=\exp(i\mathbf{k}\cdot\mathbf{x})$ has definite momentum $\mathbf{p}=\hbar\mathbf{k}$" is a slightly more elaborate restatement of de Broglie's rule, which essentially states "matter particles with momentum $p$ are associated with waves whose wavelength is $\lambda=\hbar/p$".

The only things you need to add to get the full statement is the directionality (i.e. saying the wavefronts are orthogonal to $\mathbf{p}$, which is easy), and the distinction between $e^{i\mathbf{k}\cdot\mathbf{x}}$, $e^{-i\mathbf{k}\cdot\mathbf{x}}$, $\cos(\mathbf{k}\cdot\mathbf{x})$, and so on. The latter is a bit trickier but the choice of $e^{i\mathbf{k}\cdot\mathbf{x}}$ can be justified, to within a sign convention, as the only plane-wave function which gives, by itself, an orientation for $\mathbf{k}$. The sign is simply that, a convention. (Since it's the same convention as $[x,p]=i\hbar$ vs $[p,x]=i\hbar$, it's not going to go away. So just say either convention is fine and we just choose the former.)

The important thing is that the weirdness of the plane-wave $\leftrightarrow$ definite-momentum identification is exactly the same weirdness of de Broglie's $\lambda=\hbar/p$, or that behind the commutator $[x,p]=i\hbar$. How could you possibly justify it? However you try, your argument will have a hole somewhere. It's a fundamental, distinguishing building blocks of (the weirdness of) quantum mechanics.

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No classical argument can ever give rise to the physical constant $\hbar$ and so I agree there will always be some hole if we try to give classical intuition for quantum mechanics. Nonetheless I think there is still something useful you can do. For example if you rotate a wavefunction then you rotate $k$ in the obvious way. Similarly if you rotate a classical system, you rotate $p$ in exactly the same way. Find enough similarities like this and you go some way to smoothing over the hole. –  Dan Piponi Apr 11 '13 at 15:52
    
Plane waves are weird in classical physics too. By this I mean they do not exist in nature. As a result you can draw conclusions that are formally correct but misleading. For example, in free space, finite EM beams can have nonzero longitudinal component. I.e., the E and B vectors are not purely transverse. –  user27777 Aug 16 '13 at 20:07

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