So, this is my first contact with Quantum Mechanics and I'm having trouble with this exercise. One of the steps involves calculating $[X,P]$, and I stuck there. Can anyone give me some help?

  • $\begingroup$ You need to tell us what assumptions you want to start with. Usually this is a starting assumption, though. $\endgroup$ – knzhou Sep 15 '16 at 21:34
  • $\begingroup$ My answer at physics.stackexchange.com/a/261151/4993 might help. (Though @EmilioPisanty 's answer below is great.) $\endgroup$ – WillO Sep 13 '19 at 12:28

"Calculating" $[X,P]$ is a hard thing to ask because the answer depends completely on where you start from.

If this is your first time studying QM, then odds are that you have been given $P=-i\hbar \frac{d}{dx}$ as a definition set in stone. In this case, calculating $[X,P]$ is just a matter of putting that operator in front of a wavefunction and seeing what happens: $$ [X,P]\psi(x) = x(-i\hbar)\frac{d}{dx}\psi(x)-(-i\hbar)\frac{d}{dx}\left[x\psi(x)\right]. $$ Working through the product rule &c (with the details left to you) it's easy to see that $[X,P]\psi(x)=i\hbar \psi(x)$, so therefore $[X,P]=i\hbar$.

On the other hand, starting with a definition like that can often be pretty jarring to someone that's starting out in QM, and with good reason, and it's not particularly clear what the physical content of this relationship is. This comes down to two essential ingredients:

  • One is de Broglie's relationship: matter is (associated in some manner with) a wave, and matter of momentum $p$ is associated with a wavelength $\lambda = h/p$. In our wave mechanics, a wave with well-defined wavelength $\lambda$ must have a wavefunction $\psi_\lambda(x)= e^{2\pi i x/\lambda}$, and the definition $P=-i\hbar \frac{d}{dx}$ ensures that $$P\psi_\lambda(x)=-i\hbar\frac{d}{dx}e^{2\pi i x/\lambda} = \frac{h}{\lambda}\psi_\lambda(x)=p\,\psi_\lambda(x).$$
  • The second is linearity: the above relationship holds only for wavefunctions with a well-defined wavelength, but we can express any wavefunction (through a Fourier transform) as a linear combination of those. There is only a single linear operator that coincides with $P$ on that basis (i.e. with $P\psi_\lambda(x)=\frac h\lambda \psi_\lambda(x)$ for all $\psi_\lambda(x)$), and it is $P=-i\hbar\frac d{dx}$.

On the other hand, it's important to say that once you go deeper into QM, the assignment $P=-i\hbar \frac{d}{dx}$ loses some of its prominence, giving way to the ('canonical') commutation relation $[X,P]=i\hbar$ as the central object.

This is partly because the two turn out to be just about equivalent, through a result known as the Stone-von Neumann theorem: in the same way that $P=-i\hbar \frac{d}{dx}$ implies that $[X,P]=i\hbar$, as calculated above, you can also do the converse and derive $P=-i\hbar \frac{d}{dx}$ from $[X,P]=i\hbar$. Neither is more fundamental than the other, but in abstract treatments of QM it is most often best to treat the canonical commutation relation $[X,P]=i\hbar$ as the foundational object; it is therefore most often treated as a postulate, and as such it cannot be "proved".

However, even when you make it that far, it is still important to keep an eye on the essential physics that it encodes, which is still just de Broglie's relation between momentum and wavelength.

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  • $\begingroup$ I disagree that you can "derive $P = -i\hbar\ d/dx$ from $[X, P] = i \hbar$". There are infinitely many other representations of the CCR that result in different wavefunctions but equivalent physics, as discussed in Shankar - e.g. $P = f(x) - i\hbar\ d/dx$ for an arbitrary function $f(x)$. $\endgroup$ – tparker Sep 15 '16 at 22:47
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    $\begingroup$ @tparker That's a trivial gauge change away, so in the absence of external requirements on the relationship between the canonical and kinematic momentum I stand by that as a general statement (with more mathematical refinement possible if required). In any case, I did my best to get a handle on the OP's starting position, which to be honest I feel your answer just blows straight past. $\endgroup$ – Emilio Pisanty Sep 16 '16 at 5:14

$[X, P] = i \hbar$ (times the identity operator) is a basic postulate of quantum mechanics called the canonical commutation relation, which cannot be derived. (Well, whether or not it can be derived depends on what you choose the starting postulates to be. Some people like to start from Dirac's quantization rule that classical Poisson brackets are replaced by $1/(i \hbar)$ times quantum commutators.)

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