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I've seen the 'derivation' as to why momentum is an operator, but I still don't buy it. Momentum has always been just a product $m{\bf v}$. Why should it now be an operator. Why can't we just multiply the wave function by $\hbar{\bf k}$? Why should momentum be a derivative of a wave function?

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Could you clarify what exactly you "do not buy"?And may I offer a hint: I give you a wavefunction $\psi(x)$. What $k$ are you going to use in multiplying it by $\hbar k$? –  Slaviks Aug 7 '13 at 16:33
Advice: don't fight with QM. Accept it and give it some time to settle. You'll either gradually develop some QM intuition, or not. Don't try to apply classical concepts to QM - it will burn your brain. –  Vasiliy Aug 7 '13 at 16:41
Do you accept that position and momentum don't commute? The uncertainty principle? The path integral? –  santa claus Aug 7 '13 at 16:46
If, by wavefunction, you mean $\psi(k,t)$, it works like you think: $P\psi(k,t) = k\psi(k,t)$ (in units $\hbar =1$), but it does not work with $\psi(x,t)$. To find the correct operator, just consider that $\psi(x,t)$ is the Fourier transform of $\psi(k,t)$ –  Trimok Aug 7 '13 at 16:54
Could you cite the "derivation"? @Trimok 's comment is the best answer: QM is not doing anything mysterious and indeed is agreeing with you here. So think of QM as being a generalization of your thinking: you can write the wavefunction in co-ordinates that agree with classical interpretations, but there are also other co-ordinates derived from your special momentum space ones. Unitary transformations relate all these co-ordinates, so they all hold the same information ... –  WetSavannaAnimal aka Rod Vance Aug 7 '13 at 23:57

3 Answers 3

The simplest possible solution to the Schrodinger equation is that of a plane wave: $$ \psi = e^{i(\boldsymbol{\mathbf{k}}\cdot \boldsymbol{\mathbf{r}}- \omega t )} $$ In this scenario, the position of the system is indeterminate, but the momentum is known. Next, take the gradient of this wavefunction. $$ \nabla \psi = ik \psi$$ But we know that $p = \hbar k $, from De Broglie. Substituting this into the equation and rearranging yields $$ \frac{\hbar}{i} \nabla \psi = p \psi $$ This is an example of an eigenvalue equation - an operator acts on a function or vector to produce a constant multiple of the original function. So, we see that the eigenvalue is the measured momentum. Therefore, the operator producing it, $ \frac{\hbar}{i} \nabla $, is called the momentum operator.

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I've seen this before, but what does it mean for momentum to be acting on psi? Does that mean that you're multiplying by hbar k? –  Anthony Aug 8 '13 at 4:31
@Tony No, it doesn't. Applying the momentum operator to the wavefunction is equivalent, as shown above, to taking the gradient and then multiplying by $ \hbar \over i $. There is still a momentum that you can measure - but that is an eigenvalue of the momentum operator. –  Mark M Aug 8 '13 at 4:39

As opposed to classical mechanics, where a particle is represented by a momentum and position vector that have determined directions and magnitudes, in Q.M. the particle is in a "superposition" of different positions and momentums. So it has a whole bunch of positions and momentums, all at the same time. As Slaviks pointed out, multiplying by a specific momentum value would no longer work, as you would have to choose one of the several (generally very many, mostly infinite) possible momenta. In the cases where a particle is in a pure momentum state, where it has only one momentum, you can indeed just as well multiply by the momentum instead of applying the momentum operator.

The momentum operator is useful for finding the average momentum should you repeatedly measure the momentum of particles that each have an identical wave function:

$$\overline{P}=<\psi|-i\hslash\partial_x|\psi>= \\ \sum\limits_{All \space momentum \space states \space \psi_i \\ with \space momentum \space P_i \space that \space \psi \space is\space composed\space of}<\psi_i|P_i|\psi_i> = \sum\limits_{All\space momenta \space P_i \space contained \space in \space \psi }(Probability\space of \space finding \space P_i)\cdotp(P_i)$$

Each individual measurement will not give the average value, but one specific momentum value. That's because measurement has caused the "collapse of the wave function" culling all of the $\psi_i$ except one. When you measure that specific particle's momentum again, you will again find the same momentum. It is now in a pure momentum state, for example $\psi_{123}$.

Operating on a mixed-momentum wave function with the momentum operator without bra-keting (averaging over momenta) isn't something very useful, because it will give you a new wave function than the one you started with (not corresponding to a state that you are dealing with), multiplied by a quantity with the unit of a momentum.

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Consider a theory with certain dynamical objects of interest. This can be the particle trajectory $x(t)$ in classical mechanics, or the wave fucntion $\psi(x,t)$ in the quantum theory.

By Noethers theorem, if certain conditions on the dynmaics of these objects hold, then an infinitesimal pertubation which is characterized by derivatives $Q$ of the object involved in the transformation, gives you a conserved quantities $I$. The quantity which is conserved because of time invariance is what you call "energy", that which is conserved because of translation invariance is what you call "momentum" and so on. To see why the spatial derivative pops up when taling about the momentum is to see why how fields come together with translations.

I don't know your math background, but essentially "$I\propto Q$". Still roughly speaking, but in more technical terms, you obtain $I$ by multiply but the conjugate momentum, which is different for every theory and then integrate over. This section of the wikipedia page on Noethers theorem suggest how in classical mechanics, where the dynamical object of interest is the trajectory $x(t)$, the different conserved observables are computed from the respective transformations. Now the example of translation in classical mechanics isn't particularly eluminating for our purposes here, because the momentum happens to coincides with the conjuage momentum. But in the following case, you can for example see how the cross product in the angular momentum arises from the cross product involved in the infinitesimal change if you do a rotation.

Now if you work with a field $f$, then the Taylor series at the point $x$ is given by

$f(y) = \sum_{n=0}^{\infty} \frac{1}{n!}\frac{\partial^n f(x)}{\partial x^n } (y-x)^n.$

Set $y=x-\lambda$ and you get

$f(x-\lambda)=\sum_{n=0}^{\infty} \frac{1}{n!} \frac{\partial^n f(x)}{\partial x^n } ((x-\lambda)-x)^k=\sum_{n=0}^{\infty} \frac{1}{n!}\left(-\lambda \frac{\partial }{\partial x }\right)^n f(x),$

which expresses translation of the field as an operator. The Taylor expansion uses the derivatives at a point to reconstruct the values of the function at other points and this is why $\frac{\partial }{\partial x }$'s suffice to describe translations. Here a little demonstration on the polynomial $p(x):=2x-bx^3$, the second and the last line giving the same result: http://i.imgur.com/VuWbVg3.png

Physicists also like to write the above formula as

$f(x-\lambda)=\text{e}^{-i \lambda\ (-i\frac{\partial}{\partial x})}f(x),$

because $-i\frac{\partial}{\partial x}$ is hermitean, i.e. mathematically symmetric in the space where physical states live in.

The quantity which is conserved because of translation invariance (which we like to call momentum), is given by the infinitesimal version of this, so "$I\propto Q\propto \frac{\partial}{\partial x}f(x)$".

The above is also true for a classical field theory. If you want to apply this to QM now, consider the Schödinger theory with $f=\psi$. The canonical momentum is given as $-i\psi$, giving the conserved quantity

$I=\int (-i\psi(x))\cdot \frac{\partial}{\partial x}\psi(x),$

which you can read as the expectation value $\int \psi(x)\left(-i\frac{\partial}{\partial x}\right)\psi(x)$.

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