1
$\begingroup$

This question already has an answer here:

From a video lecture on quantum mechanics at MIT OCW I found that you didn't need to know Schrödinger's equation to know the momentum operator which is $\frac{\hbar}{i}\frac{\partial}{\partial x}$. This can be derived from a 'simple' wave function of the type

$$ \psi = Ae^{i(\boldsymbol{\mathbf{k}}\cdot \boldsymbol{\mathbf{r}}- \omega t )} $$ Where we require eigenvalues for $\mathcal{\hat{p}}$ to be $\hbar k$ My questions are:

  • I understand the complex notation is for convenience. Since it's a complex exponential it'll give us a real and imaginary wave. Does the imaginary part have any physical significance? Are we to interpret this as two waves in superposition in the complex plane?
  • As we derive the expression for $\mathcal{\hat{p}}$ for this specific function, how does it guarantee that this is indeed the $\mathcal{\hat{p}}$ for every other arbitrary wave function? Can it be derived besides using the wave function I mentioned?
$\endgroup$

marked as duplicate by Qmechanic quantum-mechanics Nov 4 '16 at 16:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ This post asks two different questions. This makes it harder for people to write answers. A would-be answerer has to understand two things instead of one, and then decide to spend the time to write up explanations of two things instead of one. In general, you'll get more and better answers if you focus on one question per post. $\endgroup$ – DanielSank Nov 22 '15 at 20:56
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/45248/2451 and links therein. $\endgroup$ – Qmechanic Nov 22 '15 at 21:53
2
$\begingroup$

The momentum operator is not $-i\partial_x$, rather, that is the representation of the momentum operator on the position basis: namely $$ \langle x|\hat{p}|\psi\rangle = -i\frac{\partial}{\partial x}\psi(x). $$ Otherwise, the momentum operator is just defined by action on its eigenstates as $\hat{p}|p\rangle = p|p\rangle$.

I understand the complex notation is for convenience.

It is not for the sake of convenience. Complex functions emerge as solutions of the Schrödinger equation (plus arguments showing that quantum mechanics cannot be done on the real numbers only).

Since it's a complex exponential it'll give us a real and imaginary wave. Does the imaginary part have any physical significance? Are we to interpret this as two waves in superposition in the complex plane?

The wave terminology is somewhat unfortunate. There are no actual waves, rather the modulus square of the solution of the Schrödinger equation can be related to the probability density of the position operator for the point particle in quantum mechanics. Remember that $\psi(x) = \langle x |\psi\rangle$, where $|x\rangle$ is the position basis and $|\psi\rangle$ is the solution of the Schrödinger equation.

how does it guarantee that this is indeed the $p$ for every other arbitrary wave function

The form of the momentum operator on the position basis can be derived using the commutation relations only. That is, if the position operator $\hat{x}$ acts on a function as $(\hat{x}f)(x) = xf(x)$ then the only operator satisfying $([\hat{x},\hat{p}]f)(x) = if(x)$ can nothing but be $(\hat{p}f)(x) = -i\partial_x f(x)$.

$\endgroup$
  • $\begingroup$ Why do we require the commutator of position and momentum to be defined that way? In my textbook we arrived at the commutator relation by assuming the definition of $\mathcal{\hat{p}}$ in the position basis beforehand. $\endgroup$ – Weezy Nov 23 '15 at 13:03
  • $\begingroup$ It comes from the correspondence principle by which commutation relations in quantum mechanics are defined copying the equivalent Poisson brackets in Hamiltonian mechanics (divided by $i\hbar$). $\endgroup$ – gented Nov 23 '15 at 13:11
1
$\begingroup$

The momentum operator is the generator of shifts. In 3 dimensions ($\hbar =1$ ) \begin{equation*} (\exp [i\mathbf{a\cdot p}]f)(\mathbf{x})=f(\mathbf{x+a}) \end{equation*} Expanding in $\mathbf{a}$ \begin{equation*} i\mathbf{a\cdot p}f(\mathbf{x})=\mathbf{a\cdot }\partial _{\mathbf{x}}f(% \mathbf{x)} \end{equation*} or \begin{equation*} \mathbf{p}f(\mathbf{x})=\frac{1}{i}\partial _{\mathbf{x}}f(\mathbf{x)} \end{equation*} This does not depend on the particular $f$. In your case \begin{equation*} (\exp [i\mathbf{a\cdot p}]\psi )(\mathbf{x},t)=A\exp [i\mathbf{k\cdot (x+a}% -\omega t) \end{equation*} so \begin{equation*} \mathbf{p}\psi (\mathbf{x},t)=\mathbf{k}\psi (\mathbf{x},t) \end{equation*}

$\endgroup$
1
$\begingroup$

Does the imaginary part have any physical significance? Are we to interpret this as two waves in superposition in the complex plane?

In a sense neither the real part nor the imaginary part have physical significance, as these quantities do not directly appear in observables. One way to see this is that any solution $\left|\psi \right\rangle$ to Schroedinger's equation $$ i\hbar\frac{d}{dt}\left|\psi \right\rangle = H \left|\psi \right\rangle $$ is still a solution if multiplied by an arbitrary phase $e^{i\phi}$, where $\phi$ is a constant. Multiplying this phase also preserves the normalization condition which only depends on the mod-squared of the wave-function. Since we can choose any real valued phase that does not depend on position or time, this arbitrariness is called a continuous global phase symmetry. Though this is going a little off topic, this symmetry implies (via Noether's theorem) conservation of particle number.

As we derive the expression for $\hat{p}$ for this specific function, how does it guarantee that this is indeed the $\hat{p}$ for every other arbitrary wave function? Can it be derived besides using the wave function I mentioned?

The nice derivation you gave for the momentum operator shows that plain waves are eigenstates of the momentum operator, and in this way we specify the meaning of the momentum operator. In general, states are not eigenstates of momentum but can be expanded as a superposition of such eigenstates.

$\endgroup$
  • $\begingroup$ Are you in a sense implying Fourier expansion of arbitrary states in terms of these well defined states in your last statement? $\endgroup$ – Weezy Nov 22 '15 at 20:41
  • $\begingroup$ Also I couldn't understand the global phase symmetry part. Kindly elaborate? $\endgroup$ – Weezy Nov 22 '15 at 20:42
  • $\begingroup$ As for your first question, yes, Fourier expansion. I'll edit the post to make the phase symmetry idea completely concrete after I get back from taking my son to the park :) $\endgroup$ – dhudsmith Nov 22 '15 at 20:52
  • $\begingroup$ Yeah take your time! No problem. :) $\endgroup$ – Weezy Nov 22 '15 at 20:57
  • $\begingroup$ I made the edit. Let me know if its still not clear. $\endgroup$ – dhudsmith Nov 22 '15 at 22:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.