1
$\begingroup$

The momentum of an object is in part dependent on the change in position meaning the final position minus the initial position. The equation for momentum is $$p=\frac{m \Delta x}{t\sqrt{1-(\Delta x/tc)^2}}$$ where $\Delta x$ means change in position, $c$ means speed of light, and $t$ means time. The uncertainty principle states that the uncertainty of position and the uncertainty of momentum are inversely related. How are the uncertainty principle and the equation for momentum consistent with each other?

| cite | improve this question | | | | |
$\endgroup$
  • 3
    $\begingroup$ This site supports mathjax. I edited the post to use it. Please use mathjax for mathematical expressions as this makes posts readable. $\endgroup$ – DanielSank Jan 16 '16 at 16:53
  • $\begingroup$ Your formula isn't the formula for "the" momentum (and what appears inside isn't "the" velocity) but it is a formula for the average momentum and/or velocity along the trajectory $\Delta x$. The uncertainty principle constrains the uncertainties of $x,p$ at the same moment. But even if one ignores this thing, e.g. by taking $\Delta x$ infinitesimal, there is no contradiction.The uncertainty principle simply says that $x$ and $\Delta x$ cannot have well-defined values at the same moment. One must distinguish $\Delta x$ as the uncertainty of $x$, and as the change of $x$ between the moments. $\endgroup$ – Luboš Motl Jan 16 '16 at 17:16
  • $\begingroup$ First, one must come to grips with, if one can, that quantum particles do not, in general, have either definite momentum or definite position. It's not that the particle's position is uncertain, i.e., definite but not exactly known, it's that the particle doesn't have a definite position (unless you subscribe to the Bohmian formulation) and similarly for momentum $\endgroup$ – Alfred Centauri Jan 16 '16 at 17:35
1
$\begingroup$

They are consistent with each other because quantum mechanical momentum is not the change in position.

There is no quantum notion of velocity. Classically, velocity is the time derivative of $x(t)$ along a particular trajectory. The quantum theory has no notion of a real-valued trajectory $x(t)$. Quantum states, in general, are not position eigenstates, they don't have a position from which you could compute the velocity.

Quantum mechanical momentum is the quantization of the classical Hamiltonian momentum, not of classical velocity. It is the Hamilton equation of motion $$ \dot{x} = \frac{\partial H}{\partial p} = \{q,H\}$$ that relates $\dot{x}$ to $p$ for the usual Hamiltonian quadratic in the momenta, but this equation is purely classical. It doesn't exist in the quantum theory except as an equation of expectation values (by Ehrenfest's theorem) $$ \frac{\mathrm{d}}{\mathrm{d}t}\langle x\rangle = \frac{1}{\mathrm{i}\hbar}\langle[x,H]\rangle$$ which makes no statement about the standard deviations.

There is another sense in which $\dot{x} = p$ holds in the quantum theory, which is that the quantum mechanical equation of motion in the Heisenberg picture is really the quantum version of the classical Hamiltonian equations of motion, i.e. all operators are time-dependent, and so writing $$ \dot{x} = \frac{i}{\hbar}[H,x]$$ does makes sense. However, the object $\dot{x}$ is not the classical notion of velocity - the time derivative is here not along a classical trajectory, but along the one-parameter group of operators $\{\mathrm{e}^{-\mathrm{i}tH}x\mathrm{e}^{\mathrm{i}tH}\}$ that are the time-evolution of the operator $x$. In particular, it does not make sense to speak of the velocity being the "change in position" divided by the time, because, again, most quantum states don't have a well-defined position and because something that is a position eigenstate at initial time will usually not stay one.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Apologies, I had to downvote it. The quantity "velocity" is exactly as well-defined as the position or the momentum, it's $\hat v = \hat p / m$ in non-relativistic physics, for example, and it is equal to the time derivative of the position. This is what the Heisenberg equation of motion says - and this equation holds as an operator equation. (One may describe physics in pictures not focusing on time dependence of operators but that only introduces confusion here.) $\endgroup$ – Luboš Motl Jan 16 '16 at 17:12
  • $\begingroup$ @LubošMotl: Hm, I would not say that it is obvious that the time derivative of the position operator in the Heisenberg picture corresponds to OP's idea that $p$ should be "change in position divided by time". You're right that $\dot{x} = p$ as operators for quandratic Hamiltonians is the Heisenberg equation of motion in the Heisenberg picture, but OP's concept of "velocity" is not that, so I "answered in the Schrödinger picture". $\endgroup$ – ACuriousMind Jan 16 '16 at 17:19
  • $\begingroup$ The equation $\dot x = v$ is the definition of the velocity - first written by Newton - and it is always true. In the Heisenberg picture of quantum mechanics, it holds as an operator equation. In non-relativistic physics, one has $p = mv$, again valid as the operator equation. In relativity, we would have $p = m_0 v / \sqrt{1-v^2/c^2}$ which would act as a nonlocal operators on the wave functions and it's not what we need in relativity - one of the reasons why we need to switch to quantum field theory to describe particles in a relativistic way. $\endgroup$ – Luboš Motl Jan 16 '16 at 17:26
  • $\begingroup$ I don't think that the OP has any wrong notion of velocity. Velocity is the final position over the initial position, and this difference divided by the small duration. It's only you, not the OP, who is questioning that. The OP is totally right that one may keep on using this understanding of velocity in QM. Talking about Schr. picture doesn't change anything - one may still define the form of the operator of velocity at a given time by converting the expression from the Heisenberg picture by the Heis-Schr transformation. $\endgroup$ – Luboš Motl Jan 16 '16 at 17:28
  • $\begingroup$ That squiggly brace in the first equation could be misinterpreted as an anticommutator. Maybe explicitly say it's a Poisson bracket. Leaves before getting hit by theory crossfire $\endgroup$ – DanielSank Jan 16 '16 at 17:32
1
$\begingroup$

The velocity $\hat v$ is a well-defined operator for particles, equal to $\hat p/m$ in non-relativistic physics. The non-relativistic, non-linear correction factors could be added but that would create a can of worms because the correct way to describe relativistic particles requires quantum field theory where $\hat x,\hat v,\hat p$ aren't quite well-defined because the number of particles may change.

In non-relativistic physics, it is also true that $\hat v = d \hat x / dt$, at least in the Heisenberg picture where this equation holds as an operator equation.

Now, the equation with $m\cdot \Delta x$ in the original question only holds if $\hat p$ means the "average" momentum (and similarly velocity etc.) during the relevant interval of time.

But we may ignore these things by taking $\Delta x$ infinitesimal. Then, ignoring the non-relativistic corrections, $$ p = \frac{\Delta x\cdot m}{\Delta t} $$ Here, the $\Delta$ means the "change of the quantity" in the infinitesimal time $\Delta t$.

The key point. This equation is in no way "incompatible" with the uncertainty principle. The uncertainty principle says that if we know $x$ accurately e.g. at the initial moment of the infinitesimal interval, we can't know $p$ accurately. Because $p$ is rewritten in terms of $\Delta x$, the uncertainty principle simply means that if we know $x$ at the initial moment accurately, we can't know $\Delta x$ accumulated in the short period of time.

If $\delta$ means the statistical uncertainty, we will have $$ \delta x \cdot \delta(\Delta x) \geq \frac{\hbar\cdot \Delta t}{2m} $$ For example, the uncertainty principle says that if the initial $x(t)$ were completely sharply determined, the position $x(t+\Delta t)$ would already be completely undetermined.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ I really think the main point to make here is that $\Delta x$ is ill-defined in the quanutm theory. You can't even make it infinitesimal - a position eigenstate will evolve into a non-position eigenstate even during an infinitesimal time interval. $\endgroup$ – ACuriousMind Jan 16 '16 at 17:29
  • $\begingroup$ It is not ill-defined in the quantum theory. $\Delta x$ is simply equal to $\hat x(t+\Delta t) - \hat x(t)$. Your comment about "eigenstates" is pure confusion. Almost no states are eigenstates of a given operator and there's nothing wrong about this fact - that's why the superposition principle says that all linear superpositions are equally allowed. One only gets an eigenstate after a measurement. But if one does a measurement, he still needs to know what operator he is measuring, and $\Delta x$ is as good an operator to measure as any other operator. $\endgroup$ – Luboš Motl Jan 16 '16 at 17:32
  • $\begingroup$ Ah, but then it's an operator! Maybe you should point that out in your answer more clearly, because I thought you were using the $\Delta x$ from the question, where it is the difference between the initial and final positions - a number. $\endgroup$ – ACuriousMind Jan 16 '16 at 17:39
  • 1
    $\begingroup$ All physical quantities ("observables") are expressed as operators in quantum mechanics. But this is just a fact about the mathematical formalism. We may still talk about the measured values of the quantities, phenomenologically, and that's what the OP did, and we may talk about these things without knowing that the observables are represented by operators. The operator representation of observables like $x,p$ is needed for a clear proof of the uncertainty principle but it is not needed for the formulation of the uncertainty principle or the formulation of questions as as the OP's question. $\endgroup$ – Luboš Motl Jan 16 '16 at 17:40
1
$\begingroup$

The uncertainty principle states that product of the standard deviations of two observables is bounded below by a multiple of the absolute value of the expectation value of the commutator of the two operators. In detail, if $\Delta A=\sqrt{\langle \Psi|\left(A^2-\langle \Psi|A|\Psi\rangle^2\right)|\Psi\rangle}$ and $\Delta B=\sqrt{\langle \Psi|\left(B^2-\langle \Psi|B|\Psi\rangle^2\right)|\Psi\rangle}$ then $$\Delta A \Delta B \geq \left|\frac{\langle AB-BA\rangle}{2i}\right|=\left|\frac{\langle\Psi| AB-BA |\Psi\rangle}{2i}\right| $$

For a canonical momentum conjugate to a specific coordinate, the commutator is a simple scalar and so has an expectation value that doesn't depend on the state and you can get the product of the two standard deviations to be bounded below by $\hbar/2.$

The canonical momentum only sometimes is equal to the mechanical momentum. And your formulation with a $t$ in the denominator is just not good at all. And they don't have to be inversely related to have a lower bound. To better understand the uncertainty principle you can read my answer to https://physics.stackexchange.com/a/169757

| cite | improve this answer | | | | |
$\endgroup$

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