# Can you make a definition of velocity in QM ?

• gurhkn
In summary, there is no one clear and agreed upon definition of velocity in QM. Some suggest using the momentum operator divided by mass, while others argue that this definition is not valid for all scenarios. One common result is that the velocity operator for an electron in QED is given by c times the Dirac alpha matrix. Ultimately, the concept of velocity in QM is still an area of debate and further research is needed.
gurhkn
Can you make a definition of velocity in QM ?

I am trying to find a general definition of velocity in QM. QM is totally different world !

Why not the "same" as in classical mechanics. It is the time derivative of position. So the x-component of the velocity is dx/dt, which by Heisenberg's equations is some constant times the comutator [x,H].

gurhkn said:
Can you make a definition of velocity in QM ?

I am trying to find a general definition of velocity in QM. QM is totally different world !

$$v_{QM} \equiv \frac{dx_{QM}}{dt}$$

Sorry to maybe be a little pedantic, but are those definitions satisfying? When we write
$$\frac{d\hat{x}}{dt}=\frac{i}{\hbar}[\hat{x},\hat{H}]$$
we have some idea in our head that something, 'x', is changing by such and such amount in time, 't'. But what is 'x' talking about? It's easy when we talk about a whole set-up and system that we are describing and we say "that system is translating." Are we asking how fast the approximate location of a particle is changing? I think the question can be much deeper, and I'm not sure if the OP had this in mind, but I would love to see a good definition of velocity for QM. For me it seems that the best at hand is actually
$$v^\mu \equiv \frac{p^\mu}{m}$$
which might seem bland and not anything new to the discussion, but when it comes down to it, in QFT calculations and the study of say, the Dirac equation, momentum is the main quantity one looks at (especially in scattering), and so it seems like a good place to at least look for a velocity. But, in general I also would like to see a solid definition, or if what has been given above is satisfactory, I feel like I haven't been given the whole deal about what it means to 'move' in QM.

gurhkn said:
Can you make a definition of velocity in QM ?

I am trying to find a general definition of velocity in QM. QM is totally different world !

I basically agree with jfy4: if you're studying a particle of mass m, the velocity is p/m, where p is the momentum. I think this relation is true for both the operator $\hat{p}$ and the eigenvalue p. Note that the commutator of position and velocity operators is not zero, so there is an uncertainty relation for x and v, but since the commutator of momentum and velocity is zero, there is no uncertainty relation of momentum and velocity. In other words, velocity acts like momentum in quantum mechanics.

BBB

jfy4 said:
Sorry to maybe be a little pedantic, but are those definitions satisfying? When we write
$$\frac{d\hat{x}}{dt}=\frac{i}{\hbar}[\hat{x},\hat{H}]$$
we have some idea in our head that something, 'x', is changing by such and such amount in time, 't'. But what is 'x' talking about? It's easy when we talk about a whole set-up and system that we are describing and we say "that system is translating." Are we asking how fast the approximate location of a particle is changing? I think the question can be much deeper, and I'm not sure if the OP had this in mind, but I would love to see a good definition of velocity for QM. For me it seems that the best at hand is actually
$$v^\mu \equiv \frac{p^\mu}{m}$$
which might seem bland and not anything new to the discussion, but when it comes down to it, in QFT calculations and the study of say, the Dirac equation, momentum is the main quantity one looks at (especially in scattering), and so it seems like a good place to at least look for a velocity. But, in general I also would like to see a solid definition, or if what has been given above is satisfactory, I feel like I haven't been given the whole deal about what it means to 'move' in QM.

The definition of velocity in QM was given above.

Your is $v^i \equiv {p^i}/{m}$ is not valid for a low velocity particle in presence of a magnetic field for instance. It is also wrong for a Dirac electron. The application of the quantum mechanical definition gives the well-known result

$$v^i = c \alpha^i$$

for the velocity operator of an electron in QED.

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juanrga said:
The definition of velocity in QM was given above.

Your is $v^i \equiv {p^i}/{m}$ is not valid for a low velocity particle in presence of a magnetic field for instance. It is also wrong for a Dirac electron. The application of the quantum mechanical definition gives the well-known result

$$v^i = c \alpha^i$$

for the velocity operator of an electron in QED.

This is new news to me! I'm not even sure what $\alpha^\mu$ is!? care to expand a little for me? or where can I see a reference for what you just wrote?

jfy4 said:
This is new news to me! I'm not even sure what $\alpha^\mu$ is!? care to expand a little for me? or where can I see a reference for what you just wrote?

It is Dirac alpha matrix. Any reference in QED uses that velocity operator for the interaction of an electron with the vector potential.

Okay, thanks, I'll try and read up a little more on that.

Edit: wait! you mean the gamma matrices $\gamma^\mu$! I am completely familiar with those! your referencing $\bar{\Psi}\gamma^\mu \Psi$ as the current 4-vector with $\bar{\Psi}\Psi$ as the charge density correct? In which case we agree completely. I meant that in calculating scattering amplitudes (say the H diagram) the final amplitude $\mathcal{M}$ is expressed in terms of initial momenta and energy, and that we normally associate these with how the electron is moving. How do you feel about that statement?

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jfy4 said:
Okay, thanks, I'll try and read up a little more on that.

Edit: wait! you mean the gamma matrices $\gamma^\mu$! I am completely familiar with those! your referencing $\bar{\Psi}\gamma^\mu \Psi$ as the current 4-vector with $\bar{\Psi}\Psi$ as the charge density correct? In which case we agree completely. I meant that in calculating scattering amplitudes (say the H diagram) the final amplitude $\mathcal{M}$ is expressed in terms of initial momenta and energy, and that we normally associate these with how the electron is moving. How do you feel about that statement?

No I said alpha matrix.

The charge density is $q\Psi^\dagger \Psi$

and the associated current $q\Psi^\dagger c\alpha^i \Psi$

this is often rewritten as $cq \bar{\Psi} \gamma^i \Psi$, where $\bar{\Psi} \equiv \Psi^\dagger \beta$ and $\gamma^i \equiv \beta \alpha^i$.

$\beta$ is Dirac beta matrix and satisfies $\beta^2 =1$

The four current is $cq \bar{\Psi} \gamma^\mu \Psi$ if you introduce a $\gamma^0 \equiv \beta$

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Ok, then never heard of them, but they look like they fit in pretty similarly. Thanks again.

The alpha and beta matrices are the original matrices used by Dirac in 1928 in his famous articles. They are part of the so-called non-covariant version of Dirac's equation. Most books dealing in one way or another with the Dirac equation from a historical perspective use them.

jfy4 said:
This is new news to me! I'm not even sure what $\alpha^\mu$ is!? care to expand a little for me? or where can I see a reference for what you just wrote?

For example, Dirac's book: Principles of quantum mechanics (in the chapter where Zitterbewegung is discussed). Or http://en.wikipedia.org/wiki/Zitterbewegung

juanrga said:
No I said alpha matrix.

The charge density is $q\Psi^\dagger \Psi$

and the associated current $q\Psi^\dagger c\alpha^i \Psi$

this is often rewritten as $cq \bar{\Psi} \gamma^i \Psi$, where $\bar{\Psi} \equiv \Psi^\dagger \beta$ and $\gamma^i \equiv \beta \alpha^i$.

$\beta$ is Dirac beta matrix and satisfies $\beta^2 =1$

The four current is $cq \bar{\Psi} \gamma^\mu \Psi$ if you introduce a $\gamma^0 \equiv \beta$

m is a positive parameter in the Dirac equation. However the actual mass, i.e. the energy of a particle at rest, is -m for a negative energy solution. Hence this paradox can be avoided if we replace m by $m\beta$. Then instead of the alpha matrices, the gamma matrices appear, as they should.

Now to the problem of how to define velocity in QM: Beginners often think that it is necessary to perform two position measurements at two (infinitesimally) different times to measure velocity and then wonder how that should be possible as the velocity operator doesn't commute with the position operator. However, only the difference has to be measured which can be done without absolute measurement of position. In fact, this is what happens e.g. in the Doppler effect, which is probably the most routinely used method to measure the velocity of qm particles.

Given juanrga's definition of QM velocity is correct, that is v^i = c(alpha)^i, is it true that

<v^i> = <dx/dt> ?

I'm just interested because the other day I went through the Ehrenfest Theorem which was used to show that d/dt<p_x> = -<dV/dx> (partial for V, but I suck at Latex).

Averages follow from the operator equation, obviously.

Pages 19-->26 of B. Thaller's <The Dirac equation> describe the issues with the velocity and position operators for the Dirac equation.

juanrga said:
The definition of velocity in QM was given above.

Your is $v^i \equiv {p^i}/{m}$ is not valid for a low velocity particle in presence of a magnetic field for instance. It is also wrong for a Dirac electron. The application of the quantum mechanical definition gives the well-known result

$$v^i = c \alpha^i$$

for the velocity operator of an electron in QED.

Thanks for your clarification; I agree that this formulation is correct for the relativistic case. Risking the fallacy of "appeal to authority", I would like to offer this excerpt from Bjorken and Drell's Relativistic Quantum Mechanics, p.37:

Now we come to an important difference in the relativistic theory. In the Schrödinger theory the velocity operator appearing in the current is just p/m and is a constant of motion for free particles. The current is not, however, proportional to the momentum in the Dirac theory, and whereas the Ehrenfest relation (1.27) has shown that dp/dt =0 for free-particle motion, the velocity operator cα is not constant, since [α,H]≠0. Indeed in constructing eigenfunctions of cα we have to include both positive- and negative-energy solutions, since the eigenvalues of cαi are ±c whereas |<cαi>+|<c, according to (3.29).​

So, just to reiterate: for the (non-relativistic) Schrödinger equation, it is perfectly fine to use p/m for the velocity operator.

BBB

bbbeard said:
Thanks for your clarification; I agree that this formulation is correct for the relativistic case. Risking the fallacy of "appeal to authority", I would like to offer this excerpt from Bjorken and Drell's Relativistic Quantum Mechanics, p.37:

Now we come to an important difference in the relativistic theory. In the Schrödinger theory the velocity operator appearing in the current is just p/m and is a constant of motion for free particles. The current is not, however, proportional to the momentum in the Dirac theory, and whereas the Ehrenfest relation (1.27) has shown that dp/dt =0 for free-particle motion, the velocity operator cα is not constant, since [α,H]≠0. Indeed in constructing eigenfunctions of cα we have to include both positive- and negative-energy solutions, since the eigenvalues of cαi are ±c whereas |<cαi>+|<c, according to (3.29).​

So, just to reiterate: for the (non-relativistic) Schrödinger equation, it is perfectly fine to use p/m for the velocity operator.

BBB

It depends what you mean by the non-relativistic Schrödinger equation. If you mean strictly that the equation does not contain c, then you are right. However, usually people calls non-relativistic also to the Schrödinger equation for a charged particle in an electromagnetic field {*}

$$H = \frac{(p - eA/c)^2}{2m} + e\phi$$

In that case the velocity operator is $\frac{p - eA/c}{m}$.

{*} They mean that the electron is being treated non-relativistically with a quadratic momenta kinetic Hamiltonian

http://quantummechanics.ucsd.edu/ph130a/130_notes/node484.html

Of course rigorously it cannot be non-relativistic because contains c.

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## Question 1: What is the definition of velocity in quantum mechanics?

The definition of velocity in quantum mechanics is the rate of change of a particle's position with respect to time. It is a vector quantity that includes both the magnitude and direction of motion.

## Question 2: How does velocity differ in classical mechanics and quantum mechanics?

In classical mechanics, velocity is a well-defined value that can be measured precisely. However, in quantum mechanics, velocity is described by a wave function that represents the probability of finding a particle at a certain position and time. This means that velocity in quantum mechanics is a probabilistic concept.

## Question 3: Can velocity be measured in quantum mechanics?

No, velocity cannot be measured directly in quantum mechanics. This is because the act of measurement in quantum mechanics changes the state of the particle, making it impossible to measure its velocity accurately.

## Question 4: How is velocity related to momentum in quantum mechanics?

In quantum mechanics, momentum is defined as the product of mass and velocity. This means that velocity and momentum are closely related, and changes in one will affect the other.

## Question 5: What are the units of velocity in quantum mechanics?

The units of velocity in quantum mechanics are typically represented in terms of the Planck constant (h) and the mass of the particle. The most common unit for velocity in quantum mechanics is meters per second (m/s).

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