Velocity operator, its expression and eigenvalues

[Kashmir]

Cohen Tannoudji pp 215

Third postulate :The only possible result of the measurement of a physical quantity $\mathscr A$ is one of the eigenvalues of the corresponding observable $\mathbf A$

pp 225

...there of course exists an operator associated with the velocity of the particle...

pp 223

The observable $\mathbf A$ which describes a classically defined physical quantity $\mathscr A$ is obtained by replacing, in the suitably symmetrized expression for $\mathscr A, \mathbf r$ and $\mathbf p$ by the observables $\mathbf R$ and $\mathbf P$ respectively

From above we can say that there exists a velocity operator $\mathbf v=\frac{\mathbf p}{m}$ ,whose eigenvalues are the observed values of velocity.

1. I've seen multiple times that we can't define velocity in quantum mechanics, but here I find that the eigenvalues of $\mathbf v$ can be the values of velocity 2. I've also seen that the velocity operator in Schrodinger picture is :

the $j$th velocity operator is defined as :
$\frac{1}{i\hbar}[\hat{q}^j, \hat{H}]$
But we also have another expression for the velocity operator as $\mathbf v=\frac{\mathbf p}{m}$, so how do we reconcile the two.

Please help me, Thank you.

 

[DrClaude]

1. I've seen multiple times that we can't define velocity in quantum mechanics, but here I find that the eigenvalues of $\mathbf v$ can be the values of velocity

Where have you seen that?

2. I've also seen that the velocity operator in Schrodinger picture is :

the $j$th velocity operator is defined as :
$\frac{1}{i\hbar}[\hat{q}^j, \hat{H}]$
But we also have another expression for the velocity operator as $\mathbf v=\frac{\mathbf p}{m}$, so how do we reconcile the two.

Calculate the commutator for, e.g.,
$$
\hat{H} = \frac{\hat{p}^2}{2m} + \hat{V}(x)
$$

 

[vanhees71]

In general if you have an observable $A$, represented by a self-adjoint operator $\hat{A}$, then the operator, which represents the time derivative $\dot{A}$ of this observable is given by
$$\mathring{\hat{A}}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}].$$
For the Hamiltonian for a particle moving in an arbitrary potential, you have
$$\hat{H}=\frac{1}{2m} \hat{p}^2+V(\hat{x})$$
and then
$$\hat{v}=\mathring{\hat{x}}=\frac{1}{\mathrm{i} \hbar} [\hat{x},\hat{H}]=\frac{1}{2 m \mathrm{i} \hbar} [\hat{x},\hat{p}^2] = \frac{1}{2m \mathrm{i} \hbar} ([\hat{x},\hat{p}] \hat{p} + \hat{p} [\hat{x},\hat{p}]) =\frac{1}{m} \hat{p},$$
i.e., the same relation between velocity and momentum as in classical mechanics.

I've never seen the statement that there's no velocity observable in non-relativistic quantum mechanics. Where does this claim come from?

 

[andresB]

In addition to the previous answers,

In the construction of non-relativistic quantum mechanics from its Galilean symmetries the meaning of $V$ as the velocity operator is given even before we can assign the meaning of momentum to $p$ and energy to $H$. So, there is no problem with the velocity operator.

 

[vanhees71]

In constructing non-relativistic QM from its Galilean symmetries, what's given are the commutation relations among the generators of the symmetry group modulo central charges, and according to Noether's theorem these are the energy (Hamilton), momentum, angular-momentum, and center-of-mass velocities. A careful analysis leads to the conclusion that there's one non-trivial central charge, which must not be set to 0 in order to achieve a physically interpretable dynamics, and that's the mass. In addition the only change to the classical Galilei group is that the rotation subgroup is substituted by its covering group, the SU(2).

What has to be constructed is the position operator, but this is no problem in non-relativistic physics. In the analogous case of the Poincare group in special relativity it's on the one hand a bit simpler, because there are no non-trivial central charges, and $m^2$ is a Casimir operator of the Lie algebra. On the other hand the construction of a position operator is not that simple. Of all the representations, which lead to physically useful (interacting) QFTs (i.e., admit a microcausality condition and thus a unitary S-matrix) the ones with $m^2>0$ ("massive particles") and $m^2=0$ ("massless particles") are to be chosen. For the massive case one can always construct a position operator. For the massless case that's only possible for particles with spin 0 or 1/2, but not for particles with spin $s \geq 1$.

 

[andresB]

In constructing non-relativistic QM from its Galilean symmetries, what's given are the commutation relations among the generators of the symmetry group modulo central charges, and according to Noether's theorem these are the energy (Hamilton), momentum, angular-momentum, and center-of-mass velocities.

That identification of the generators of the Galilei algebra with the usual physical quantities is way too hasty. I can give you a physically relevant Hilbert-space example where the generators of space-time transformations =/= the usual physical quantities: https://arxiv.org/pdf/2004.08661.pdf

 

[vanhees71]

Interesting, I'll have a look at the paper. I guess it's related to the case for a particle in a magnetic field, where the canonical momentum (the generator of translations) is not identical with the kinetical momentum, which is $m \vec{v}$.

 

[Demystifier]

I've seen multiple times that we can't define velocity in quantum mechanics

No, you have seen that we can't simultaneously define velocity and position in quantum mechanics. Eigenstates of the velocity operator are not eigenstates of the position operator, and vice versa.

 

[Kashmir]

How do we define velocity( not the operator) in quantum mechanics?
What do the eigenvalues of velocity operator mean? Is it $\frac {dr}{dt}$?

 

[vanhees71]

I've tried to answer this in #3 of this thread? What's still not satisfying you with this answer?

 

[Kashmir]

I've tried to answer this in #3 of this thread? What's still not satisfying you with this answer?

Why does position operator change with time? Isn't position operator independent of time?

 

[Kashmir]

I've tried to answer this in #3 of this thread? What's still not satisfying you with this answer?

Also one of the rules to get the quantum expression of a physical quantity is to replace r,p by the corresponding operators R, P.

So one expression for velocity should be $\frac{(d\hat R)}{dt}$. We dont use it, why is that?

 

[LittleSchwinger]

Also one of the rules to get the quantum expression of a physical quantity is to replace r,p by the corresponding operators R, P.

So one expression for velocity should be $\frac{(d\hat R)}{dt}$. We dont use it, why is that?

Note that in Classical Mechanics $p = m\frac{dx}{dt}$, but upon quantisation momentum is not simply "replaced" in that manner.

 

[vanhees71]

We had this question recently in the following thread:

Also one of the rules to get the quantum expression of a physical quantity is to replace r,p by the corresponding operators R, P.

So one expression for velocity should be $\frac{(d\hat R)}{dt}$. We dont use it, why is that?

That's true in the Heisenberg picture of time evolution. In this picture the "covariant time dervivate" is identical with taking the derivative of the observables wrt. time,
$$\mathring{\hat{A}(t)}=\frac{\mathrm{d}}{\mathrm{d} t} \hat{A}(t).$$
Note that in the usually used Schrödinger picture the observables are usually independent of time. That's why you need to introduce the "covariant time derivative" via
$$\mathring{\hat{A}}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}].$$

 

[LittleSchwinger]

That's true in the Heisenberg picture of time evolution

I think they might more be asking something like why is $v = -i\frac{\hbar}{m}\frac{\partial}{\partial x}$ and not $\frac{dx}{dt}$, even in the Schrodinger picture. In other words the OP might think quantisation allows direct replacement in classical expressions. Not sure though.

 

[vanhees71]

I see! Indeed, the main obstacle to teach/learn QM for the first time is that it is a pretty abstract description of Nature, and that's why we have all these (partiall heated) debates in the "interpretation forum" (and in the real world since the first formulation of QT and Born's probability interpretation of the quantum state). I've not yet an answer, what helps this first step into the "quantum world".

My first encounter with modern QM was in high school, where we learnt about the Schrödinger equation after having a big load of "old quantum theory" and "wave-particle duality". There we got heuristically the idea that momentum is $\hat{p}=-\mathrm{i} \hbar \partial_x$ and $\hat{x}=x$, but I was not really convinced. Anyway, the argument went as follows: From de Broglie's and Einstein's "wave-particle dualism" one got that $p=\hbar k=h/\lambda$ and $E=\hbar \omega=h \nu$. Since for a free particle $E=p^2/(2m)$ the dispersion relatio for the wave reads $\hbar \omega=\hbar^2 k^2/(2m)$. Now the wave equation that delivers this dispersion relation, when plugging in the plane-wave ansatz $\psi=\psi_0 \exp(\mathrm{i} k x-\mathrm{i} \omega t)$ is
$$\mathrm{i} \hbar \partial_t \psi=-\frac{\hbar^2}{2m} \partial_x^2 \psi,$$
and thus
$$\hat{H}=\frac{\hat{p}^2}{2m} \quad \text{with} \quad \hat{p}=-\mathrm{i} \hbar \partial_x.$$
That's very hand-waving, and one just got used to these association with operators and observables and solving of (of course only the simplest) eigenvalue problems.

Then at university, fortunately I had a professor who used Dirac's approach. The heuristics is of course much more abstract but quite straigh-forward. From classical mechanics one knows the Poisson brackets from Hamilton's canonical formalism, and the entire theory for a particle moving in one dimension came from the canonical Poisson brackets with the Poisson bracket written as a commutator, i.e., $\{A,B \} \rightarrow [\hat{A},\hat{B}]/(\mathrm{i} \hbar)$. This "canonical quantization" is most natural in the Heisenberg picture with the operators carrying the entire time dependence and the states are simply constant in time. This made the entire thing much more plausible then the hand-waving wave-mechanics heuristics, but of course it leads to the same theory at the end.

Today, I think the most convincing approach is to argue with the symmetries of the classical theory, i.e., with the Galilei group of classical mechanics and its unitary ray representations in Hilbert space. However, for this approach you already must have the idea of the Hilbert space formulation, and thus you can't use it in the introductory QM lecture or at least not in the beginning.

 

[LittleSchwinger]

Then at university, fortunately I had a professor who used Dirac's approach

My university often didn't offer QM outside of a hand-wavy "Modern Physics" course which simply presented some factoids. I did get it as a special course, but it involved a professor telling me to read Dirac and come back to him in a few months and take a bespoke exam.

So my professor also used the "Dirac Approach"

 

[vanhees71]

This is more an application of Russian didactics using Dirac's approach as the material to learn ;-)).

 

[andresB]

How do we define velocity( not the operator) in quantum mechanics?
What do the eigenvalues of velocity operator mean? Is it $\frac {dr}{dt}$?

Let us consider 1-dimensional motion and a state $\left|\psi(t)\right\rangle$ that obeys the Schodinger equation $$\frac{d}{dt}\left|\psi(t)\right\rangle =-i\hat{H}\left|\psi(t)\right\rangle. $$

At time t, the position expectation value is $\left\langle \psi(t)\right|\hat{X}\left|\psi(t)\right\rangle .$
This is a real number that gives a (expected) position in the axis. Now, this number will change with time since $\left|\psi(t)\right\rangle$ changes with time. The velocity operator $\hat{V}$ is defined such that the following equation holds

$$\left\langle \psi(t)\right|\hat{V}\left|\psi(t)\right\rangle =\frac{d}{dt}\left\langle \psi(t)\right|\hat{X}\left|\psi(t)\right\rangle. $$

 

[vanhees71]

Yes, and now you can use the first equation and its adjoint,
$$\mathrm{d}_t \langle \psi(t)|=+\mathrm{i} \langle \psi(t) \hat{H},$$
to work out the time derivative
$$\mathrm{d}_t \langle \psi(t)|\hat{V}|\psi(t) \rangle = \frac{1}{\mathrm{i}} \langle \psi(t)|[\hat{X},\hat{H}]|\psi(t) \rangle,$$
and accordingly we define the operator that represents the time derivative of $X$ by the commutator in the expectation value on the right-hand side,
$$\hat{V}=\mathring{\hat{X}} = \frac{1}{\mathrm{i}} [\hat{X},\hat{H}].$$
Note that we used natural units with $\hbar=1$ in this calculation.

 

[Kashmir]

In other words the OP might think quantisation allows direct replacement in classical expressions

The book writes how to obtain a quantum expression from a classical one as :
"The observable $\mathbf A$ which describes a classically defined physical quantity $\mathscr A$ is obtained by replacing, in the suitably symmetrized expression for $\mathscr A, \mathbf r$ and $\mathbf p$ by the observables $\mathbf R$ and $\mathbf P$ respectively "From above why cant we say that there exists a velocity operator given as $\mathbf v=\mathbf {\frac{d\hat R}{dt}}$

 

[andresB]

The book writes how to obtain a quantum expression from a classical one as :
"The observable $\mathbf A$ which describes a classically defined physical quantity $\mathscr A$ is obtained by replacing, in the suitably symmetrized expression for $\mathscr A, \mathbf r$ and $\mathbf p$ by the observables $\mathbf R$ and $\mathbf P$ respectively "From above why cant we say that there exists a velocity operator given as $\mathbf v=\mathbf {\frac{d\hat R}{dt}}$

You can, but only in the Heisenberg picture.

 

[Kashmir]

You can, but only in the Heisenberg picture.

So you say the quantization rules apply only in Heisenberg picture??

 

[LittleSchwinger]

So you say the quantization rules apply only in Heisenberg picture??

You have to keep in mind the Poisson bracket goes to commutator aspect of quantisation and express everything in terms of the canonically conjugate quantities of position and momentum.
I could similarly say momentum is $p = m\frac{dx}{dt}$. Making a direct substitution like that would give me the wrong momentum operator since it doesn't have commutator that matches the classical Poisson bracket of position and momentum.

Essentially quantisation takes place in the Hamiltonian framework. So before you make any replacements make sure your function is expressed purely in terms of $x,p$. This will eliminate $v$ and replace it with $\frac{1}{m}p$. Only then do you make direct replacements.

Even then you should know that this kind of simple direct replacement will only work for operators at most quadratic in $p$ due to Groenewold's theorem.

 

[vanhees71]

The book writes how to obtain a quantum expression from a classical one as :
"The observable $\mathbf A$ which describes a classically defined physical quantity $\mathscr A$ is obtained by replacing, in the suitably symmetrized expression for $\mathscr A, \mathbf r$ and $\mathbf p$ by the observables $\mathbf R$ and $\mathbf P$ respectively "From above why cant we say that there exists a velocity operator given as $\mathbf v=\mathbf {\frac{d\hat R}{dt}}$

That's right in the Heisenberg picture. In general the time derivative of an observable $A$, represented by a self-adjoint operator $\hat{A}(\hat{\vec{x}},\hat{\vec{p}},t)$, is represented by
$$\mathring{\hat{A}}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}]+\partial_t \hat{A}.$$
The $\partial_t$ only refers to the explicit time-dependence (if there is any). In the Schrödinger picture $\hat{\vec{x}}$ and $\hat{\vec{p}}$ do not depend on time at all. Only in the Heisenberg picture they carry the "full time dependence" of states and observable-operators.

 

[LittleSchwinger]

Maybe it might help Kashmir to say there isn't really a separate Schrödinger picture* for classical mechanics. Formulae like $v = \frac{dx}{dt}$ only have an analogue in the Heisenberg picture. If you try to use them to guess at what operators should be in the Schrödinger picture you'll get nonsense.

*Since states and observables are basically identified in classical mechanics.

 

[andresB]

So you say the quantization rules apply only in Heisenberg picture??

That's not any quantization rule I'm aware of. The usual procedure consists in changing the Poisson brackets for commutators, in the case of the velocity it reads

$$\dot{q}=\left\{ q,H\right\} \rightarrow\hat{V}=-i\left[\hat{X},\hat{H}\right].$$

The last expression is true in any picture.
Notice that we recover the expected relation between the velocity and the momentum (no vector potential for simplicity)
$$\hat{H}=\frac{\hat{P}^{2}}{2m}+\phi \rightarrow \hat{V}=\frac{\hat{P}}{m}.$$

 

[vanhees71]

QT is invariant under change of the picture of time evolution. The observable algebra is guessed from classical analogues, as far as they exist, using the symmetry principles defining these observables in the classical theory.

Indeed from this you can construct the position operator together with the momentum, angular-momentum, boost generators, Hamiltonian, and mass, whose commutator relations are determined by the structure of the "quantum Galilei symmetry group" (Bargmann group). For a single particle with spin $1/2$, it's sufficient to define the algebra for position and momentum, and all other observables are then given by the Bargmann algebra, i.e.,
$$[\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \hat{1}, \quad [\hat{x}_k,\hat{x}_j]=[\hat{p}_k,\hat{p}_j]=0.$$

There are certain rules. Among them is that the time derivative of an observable $A$ is given in any picture by
$$\mathring{\hat{A}}(\hat{\vec{x}},\hat{\vec{p}},t)=\frac{1}{\mathrm{i} \hbar} [\hat{A}(\hat{\vec{x}},\hat{\vec{p}},t),\hat{H}]+ \partial_t \hat{A}(\hat{\vec{x}},\hat{\vec{p}},t).$$
For
$$\hat{H}=\frac{\hat{p}^2}{2m}+\phi(\hat{x})$$
it follows indeed
$$\hat{V}=\mathring{\hat{x}}=\frac{1}{\mathrm{i} \hbar}[\hat{x},\hat{p}]=\frac{1}{m} \hat{p},$$
and that's independent of the picture of time evolution chosen.

The defining feature of the Heisenberg picture is that for all observable-operators
$$\mathring{\hat{A}}_{\text{H}}=\equiv \frac{\mathrm{d}}{\mathrm{d} t} \hat{A}_{\text{H}}.$$