I Fourier conjugates and momentum

[redtree]

Given that position and momentum are Fourier conjugates, what is the derivation for the equation $\hbar \vec{k} = m \vec{v}$, where momentum-space momentum is defined as $\hbar \vec{k}$ and position-space momentum is defined as $m \vec{v}$?

 

[PeterDonis]

Write down an eigenfunction in momentum space, i.e., a plane wave with a single value of $\vec{k}$. Then Fourier transform it into the position representation. What is the phase velocity of the resulting wave function in position space?

(Another way of seeing it: what is the representation of the momentum operator in position space?)

 

[redtree]

Just to make sure I understand: are you telling me to Fourier transform the following?:
$\langle \vec{k} |\phi \rangle = \phi(\vec{k})$

In general, I prefer to work in standard mathematical notation (as opposed to Dirac notation) so an explanation in that notation would be much better from my perspective. I hope that doesn't inconvenience you too much.

 

[redtree]

The momentum operator in position space is $-i \hbar \frac{\partial}{\partial \vec{x}}$ while in momentum space it is $\vec{k}$. How does that lead to $\hbar \vec{k} = m \vec{v}$?

 

[PeterDonis]

are you telling me to Fourier transform the following?

No, because you didn't follow your own advice:

I prefer to work in standard mathematical notation (as opposed to Dirac notation)

Which is a good idea for this problem. So write down the momentum eigenfunction in standard notation, then Fourier transform it to position space--or, equivalently, write down the momentum eigenfunction in the position representation. It will describe a plane wave. What is the phase velocity of that plane wave?

 

[redtree]

\begin{equation}
\begin{split}
\hat{\vec{p}} \psi(\vec{x})&= -i \hbar \frac{\partial}{\partial \vec{x}} \psi(\vec{x})
\end{split}
\end{equation}

Such that:
\begin{equation}
\begin{split}
\mathcal{F}\left[ -i \hbar \frac{ \partial}{\partial \vec{x}} \psi(\vec{x})\right]&=-i \hbar \frac{ \partial}{\partial \vec{x}} \int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i \vec{k} \vec{x}} \phi(\vec{k})
\end{split}
\end{equation}

Not sure what I should do next?

 

[redtree]

By convention: the phase velocity ($\vec{v}_P$) is defined as follows, where $\vec{p}= \hbar \vec{k}$ and $E = \hbar \omega$
\begin{equation}
\begin{split}
\vec{v}_P &\doteq \frac{\omega}{\vec{k}}
\\
&=\frac{\omega}{\frac{\vec{p}}{\hbar}}
\\
&=\frac{\hbar \omega}{\vec{p}}
\\
&=\frac{E}{\vec{p}}
\end{split}
\end{equation}

 

[PeterDonis]

By convention: the phase velocity ($\vec{v}_P$) is defined as follows

Yes, the phase velocity is $E / p$. You know $p$ for the plane wave. What is $E$ for the plane wave? (Note: since you've defined momentum in position space as $m v$, you are implicitly assuming non-relativistic QM. So the energy of the plane wave should be the non-relativistic energy.)

 

[PeterDonis]

Not sure what I should do next?

In momentum space, what is $\phi(k)$? Bear in mind that we are talking about a plane wave, which only has nonzero amplitude for a single value of $k$.

 

[redtree]

Yes, the phase velocity is $E / p$. You know $p$ for the plane wave. What is $E$ for the plane wave? (Note: since you've defined momentum in position space as $mv$, you are implicitly assuming non-relativistic QM. So the energy of the plane wave should be the non-relativistic energy.)

I defined momentum in wavenumber (momentum) space as $\hbar \vec{k}$. I haven't actually defined momentum in position space at all. I am trying to understand how defining momentum in wavenumber space as I have leads to a measure of momentum space where $\vec{p} = m v$

Given:
\begin{equation}
\begin{split}
\hat{E}&= i \hbar \frac{\partial}{\partial t}
\end{split}
\end{equation}

Such that:
\begin{equation}
\begin{split}
\frac{\hat{E}}{\hat{\vec{p}}}&=\frac{\left(i \hbar \frac{\partial}{\partial t}\right)}{\left(-i \hbar \frac{\partial}{\partial \vec{x}} \right)}
\\
&=-\left( \frac{\partial }{\partial t} \right)\left( \frac{\partial \vec{x}}{\partial} \right)
\end{split}
\end{equation}

I don't think the partials $\partial$ cancel.

 

[PeterDonis]

Given:

^E​

$$\begin{equation} \begin{split} \hat{E}&= i \hbar \frac{\partial}{\partial t} \end{split} \end{equation}$$​

Where are you getting that from? Bear in mind that we are considering a momentum eigenstate; such a state is time-independent, so $\partial / \partial t$ of anything is zero. But it has a well-defined, nonzero energy. What is that energy? (Hint: look at the time-independent, non-relativistic Hamiltonian for a free particle.)

 

[PeterDonis]

I haven't actually defined momentum in position space at all.

Sure you have. Defining it in momentum space automatically defines it in position space, via the Fourier transform. The Fourier transform of a momentum eigenstate to position space is found in most QM textbooks. Or you could even try solving for the momentum space eigenstate in position space directly: just look for some function $f(x)$ that satisfies the eigenvalue equation $\hat{p} f(x) = - i \hbar \partial f / \partial x = \hbar \vec{k} f(x)$.

 

[redtree]

Sure you have. Defining it in momentum space automatically defines it in position space, via the Fourier transform. The Fourier transform of a momentum eigenstate to position space is found in most QM textbooks. Or you could even try solving for the momentum space eigenstate in position space directly: just look for some function $f(x)$ that satisfies the eigenvalue equation $\hat{p} f(x) = - i \hbar \partial f / \partial x = \hbar \vec{k} f(x)$.

That is true. Defining momentum in momentum space does imply a definition in position space via the Fourier transform. I misstated the problem. It's the equivalence between $\hbar \vec{k}$ and $m \vec{v}$ that I don't see.

 

[PeterDonis]

It's the equivalence between $\hbar \vec{k}$ and #m \vec{v}## that I don't see.

See my question in post #11. Also, I misstated my earlier question about phase velocity, I should have asked about group velocity, which is $dE / dp$, not $E / p$. But either way you want a formula for $E$ in terms of $p$ in order to answer the question.

 

[redtree]

In momentum space, what is $\phi(k)$? Bear in mind that we are talking about a plane wave, which only has nonzero amplitude for a single value of $k$.

My understanding of a plane wave is that it has non-zero amplitudes for many values of $\vec{k}$. Given:
\begin{equation}
e^{i x}=\cos(x) + i \sin(x)
\end{equation}If a wavefunction is zero for all values of $\vec{k}$, except a single value $\vec{k}_0$, then the I would assume the following is true:
\begin{equation}
\begin{split}
\phi(\vec{k})&= \delta(\vec{k}-\vec{k}_0)
\end{split}
\end{equation}

Where:
\begin{equation}
\begin{split}
\mathcal{F}\left[\delta(\vec{k}-\vec{k}_0) \right]&=e^{ i \vec{k}_0 \vec{x}}
\\
&=\psi(\vec{x})
\end{split}
\end{equation}

Such that:
\begin{equation}
\begin{split}
-i \hbar \frac{ \partial}{\partial \vec{x}} \psi(\vec{x})&=-i \hbar \frac{ \partial}{\partial \vec{x}} e^{ i \vec{k}_0 \vec{x}}
\\
&= \vec{k}_0 e^{ i \vec{k}_0 \vec{x}}
\\
&= \vec{k}_0 \psi(\vec{x})
\end{split}
\end{equation}

 

[PeterDonis]

My understanding of a plane wave is that it has non-zero amplitudes for many values of $\vec{k}$.

I don't know where you're getting that from. A plane wave is a momentum eigenstate. That means it has only one value for momentum, hence only one value for $\vec{k}$.

I would assume the following is true:

ϕ(⃗k)​

$$\begin{equation} \begin{split} \phi(\vec{k})&= \delta(\vec{k}-\vec{k}_0) \end{split} \end{equation}$$​

Yes.

Such that

Not quite. The eigenvalue equation, as I said in my previous post, is

$$
\hat{p} \psi(\vec{x}) = - i \hbar \frac{\partial}{\partial \vec{x}} \psi(\vec{x}) = \hbar \vec{k} \psi(\vec{x})
$$

This is from the definition of momentum as $\hbar k$, the definition of an eigenstate/eigenvalue, and the Fourier transform. The function you wrote down does not solve that equation. But it is close.

 

[redtree]

As you may be noticing, I am trying to avoid utilizing the Hamiltonian in the derivation. It seems to me that the Hamiltonian assumes the very momentum equivalence (between $\hbar \vec{k}$ and $m \vec{v}$) that I am trying to prove.

As to the following:
\begin{equation}
\begin{split}
\hat{E}&= i \hbar \frac{\partial}{\partial t}
\end{split}
\end{equation}

For the purposes of our discussion, I derive that from the wave equation for a free particle as follows:
\begin{equation}
\begin{split}
\psi(\vec{x},t)&=A e^{i (\vec{k}\vec{x}-\omega t)}
\end{split}
\end{equation}

Such that:
\begin{equation}
\begin{split}
i \hbar \frac{\partial}{\partial t} \left[\psi(\vec{x},t)\right]&= i \hbar \frac{\partial}{\partial t}\left[A e^{i (\vec{k}\vec{x}-\omega t)} \right]
\\
&=\hbar \omega \psi(\vec{x},t)
\\
&=E \psi(\vec{x},t)
\end{split}
\end{equation}

 

[PeterDonis]

It seems to me that the Hamiltonian assumes the very momentum equivalence (between $\hbar \vec{k}$ and $m \vec{v}$) that I am trying to prove.

I don't know why you would think that. The Hamiltonian for a free particle doesn't mention anything at all about waves.

For the purposes of our discussion, I derive that from the wave equation for a free particle

What wave equation for a free particle? A free particle isn't a wave. At least, you can write down its Hamiltonian without having to say anything about waves, as I said above. And if you're looking for a way to derive $\hbar \vec{k} = m \vec{v}$, how do you expect to do it if you refuse to look at any formula that doesn't have $\vec{k}$ in it?

 

[redtree]

If you're going to use the Hamiltonian as a measure of $\hbar \omega$, don't you have to prove that? It's the same problem as assuming $\hbar \vec{k}= m \vec{v}$, i.e., $\hbar \omega = \frac{m \vec{v}^2}{2} + \vec{V}$.

Any formula that doesn't have $\vec{k}$ (or $\omega$ for that matter) needs to be related to $\vec{k}$ (or $\omega$), presumably via the Fourier transform.

 

[PeterDonis]

If you're going to use the Hamiltonian as a measure of $\hbar \omega$,

Who said we are going to do that? You keep on reading things into my posts that I haven't said. Write down the Hamiltonian for a free particle in terms of the momentum $p$, and then take its derivative with respect to $p$. What do you get?

 

[PeterDonis]

Also, with regard to the formula $E = i \hbar \partial / \partial t$ that you wrote down. When you write down the Hamiltonian for a free particle, it is an operator, and since you have it in terms of $p$, you can write down what the operator is in the position representation. What do you get? Hint: it's not $i \hbar \partial / \partial t$.

 

[redtree]

The Hamiltonian:
\begin{equation}
\begin{split}
\mathcal{H}&= \frac{\vec{p}^2}{2 m}+ \vec{V}
\end{split}
\end{equation}

Such that:
\begin{equation}
\begin{split}
\frac{\partial}{\partial \vec{p}}\left[\mathcal{H}\right]&= \frac{\vec{p}}{m}+ \frac{\partial \vec{V}}{\partial \vec{p}}
\end{split}
\end{equation}

 

[redtree]

You are right. Sorry. I was rushing.

 

[redtree]

For a free particle:
$\frac{\partial \mathcal{H}}{\partial \vec{p}}=\frac{\vec{p}}{m}$

 

[PeterDonis]

You are right.

Ok, I see you edited the post. So, we have that $dE / d\vec{p} = \vec{p} / m$. And this describes a free particle; we haven't said anything about waves. But if we treat $dE / d\vec{p}$ as the group velocity of a wave associated with the particle--such as a quantum wave function describing the particle--then we have that $\vec{v} = \vec{p} / m$, or $\vec{p} = m \vec{v}$. Now just equate that with the other formula for $\vec{p}$ that you already know from the rest of the stuff we did above.

 

[redtree]

Yes; that's good, but in this derivation, group velocity has been defined in terms of $\omega$ and $\vec{k}$ not $\vec{x}$ and $t$, where $\vec{v}_P=\frac{\omega}{\vec{k}}$ and $\vec{v}_G=\frac{\partial \omega}{\partial \vec{k}}$.

 

[redtree]

This derivation also utilizes two separate definitions of total energy $E = \hbar \omega$ and $E =\mathcal{H}$ and equates them without deriving the equality, i.e. $\hbar \omega = \mathcal{H}$.

 

[PeterDonis]

in this derivation, group velocity has been defined in terms of $\omega$ and $\vec{k}$

No, it hasn't, it's been defined in terms of $E$ and $p$. Those have representations in both momentum space and position space. You are focusing on the momentum space representations, but actually what I said is independent of any representation. Properly viewed, it's a relationship between operators; it's telling you that the wave number operator $\hat{k}$ must have a certain relationship to the group velocity operator $\hat{v}$. (A more rigorous derivation would also show that the group velocity operator, in the appropriate classical limit, gives the classical velocity of the free particle. A really rigorous derivation would also not use plane waves but wave packets, and would express the relationship in terms of expectation values.)

This derivation also utilizes two separate definitions of total energy $E = \hbar \omega$ and $E =\mathcal{H}$

No, it doesn't. It just defines "energy" as the Hamiltonian operator. It doesn't say anything about any particular representation of it. See above.

 

[redtree]

I get your point about operators having representations in position and momentum space, such that in whatever space one is working, the operator produces the observable. Thus, the wavenumber operator in position space is $-i \frac{\partial}{\partial \vec{x}}$ while in momentum space it is simply $\vec{k}$. The momentum operator is simply $\hbar$ times the wavenumber operator $\hat{\vec{k}}$. Thus, in either momentum or position space, the momentum operator is a function of the wavenumber operator. The position operator is essentially the opposite of the wavenumber operator, such that in position space $\hat{\vec{x}}=\vec{x}$ and in momentum space $\hat{\vec{x}}=i \frac{\partial}{\partial \vec{k}}$. I still don't see how momentum can be a function of group velocity unless group velocity is defined in terms of $\vec{k}$ and $\omega$. Where is there a form of the momentum operator where $\hat{\vec{p}}$ is a function of the position operator $\hat{\vec{x}}$?

 

[PeterDonis]

Where is there a form of the momentum operator where $\hat{\vec{p}}$ is a function of the position operator $\hat{\vec{x}}$?

There isn't. Neither of those operators is a function of the other.

I still don't see how momentum can be a function of group velocity

That's not what we showed. We showed that group velocity is a function of momentum--specifically, it's $\hat{p} / m$. Again, this is independent of the representation we choose for the operators.

 

[vanhees71]

Concerning the meaning of group velocity, see this posting:
https://www.physicsforums.com/threads/propagation-of-de-broglie-waves.789751/page-2#post-5854853

 

[redtree]

My understanding of group velocity. Given a Minkowski metric, where $\textbf{k}$ denotes 4-wavenumber and $\vec{k}$ denotes 3-space wavenumber:

\begin{equation}

\begin{split}

\textbf{k}^2&= \vec{k}^2 - \omega^2

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

\omega&=\sqrt{\vec{k}^2-\textbf{k}^2}

\end{split}

\end{equation}Given a definition of phase velocity $\vec{v}_P$:

\begin{equation}

\begin{split}

\vec{v}_P &\doteq \frac{\omega}{\vec{k}}

\end{split}

\end{equation}Such that group velocity $\vec{v}_G$:

\begin{equation}

\begin{split}

\vec{v}_G&=\frac{\partial }{\partial \vec{k}}\left[\omega \right]

\\

&=\frac{\partial }{\partial \vec{k}}\left[\sqrt{\vec{k}^2-\textbf{k}^2} \right]

\\

&=\frac{\vec{k}}{\sqrt{\vec{k}^2-\textbf{k}^2}}

\\

&=\frac{\vec{k}}{\omega}

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

\vec{v}_P \vec{v}_G &=1

\end{split}

\end{equation}

 

[PeterDonis]

Given a Minkowski metric

This is all irrelevant for this discussion since we have been using non-relativistic QM. We are not talking about QFT.

 

[redtree]

Yes; it is largely (though not totally) irrelevant; it was merely an aside based on the other post from vanhees71.

 

[redtree]

An operator is useful because in whatever state space one works, the operator will produce its associated vector (or scalar, vector combination, etc.). In this sense, it allows one to write equations independent of a given state space. However, the operator remains associated with its particular vector (or scalar, vector combination, etc.). Thus, $\hat{k}$ produces $\vec{k}$ in all state spaces. The same is true for $\hat{x}$ (producing $\vec{x}$). In terms of momentum, instead of defining momentum as $\vec{p}\doteq \hbar \vec{k}$, a better definition is probably the state-space independent definition: $\hat{p}\doteq\hbar \hat{k}$. In this context, the derivation we have been discussing is as follows:Given, where the operator $\hat{v_P}$ produces a ratio of a scalar over a vector, i.e., $\frac{\omega}{\vec{k}}$:

\begin{equation}

\begin{split}

\hat{v_P} &\doteq \hat{\left( \omega\frac{1}{\vec{k}}\right)}

\end{split}

\end{equation}Where:

\begin{equation}

\begin{split}

\vec{p}&=\hbar \vec{k}

\end{split}

\end{equation}And:

\begin{equation}

\begin{split}

E&=\hbar \omega

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

\hat{v_P}&= \hat{\left( \hbar \omega\frac{1}{\hbar \vec{k}}\right)}

\\

&= \hat{\left( E\frac{1}{\vec{p}}\right)}

\end{split}

\end{equation}Thus, the group velocity operator $\hat{v_G}$ acts to produce the ratio as follows:

\begin{equation}

\begin{split}

\hat{v_G}&=\hat{\left( \frac{\partial \omega}{\partial \vec{k}}\right)}

\\

&=\hat{\left( \frac{\hbar \partial \omega}{\hbar \partial \vec{k}}\right)}

\\

&=\hat{\left( \frac{\partial E}{\partial \vec{p}}\right)}

\end{split}

\end{equation}Assuming:

\begin{equation}

\begin{split}

E&=\mathcal{H}

\\

&=\frac{\vec{p}^2}{2 m}+V

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

\frac{\partial E}{\partial \vec{p}} &=\frac{\partial \mathcal{H}}{\partial \vec{p}}

\\

&=\frac{\vec{p}}{m}+\frac{\partial V}{\partial \vec{p}}

\end{split}

\end{equation}Assuming $\frac{\partial V}{\partial \vec{p}}=0$

\begin{equation}

\begin{split}

\frac{\partial E}{\partial \vec{p}} &=\frac{\vec{p}}{m}

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

\hat{v_G} &=\frac{\hat{p}}{m}

\\

\hat{p}&=m \hat{v_G}

\end{split}

\end{equation}I note the following issues. First, $\hat{v_G}=\hat{\left( \frac{\partial \omega}{\partial \vec{k}}\right)}$ and NOT $\hat{\left( \frac{\partial \vec{x}}{\partial t}\right)}$. The same definition for $\hat{v_G}$ must be maintained throughout the derivation, i.e. consistency in definitions. In this context, I note the following:

\begin{equation}

\begin{split}

\hat{p}&=m \hat{v_G}

\end{split}

\end{equation}Can be rewritten as follows, given $\hat{p}=\hbar \hat{k}$ and $\hat{v_G}=\frac{\partial \omega}{\partial \vec{k}}$ (again, consistency in definitions):

\begin{equation}

\begin{split}

\hbar \hat{k}&=m \frac{\partial \omega}{\partial \vec{k}}

\end{split}

\end{equation}In order to complete the derivation, one must show the following:

\begin{equation}

\begin{split}

\frac{\partial \omega}{\partial \vec{k}}&=\frac{\partial \vec{x}}{\partial t}

\end{split}

\end{equation}Unfortunately, I don't see how this can be done.

 

[PeterDonis]

In this context, the derivation we have been discussing is as follows

No, it isn't. You pulled $\omega$ out of nowhere, and you switched from operators back to vectors halfway through. Here is the derivation I gave earlier, restated for clarity (I've left the vector arrows out, but it is assumed that $\hat{p}$, $\hat{k}$, and $\hat{v_G}$ are vector operators, while $\hat{H}$ is a scalar operator):

$$
\hat{p} = \hbar \hat{k}
$$

$$
\hat{H} = \frac{\left( \hat{p} \right)^2}{2 m}
$$

$$
\hat{v_G} = \frac{d \hat{H}}{d \hat{p}} = \frac{\hat{p}}{m}
$$

Therefore,

$$
\hat{p} = m \hat{v_G} = \hbar \hat{k}
$$

Now, the above is just operators, and all of those equations will hold in any representation and for any state; the only assumption we have made is that we are dealing with a free particle, since there is no potential term in the Hamiltonian. Previously in this thread, we were specifically talking about momentum eigenstates, but as the above shows, we don't actually need to do that to derive $\hat{p} = m \hat{v_G}$.

However, since the above should apply to any state, we should be able to see how it applies to a momentum eigenstate. In the momentum representation, this is of course trivial: for a momentum eigenstate $\phi(p)$, we have $\hat{p} \phi(p) = \hbar \hat{k} \phi(p) = \hbar \vec{k} \phi(p)$, where now I've put the vector arrow back on in the last step to make it clear that $k$ there is an eigenvalue, not an operator. Therefore, since $\hat{v_G} = \hat{p} / m$, we have $\hat{v_G} \phi(p) = \left( \hbar \vec{k} / m \right) \phi(p)$.

In the position representation, the same operator and eigenvalue equations should hold, and we know that they do hold for a state $\psi(x) = exp \left( i \vec{k} \cdot \vec{x} \right)$, because we have $\hat{p} = - i \hbar \partial / \partial \vec{x}$, and therefore $\hat{H} = - \left( \hbar^2 / 2 m \right) \partial^2 / \partial \vec{x}^2$.

In order to complete the derivation, one must show the following

Before one can even talk about $\omega$ and $\vec{x}$, one needs to first show what operators correspond to them. For $\vec{x}$, this is easy: in the position representation, we have $\hat{x} \psi(\vec{x}) = \vec{x} \psi(\vec{x})$. But what operator corresponds to $\omega$?

Your suggestion, as far as I can see, is that we should define $\hat{H} = \hbar \hat{\omega}$, by analogy with $\hat{p} = \hbar \hat{k}$. If we do that, then we have $d \hat{\omega} / d\hat{k} = d \hat{H} / d \hat{p}$, and the expression for $\hat{v_G}$ stays the same as in the above derivation, which is still perfectly valid.

You then, however, say that we should have $d \hat{\omega} / d\hat{k} = d \hat{x} / dt$. But for a free particle, none of the operators we have given are time-dependent, so $d\hat{x} / dt$ is zero trivially; yet we still have a well-defined group velocity operator, as has been demonstrated above. So I'm not sure where this claim of yours is coming from.

To put this another way: you seem to be saying that we should have a "velocity operator" $\hat{v} = d \hat{x} / dt$. But what we really want is to justify the idea that, in the classical limit, whatever "velocity" we are getting from quantum mechanics should become the ordinary classical velocity we are used to for a free particle, which is $d\vec{x} / dt$. But that's not a statement about operators; it's a statement about expectation values, since those are what should correspond to classical quantities in the classical limit. So what we really want to show is that $\langle \hat{v_G} \psi \rangle = d \langle \hat{x} \psi \rangle / dt$.

We can't actually do this for a momentum eigenstate, because it doesn't have a well-defined expectation value for position. But we can do it for a wave packet, and I think (based on what I remember of reading in QM textbooks quite a while back, but I have not had time to check) that for a wave packet, the expectation value equation comes out as I have just said.

 

[vanhees71]

There is the following postulate in QT: If $\hat{A}$ represents the observable $A$ and $\hat{H}$ is the Hamilton operator of the system, then
$$\mathring{\hat{A}}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}]$$
represents the time derivative $\dot{A}$.

For a non-relativistic particle you usually have
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m} + V(\vec{x}),$$
and thus the operator representing velocity is
$$\hat{\vec{v}}=\mathring{\hat{\vec{x}}}=\frac{1}{\mathrm{i} \hbar} [\hat{\vec{x}},\hat{H}]=\frac{1}{m} \hat{\vec{p}},$$
which hardly is a surprise ;-)).

 

[redtree]

I think a lot of the misunderstanding here comes down to problems with notation.For example: $\hat{v_G}=\hat{\left( \frac{\partial \omega}{\partial \vec{k}}\right)}$ does NOT mean $\partial \hat{\omega}/\partial \vec{k}$. Rather it denotes an operator that produces a ratio of Eigenvalues (or equivalently, the derivative one Eigenvalue, i.e., $\omega$, with respect to another Eigenvalue, i.e., $\vec{k}$). Given that most operators are written in forms that evoke the Eigenvalue that they produce, i.e., $\hat{k} \rightarrow \vec{k}$, I was merely trying to represent his particular ratio (or derivative) operator $\hat{v_G}$ in a form that evokes the ratio (or derivative) it produces in generic state space, i.e., $\frac{\partial \omega}{\partial \vec{k}}$.The point remains. A ratio (or derivative) operator (``group velocity'') has been defined such that it produces the ratio (or derivative) $\frac{\partial \omega}{\partial \vec{k}}$ (which incidentally $=\frac{\vec{k}}{\omega}$) in a state space. Just as $\hat{k}$ produces $\vec{k}$ in all state spaces. One cannot assert that $\hat{v_G}$ produces $\frac{\partial \omega}{\partial \vec{k}}$ in some state spaces and $\frac{\partial \vec{x}}{\partial t}$ in others.Therefore, assuming $\hat{E}=\hat{\mathcal{H}}$, where $\hat{\mathcal{H}}=\frac{\hat{p}^2}{2m}$ for a free particle, we agree:

\begin{equation}

\begin{split}

\hat{p}&=\hbar \hat{k}

\\

&=m \hat{v_G}

\end{split}

\end{equation}Where, as I assert, the operator $\hat{v_G}$ ALWAYS produces the ratio (or derivative) $\frac{\partial \omega}{\partial \vec{k}}$ in state space.Perhaps my understanding is incorrect, but I have always understood that in stating $\vec{p}= m \vec{v}$, velocity generally refers to the ratio (or derivative) $\frac{\partial \vec{x}}{\partial t}$, which is NOT what we have derived, unless one assumes (which I do NOT assume) that $\frac{\partial \omega}{\partial \vec{k}}=\frac{\partial \vec{x}}{\partial t}$.

 

[redtree]

Parenthetically:

You then, however, say that we should have $d \hat{\omega} / d\hat{k} = d \hat{x} / dt$. But for a free particle, none of the operators we have given are time-dependent, so $d\hat{x} / dt$ is zero trivially; yet we still have a well-defined group velocity operator, as has been demonstrated above. So I'm not sure where this claim of yours is coming from.

$\omega$ is the Fourier conjugate of time $t$, just as $\vec{k}$ is the Fourier conjugate of position $\vec{x}$, which means that any equation with $\omega$ is not time independent. This is clear given: $e^{i \textbf{k} \textbf{r}} = e^{i (\vec{k}\vec{x}-\omega t)}$, where $\textbf{k}=[\vec{k}, i \omega]$ and $\textbf{r}=[\vec{x}, i t]$.Yes, I define energy $E \doteq \hbar \omega$ in a similar fashion to momentum $\vec{p}\doteq \hbar \vec{k}$.

 

[PeterDonis]

For example: $\hat{v_G}=\hat{\left( \frac{\partial \omega}{\partial \vec{k}}\right)}$ does NOT mean $\partial \hat{\omega}/\partial \vec{k}$.

That's correct, because you wrote $\vec{k}$ instead of $\hat{k}$. Note that when I wrote it earlier, I specifically wrote $\hat{k}$.

Rather it denotes an operator that produces a ratio of Eigenvalues (or equivalently, the derivative one Eigenvalue, i.e., $\omega$, with respect to another Eigenvalue, i.e., $\vec{k}$).

I'm not sure this is correct; I think it might be if you said "expectation value" instead of "eigenvalue".

$\omega$ is the Fourier conjugate of time $t$, just as $\vec{k}$ is the Fourier conjugate of position $\vec{x}$,

This is not correct. First, you should be writing operators, not vectors. Second, time $t$ doesn't correspond to any operator, so it can't have a Fourier conjugate. Even though the formal analogy you give is tempting, it's not valid.

This is clear given: $e^{i \textbf{k} \textbf{r}} = e^{i (\vec{k}\vec{x}-\omega t)}$, where $\textbf{k}=[\vec{k}, i \omega]$ and $\textbf{r}=[\vec{x}, i t]$.

To the extent this makes sense, it only makes sense in relativistic QM, not non-relativistic QM. In non-relativistic QM, there is no such thing as 4-vectors (i.e., vectors that combine "time" and "space" components), but that's what you're writing down.

I define energy $E \doteq \hbar \omega$ in a similar fashion to momentum $\vec{p}\doteq \hbar \vec{k}$.

This is OK as far as operators go, but, as noted above, it doesn't mean the $\hat{\omega}$ operator has a Fourier conjugate in non-relativistic QM, because there is no "time operator" the way there is a position operator.

 

[PeterDonis]

I have always understood that in stating $\vec{p}= m \vec{v}$, velocity generally refers to the ratio (or derivative) $\frac{\partial \vec{x}}{\partial t}$,

Where did you get this understanding from? Can you give a reference? (It should be evident from this discussion that my understanding is different, at least with regard to non-relativistic QM, as opposed to classical mechanics.)

 

[vanhees71]

In classical physics, indeed $\vec{v}=\dot{\vec{x}}$. I explained in #37 the relation to the operator valued representants of observables in quantum theory.

One should, however caution the OP, that $\hat{\vec{p}}$ refers to canonical, not mechanical, momentum. This becomes clear when using the formalism for a particle moving in a magnetic field, but we should first try to understand the formalism, before discussing such subtleties!

 

[redtree]

In classical physics, indeed $\vec{v}=\dot{\vec{x}}$. I explained in #37 the relation to the operator valued representants of observables in quantum theory.

One should, however caution the OP, that $\hat{\vec{p}}$ refers to canonical, not mechanical, momentum. This becomes clear when using the formalism for a particle moving in a magnetic field, but we should first try to understand the formalism, before discussing such subtleties!

The generalized coordinates of Hamiltonian/Lagrangian mechanics are generalized momentum, generalized position, generalized velocity (the time derivative of generalized position) and time. It's not clear to me how this solves the question I am asking. No specific position frame has ever been assumed in the preceding discussion.

 

[redtree]

Where did you get this understanding from? Can you give a reference? (It should be evident from this discussion that my understanding is different, at least with regard to non-relativistic QM, as opposed to classical mechanics.)

My understanding of this relationship comes from Hamiltonian/Lagrangian mechanics and the Legendre transform.

 

[vanhees71]

The generalized coordinates of Hamiltonian/Lagrangian mechanics are generalized momentum, generalized position, generalized velocity (the time derivative of generalized position) and time. It's not clear to me how this solves the question I am asking. No specific position frame has ever been assumed in the preceding discussion.

Of course, to be able to write down the components of the position vector you need to define a frame, and in QT you assume an inertial frame. The operators $\hat{\vec{x}}$ are representing the components of the position vector with respect to such a frame (involving a point and a Cartesian basis in the 3D affine Euclidean manifold defining "space" in Newtonian physics.

Maybe I misunderstood your question, which I read as if you were asking, how to represent velocities of particles in non-relativistic quantum theory (it's non-relativistic, because in relativistic physics the construction of position observables is much less trivial, and I thought it's good to stick to non-relativsitic physics first).

 

[PeterDonis]

The generalized coordinates of Hamiltonian/Lagrangian mechanics are generalized momentum, generalized position, generalized velocity (the time derivative of generalized position) and time.

No, they aren't. They're generalized position and generalized momentum, period. Velocity is not a coordinate, it's a time derivative of one; and time is not a coordinate at all, it's a parameter.