Fourier conjugates and momentum

[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.