The definition of velocity in the de Broglie relation

[redtree]

I apologize ahead of time for the simplicity of the question, but this has really been bothering me.Given the de Broglie relation, assuming natural units, where $\hbar = 1$:

\begin{equation}

\begin{split}

\vec{k} &= M \vec{v}

\end{split}

\end{equation}My question regards velocity and what is meant by $\vec{v}$. In classical physics, $\vec{v}= \frac{d \vec{x}}{dt}$. However, in quantum physics, position and momentum are Fourier conjugates, whose variance is related by the Uncertainty Principle. Mathematically, I don't see how it is possible that a variable and its derivative can be Fourier conjugates. Therefore, it seems $\vec{v}$ represents a different type of velocity. My understanding is that it represents group velocity ($\vec{v}_G$) related to phase velocity $\vec{v}_P$ as follows:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}And:

\begin{equation}

\begin{split}

\vec{v}_G &= \frac{d \omega}{d \vec{k}}

\end{split}

\end{equation}Such that, in this case, ``velocity'' is a function of momentum (rather than position), and there is no problem with velocity and position as Fourier conjugates.Unfortunately, I don't think I have ever seen this distinction in velocity mentioned in discussions of quantum physics, and worse, I have often seen classical and group velocity used interchangeably in the mathematics of quantum physics.What am I missing?

 

[stevendaryl]

Yes, you're right, that the velocity $v = \frac{p}{m}$ in quantum mechanics is the group velocity, rather than the phase velocity. Are you asking how to understand that?

I think that you can see it by thinking of velocity as an operator. In quantum mechanics, observables such as momentum, position, etc., are associated with operators, and the expectation value of the observable depends on the wave function as follows:

$\langle O \rangle = \int \psi^* \hat{O} \psi\ dx$

So if you have a wave function that varies with time, then the expectation value of position is given by:

$\langle x(t) \rangle = \int \psi^*(x,t) \hat{x} \psi(x,t)\ dx$

If you want to have a velocity operator $\hat{v}$, then presumably you would want to have its expectation value relate to that of $\hat{x}$ as follows:

$\langle v(t) \rangle = \frac{d}{dt} \langle x(t) \rangle$

This implies that

$\langle v(t) \rangle = \int [(\frac{d \psi}{dt})^*\ \hat{x} \psi + \psi^*\ \hat{x}\ (\frac{d \psi}{dt})]\ dx$ (using the convention that the operators are constants, while the time dependence is in the wave function---that's the Schrodinger picture).

That isn't of the form $\int \psi^* \hat{O} \psi\ dx$, so it's not obvious what the velocity operator should be. However, we can manipulate the integral using Schrodinger's equation:

$i \hbar \frac{d\psi}{dt} = H \psi$

$-i \hbar \frac{d \psi^*}{dt} = H \psi^*$

where $H$ is the hamiltonian operator. Rewriting the integral in terms of $H$, we have:

$\langle v(t) \rangle = \int [\frac{1}{-i \hbar} (H \psi^*)\ \hat{x} \psi + \frac{1}{i\hbar} \psi^*\ \hat{x}\ (H \psi)]\ dx$

At this point, we use a fact about Hermitian operators (and $H$ is one):

$\int (H \psi^*) \phi\ dx = \int \psi^* (H \phi)\ dx$

(assuming $\psi$ and $\phi$ are well-behaved). So letting $\phi = \hat{x} \psi$, we have:

$\int (H \psi^*)\ \hat{x} \psi\ dx = \int \psi^*\ H (\hat{x}\ \psi)]\ dx$

Plugging this into our equation, we get:

$\langle v(t) \rangle = \int [\frac{1}{-i \hbar} \psi^*\ (H \hat{x} \psi) + \frac{1}{i\hbar} \psi^*\ \hat{x}\ (H \psi)]\ dx$
$= \int \psi^*\ \frac{1}{-i \hbar} (H \hat{x} - \hat{x}\ H) \psi\ dx$

So now we have our $\hat{v}$ operator:

$\langle v(t) \rangle = \int \psi^*\ \hat{v}\ \psi\ dx$

where $\hat{v} = \frac{1}{-i \hbar} (H \hat{x} - \hat{x}\ H)$

That combination, $H \hat{x} - \hat{x} H$ is called a commutator, and is written: $[H,\hat{x}]$.

Obviously, the same reasoning goes for any operator other than $\hat{x}$. For any time-independent operator $\hat{O}$, we have:

$\hat{\frac{dO}{dt}} = \frac{1}{-i \hbar} [H,\hat{O}]$

For a free particle, we can use $H = \frac{- \hbar^2}{2m} \frac{\partial^2}{\partial x^2}$, and the commutator gives: $\hat{v} = \frac{1}{m} (-i \hbar \frac{\partial}{\partial x})$

 

[stevendaryl]

I guess my long-winded explanation doesn't actually explain the original ideas behind the De Broglie relations. That was before Schrodinger's equation and the idea of observables as operators.

I think that De Broglie was just trying to understand quantum phenomena in terms of waves, and assumed that a particle traveling at a constant velocity had an associated wave $\psi(x,t) = e^{i (kx - \omega t)}$. Generalizing from Planck's and Einstein's quantization condition for monchromatic light, $E = \hbar \omega$, De Broglie assumed that momentum had a similar rule: $p = \hbar k$. Then I think it was just a matter of assuming that the classical velocity-momentum relation held: $v = \frac{p}{m} = \frac{\hbar k}{m}$. The fact that this turns out to be the group velocity, $\frac{d\omega}{dk}$ doesn't seem immediately obvious to me. I don't know, historically, whether De Broglie assumed $v = \frac{p}{m}$ and then showed that this was the group velocity, or assumed that $v$ was the group velocity and then showed that this was $\frac{p}{m}$.

 

[redtree]

My real question is why in Quantum Mechanics it so often seems that velocity as $\vec{v}=\frac{d \vec{x}}{dt}$ and as $\vec{v}=\frac{d \omega}{d \vec{k}}$ seem to be used interchangeably, when it seems so obvious that they cannot be. One blatant example is the time slicing derivation of the path integral.

 

[stevendaryl]

My real question is why in Quantum Mechanics it so often seems that velocity as $\vec{v}=\frac{d \vec{x}}{dt}$ and as $\vec{v}=\frac{d \omega}{d \vec{k}}$ seem to be used interchangeably, when it seems so obvious that they cannot be. One blatant example is the time slicing derivation of the path integral.

$v = \frac{dx}{dt}$

is true by definition of velocity. To interpret the equation in quantum mechanics is not immediate, because $v$ and $x$ become operators, not values, but you can recover that definition in quantum mechanics by using the Heisenberg picture (where the state is considered time-independent and the operators are considered time-dependent). In the Heisenberg picture, you have the time evolution for an operator given by: $\frac{dx}{dt} = \frac{1}{-i\hbar} [H,x] = \frac{1}{+i\hbar} [x,H]$

So this definition leads to $v = \frac{dx}{dt} = \frac{1}{i\hbar} [x,H]$

If $H = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)$, then this leads to:

So: $\v = \frac{-i \hbar}{m} \frac{\partial}{\partial x} = \frac{\hat{p}}{m}$

The fact that $v = \frac{d\omega}{dk}$ is not true by definition, but instead is derivable. If you write $E = \hbar \omega$ and $p = \hbar k$, then it is equivalent to $v = \frac{\partial E}{\partial p}$, which is true classically. Quantum-mechanically, you can make that into an operator equation by replacing $E$ by the operator $H$ and by using the following fact: If $A(p,x)$ is an operator that involves both momentum and position, then

$\frac{\partial A}{\partial p} = \frac{1}{i \hbar} [\hat{x}, A]$
$\frac{\partial A}{\partial x} = \frac{1}{-i \hbar} [\hat{x}, A]$

So: $\frac{d\omega}{dk} \rightarrow \frac{dE}{dp} \rightarrow \frac{\partial H}{\partial p} \rightarrow \frac{1}{i \hbar} [x,H]$

So the two equations for $v$ agree as operator equations.

The path integral approach to quantum mechanics is very different. You don't start out with $x$ and $v$ and $p$ as operators, but instead are considering paths of the form $x = f(t)$. Along a path, $v$ is just defined to be $\frac{df}{dt}$ and $p$ is just defined to be $m v$ (or more generally, $p = \frac{\partial L}{\partial x}$ where $L$ is the Lagrangian). I'm not sure where $v = \frac{d\omega}{dk}$ comes into play in path integrals.

 

[bobob]

$p = \hbar k$, $E = \hbar\omega$ and $E = p^2/2m$,

So, $\hbar\omega = \hbar^2 k^2/2m$
$\omega = \hbar k^2/2m$
$\frac {d\omega} {dk} = 2\hbar k/2m = p/m = v$

 

[redtree]

Time slicing derivation of the path integral: http://hitoshi.berkeley.edu/221a/pathintegral.pdf

Momentum used as $m \vec{v}$ and $\hbar \vec{k}$ interchangeably, where $\vec{v} = \frac{d\vec{x}}{dt}$.

 

[redtree]

$p = \hbar k$, $E = \hbar\omega$ and $E = p^2/2m$,

So, $\hbar\omega = \hbar^2 k^2/2m$
$\omega = \hbar k^2/2m$
$\frac {d\omega} {dk} = 2\hbar k/2m = p/m = v$

I like this. I've got to think on this one. Thanks!

 

[PeroK]

Time slicing derivation of the path integral: http://hitoshi.berkeley.edu/221a/pathintegral.pdf

Momentum used as $m \vec{v}$ and $\hbar \vec{k}$ interchangeably, where $\vec{v} = \frac{d\vec{x}}{dt}$.

I only mention this because no one else has. You do, in fact, have the relation:

$\langle p \rangle = m\frac{d\langle x \rangle}{dt}$

In terms of the path-integral formulation, the link above says "by identifying $\dot{x} = \frac{x_1 - x_0}{\Delta t}$". At each step in the limiting process, they take the classical path as the best guess for filling in the gaps.

The book I have, Sakurai, justifies this to some extent by showing that in this way he gets the same propagator for the free particle as he has already derived directly from the Schroedinger equation.

Sakurai also says: "... this is not meant to be a derivation. Rather we (or Feynman) have attempted a new formulation of QM based on paths ...".

He then shows that the path integral formulation so generated is equivalent to the Schroedinger equation.

So, I think Sakurai is clear that he has borrowed an idea from classical mechanics to motivate the path integral formulation. But, once he has his formulation, he then proves that is is equivalent to Schroedinger. Whatever dirty tricks he used to get the formulation in the first place don't matter then!