# A Total derivative of momentum in quantum mechanics

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1. Apr 10, 2017

### kd6ac

In quantum mechanics, the velocity field which governs phase space, takes the form

\boldsymbol{\mathcal{w}}=\begin{pmatrix}\partial_tx\\\partial_tp\end{pmatrix}
=\frac{1}{W}\begin{pmatrix}J_x\\J_p\end{pmatrix}
=\begin{pmatrix}\frac{p}{m}\\-\frac{dV}{dx}-\sum\limits_{l=1}^\infty
\frac{(i\hbar/2)^{2l}}{(2l+1)!}\frac{\partial_p^{2l} W}{W}\frac{d^{2l+1}
V}{dx^{2l+1}}\end{pmatrix}\

For reference see Eq.(17) of https://journals.aps.org/pra/abstract/10.1103/PhysRevA.95.022127
The total derivative of p which is now a function in x, p and t, takes the form

\label{eq:Total_Derivative}
\frac{dp}{dt}=\partial_tp+\frac{dx}{dt}\partial_xp+\frac{dp}{dt}\partial_pp
=\partial_tp+\frac{p}{m}\partial_xp+\frac{dp}{dt}\; .

In the above equation

\frac{dx}{dt}=\frac{p}{m}\text{ and }\partial_pp=1\; .

My question is whether Eq.(\ref{eq:Total_Derivative}) is correct or not, and can I therefore conclude that

\partial_tp=-\frac{p}{m}\partial_xp=\frac{J_p}{W}\; .

2. Apr 10, 2017

### andrewkirk

It looks like equation 2 is suppose to be a version of a Total Derivative of p wrt t, where p is written as a function of p, x and t. That would be strange to express p as a function of itself. Are you sure that's what your source intended?

If it is then what we will actually have is $p=f(x,p,t)$ for some function $f:\mathbb R^3\to\mathbb R$ and the correct form of the total derivative will be:
$$\frac{dp}{dt}(x,p,t) \frac{df}{dt}(x,p,t)=\partial_tf(x,p,t)+\frac{dx}{dt}\partial_xf(x,p,t)+\frac{dp}{dt}\partial_pf(x,p,t)$$
which is different from what you wrote. In particular note that $\partial_pf(x,p,t)$ is unlikely to equal 1.

We can however deduce from the equation that
$$\frac{dp}{dt}(x,p,t)=\left(\partial_tf(x,p,t)+\frac{dx}{dt}\partial_xf(x,p,t)\right)\ /\ \left(1-\partial_pf(x,p,t)\right)$$
provided $\partial_pf(x,p,t)\neq 1$.

3. Apr 11, 2017

### kd6ac

Dear andrewkirk, thank you so much for looking into this.

I feel I should add a bit more background to the problem:

Wigner's representation of quantum mechanics combines Schr\"odinger's position and momentum representations. The respective continuity equation takes the form,

\label{eq:Continuity_Equation}
\partial_t W + \nabla\cdot\boldsymbol{J} = \partial_t W + \boldsymbol{w}\cdot\nabla W +
W\nabla\cdot\boldsymbol{w} = 0\; ,

where $W$ is the Wigner function (analogous to $\Psi$ in Schr\"odinger's representation) and the vector field $\boldsymbol{J} = W\boldsymbol{\mathcal{w}}$ is the Wigner current, which is the rate of flow per unit area of $W$. Rewriting Eq.(\ref{eq:Continuity_Equation}), yields the total derivative of $W$,

\label{eq:Total_Derivative_of_W}
\frac{dW}{dt} = \partial_t W + \boldsymbol{w}\cdot\nabla W = - W\nabla\cdot\boldsymbol{w}\; .

Therefore, the velocity field $\boldsymbol{w}=\boldsymbol{J}/W$ transports the values of $W$ and it takes the following form

\label{eq:Quantum_Mechanics_w}
\boldsymbol{\mathcal{w}}=\begin{pmatrix}\partial_tx\\\partial_tp\end{pmatrix}
=\frac{1}{W}\begin{pmatrix}J_x\\J_p\end{pmatrix}
=\begin{pmatrix}\frac{p}{m}\\-\frac{dV}{dx}-\sum\limits_{l=1}^\infty
\frac{(i\hbar/2)^{2l}}{(2l+1)!}\frac{\partial_p^{2l} W}{W}\frac{d^{2l+1}
V}{dx^{2l+1}}\end{pmatrix}\; .

For reference can see the article https://journals.aps.org/pra/abstract/10.1103/PhysRevA.95.022127

Note that the Wigner function $W$ is a function of $(x,p,t)$ and so in Eq.(\ref{eq:Quantum_Mechanics_w}), $\partial_t p$ is also a function of $(x,p,t)$. This is why I am lead to believe that $p=f(x,p,t)$, as you also suggested in your reply.

In Eq.(2) of my original post it is indeed the total derivative of $p=f(x,p,t)$ that I wanted to express but you are suggesting that my definition is wrong. Please be so kind and elaborate on this. As far as I can tell, if I replace $f(x,p,t)$ with $p$ in your definition of the total derivative then I get the left hand side equation of Eq.(2) in my original post.

Also the partial derivative of $f(x,p,t)$ wrt $p$ is treating $x$ and $t$ as constant values, i.e. they are fixed. So effectively when I partially differentiate $p=f(x,p,t)$ wrt $p$ then $p=f(p)$, but it is equal to itself, so $\partial_p f(x,p,t)=\partial_p f(p)=\partial_p p=1$. But you also disagree with this. Please be so kind and elaborate on this as well.

Last edited: Apr 11, 2017
4. Apr 11, 2017

### andrewkirk

The error is in that last piece of reasoning. The statement $\partial_p f(x,p,t)=\partial_p f(p)$ is meaningless, because $f(p)$ does not mean anything. Put differently, it is an illegal symbol string. $f$ is a function from $\mathbb R^3$ to $\mathbb R$, so it requires three inputs. The symbol string $f(a,b,c)$ indicates the real number (or complex number if that is the domain we are working in, which seems likely since it is QM) we get when we apply $a,b$ and $c$ as the 1st, 2nd and 3rd inputs to the function. The symbol string $f(p)$ has no interpretation.

A simple demonstration that $\partial_pf(x,p,t)$ is not necessarily equal to 1 is where $f$ is the constant function that always returns constant $C$, ie $\forall x\forall p\forall t:\ f(x,p,t)=C$. All three partial derivatives of this function are everywhere zero: $\partial_xf=\partial_pf=\partial_tf=0$.

You may find it helpful to read through this Insights article on partial differentiation. Partial differentiation is, in my experience, often taught poorly, and misunderstandings of how it works and what it means abound.

5. Apr 13, 2017

### kd6ac

Thank you for the link to partial derivatives. After reading it I realize that I might be wrong to think that $p=f(x,p,t)$. I think it will help if I show you what I am trying to do and for that reason allow me to talk about an equivalent case in classical mechanics.

Newtonian mechanics states that $F=\frac{dp}{dt}=-\frac{dV}{dx}$, which is the momentum component of the classical phase space velocity field,

\boldsymbol{u}=\begin{pmatrix}\frac{dx}{dt}\\\frac{dp}{dt}\end{pmatrix}
=\begin{pmatrix}\partial_p\mathcal{H}\\-\partial_x\mathcal{H}\end{pmatrix}
=\begin{pmatrix}\frac{p}{m}\\-\frac{dV}{dx}\end{pmatrix}\; .

Using the chain rule,

\label{eq:Chain_Rule_dpdt}
\frac{dp}{dt}=\frac{dx}{dt}\frac{dp}{dx}=\frac{p}{m}\frac{dp}{dx}=-\frac{dV}{dx}\; ,

where $m$ is the particle mass. Integrating in $x$, yields

\int dp\frac{p}{m}=\frac{p^2}{2m}=-V + \mathcal{H}\; ,

where $\mathcal{H}=\frac{p^2}{2m}+V$ is an integration constant and is known as the total energy of the system, referred to as the Hamiltonian.

Firstly do you agree with the above? Is my derivation of the Hamiltonian valid?

I want to do something similar in quantum mechanics and I now believe, instead of using some complicated expression for the total derivative of p, I can instead use Eq. (\ref{eq:Chain_Rule_dpdt}). What do you think? Am I allowed to?

6. Apr 13, 2017

### vanhees71

In usucal quantum mechanics ("first quantization") you have just the Heisenberg algebra (descrbing a particle neglecting its spin or a scalar particle), i.e.,
$$\frac{1}{\mathrm{i}}[\hat{x}_j,\hat{p}_k]=\delta_{jk}, \quad [\hat{x}_j,\hat{x}_k]=0, \quad [\hat{p}_j,\hat{p}_k]=0,$$
where I've used units with $\hbar=1$ for convenience. Then you have a Hamilton operator (representing energy). Usually one uses
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m} + V(\hat{\vec{x}}).$$
The time evolution of operators and states is quite arbitrary (it's known as the choice of the picture of time evolution). If you want to have the closest analogy to classical Hamiltonian mechanics, you can put all time dependence to the operators that represent observables. Then you get equations of motion as in the Poisson-bracket formalism of the classical theory by writing commutators (multiplied by $1/\mathrm{i}$) instead of poisson brackets and operators instead of phase-space functions, i.e., for the equations of motion you have
$$\frac{\mathrm{d}}{\mathrm{d} t} \hat{A}(t) = \frac{1}{\mathrm{i}} [\hat{A}(t),\hat{H}],$$
where $\hat{A}$ is any self-adjoint operator that represents an observable.

This is, of course a hand-waving rule, but a great heuristics to get started with QM with the least mathematical effort. It's called "canonical quantization". You come amazingly far with it, but one should be aware that a true understanding of the commutation relations are symmetry principles, which you should study after being somewhat familiar with the formalism.