Maybe the title is not accurately stated.
There are few doubts about my question itself, these are-

  1. Is Kinetic Momentum an operator?
  2. I said "A Charged Particle", not "A Moving Charged Particle".
    The Kinetic Momentum is defined as $\vec{\pi}=\vec{p}-q\vec{A}$. But what if the Charged Particle is NOT MOVING? Will $\vec{p}=0$ then? So that $\vec{\pi}=-q\vec{A}$?

$\bullet$ Here is a question that asks: Find the $z-$component of $\mathbf{(\vec{\pi} \times \vec{\pi})}$, only information provided - the vector potential $\vec{A}=(A_{x},A_{y},A_{z})$ and magnetic field $\vec{B}=(B_{x},B_{y},B_{z})$.
Since this is a quantum mechanics question paper (so presumably the Hamiltonian ($H$), $\vec{p}$, $\vec{\pi}$ all are operators (but $\vec{A}$ is not; strange?)), I was attempting it by applying a test wave-function $\psi(\vec{r})$ on the right. So then- \begin{align} (\vec{\pi} \times \vec{\pi}) \psi({\vec{r})} &= [(\vec{p}-q\vec{A})\times(\vec{p}-q\vec{A})]\psi\\ &= (\vec{p} \times\vec{p})\psi - q(\vec{p}\times\vec{A})\psi-q(\vec{A}\times\vec{p})\psi+q^2(\vec{A}\times\vec{A})\psi \end{align}

Now $1^{st}$ term: $\vec{\nabla}\times\vec{\nabla}= \vec{0}$, $4^{th}$ term: $\vec{0}$.
$2^{nd}$ term gives (ignoring $(-q)$): $(\vec{p}\times\vec{A})= -i\hbar\vec{B}$,
And $3^{rd}$ term (ignoring $(-q)$): $(\vec{A}\times\vec{p})=\space \mathbf{??????}$

I am aware of the relations: $\vec{B}=\vec{\nabla}\times\vec{A}$, $\vec{\nabla}\cdot\vec{A}=0$ and $\vec{A}=-\frac{1}{2}(\vec{r}\times\vec{B})$. But I am unable to proceed any further.
The positioning of $\vec{\nabla}$ in the grad., div., curl identites may be one confusion. Maybe some other physics is involved?

What am I not understanding, or getting wrong?
Is $\mathbf{\vec{\pi}}$ an operator at all? Do I even need to consider the smooth function $\psi(\vec{r})$?

The answer, i.e. the $z-$component of $(\vec{\pi}\times\vec{\pi})$, is given as: $(\vec{\pi}\times\vec{\pi})_z=iq\hbar B_z$.

TL;DR: what is the value of $(\vec{\pi}\times\vec{\pi})$ mathematically? And what this physically is?


1 Answer 1


Yes it is an operator which is well defined by your formula by: $$ \vec \pi = \vec p-q\vec A $$ Btw, $\vec A$ is also an operator in the quantum theory, it's a function of the position operator (in QFT things get more complicated though).

The reason why it is natural to consider is first off that it is gauge invariant, so has more chances of being physically relevant compared to $\vec A, \vec p$ which are not. Second, a charged (non relativistic) particle has the following Hamiltonian (with $V$ the electric potential): $$ H = \frac{1}{2m}\vec \pi^2+qV $$ so in particular, from the equations of motion for velocity: $$ \vec v = \frac{1}{m}\vec \pi $$ (note that all of this is valid in a classical and quantum theory)

This answers your first question, when your particle is not moving, i.e. $\vec v = 0$, then $\vec \pi=0$.

The cross product is a more compact way of writing the commutators: $$ (\vec \pi\times\vec \pi)_k = \frac{1}{2}\epsilon_{ijk}[\pi_i,\pi_j] $$ which you can calculate directly using the canonical commutation relations (setting $\hbar=1$): $$ [x_i,p_j] = i\delta_{ij} \\ [x_i,A_j] = 0 \\ [p_i,A_j] = -i\partial_iA_j $$ so $$ [\pi_i,\pi_j] = -q([p_i,A_j]+[A_i,p_j])\\ =iq(\partial_iA_j-\partial_jA_i) \\ =iq\epsilon_{ijk}B_k $$ hence: $$ (\vec \pi\times\vec \pi)_k = iqB_k $$

These commutators are typically useful to know in order to deduce the equations of motion of $\vec \pi$.

Hope this helps.

  • $\begingroup$ So we can write any commutator, say [x,p], using cross product? Like, $(\vec{r}\times\vec{p})_k=\frac{1}{2}\epsilon_{ijk}[r_i,p_j]$? $\endgroup$ Sep 4, 2022 at 13:48
  • $\begingroup$ No it only works on a vector on itself. For example: $$[\vec L,\vec L] = i\vec L$$ In your example: $$(\vec r\times \vec p)_k = \epsilon_{ijk}r_ip_j$$ so there is no commutator (note that when $i\neq j$, $r_i,p_j$ commute so there is no ordering ambiguity) $\endgroup$
    – LPZ
    Sep 4, 2022 at 14:24
  • $\begingroup$ Everything is very understandable. Thanks!! Still I am wandering if you have written $[L_i,L_j]=i\epsilon_{ijk}L_k\space(\hbar=1)$ as $[\vec{L},\vec{L}]\stackrel{?}{=}i\vec{L}$? I mean, is it a standard way of writing these stuff? Thanks again! $\endgroup$ Sep 4, 2022 at 15:23
  • $\begingroup$ Yes sorry my bad, I meant: $$\vec L \times \vec L =i\vec L$$ $\endgroup$
    – LPZ
    Sep 4, 2022 at 16:04
  • $\begingroup$ In QM only, not classical mechanics (telling myself!). Many thanks!! $\endgroup$ Sep 4, 2022 at 16:26

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