1
$\begingroup$

I was given the equation of the Lagrangian: \begin{equation} L~=~\frac{1}{2}m \dot{x}^2+\frac{e}{c}\vec{\dot{x}}\cdot \vec{A}(\vec{x},t)-e\phi (\vec{x},t). \end{equation} I proceeded to use the equation: \begin{equation} H~=~\sum_{i} \dot{q_i} \frac{\partial L}{\partial \dot{q_i}} -L \end{equation} to get the Hamiltonian as: \begin{equation} H~=~\frac{1}{2}m \dot{x}^2+e\phi (\vec{x},t), \end{equation} but, in the text, the Hamiltonian is given as: \begin{equation} \hat{H}~=~\frac{1}{2m}(\frac{\hbar}{i}\nabla - \frac{e}{c}\vec{A})\cdot(\frac{\hbar}{i}\nabla - \frac{e}{c}\vec{A})+e\phi \end{equation} So,why and where did I go wrong?

$\endgroup$
  • $\begingroup$ Remember that $H$ is a function of $p$, not $\dot{x}$. Your solution is not written in that form. The answer you approved does this. $\endgroup$ – garyp Mar 26 '15 at 13:57
1
$\begingroup$

To make the correct answer clearer, allow me to introduce the canonical momentum $\vec{p}$, given by: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}$$ This way we can rewrite the Hamiltonian as: $$H=\vec{p}\cdot\vec{\dot{x}}-L$$ Let's start by computing $\vec{p}$: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}=m\vec{\dot{x}}+\dfrac{e}{c}\vec{A}(\vec{x},t)$$ And you get: \begin{align}H&=m\dot{x}^2+\dfrac{e}{c}\vec{x}\cdot\vec{A}-\dfrac{1}{2}m\dot{x}^2-\dfrac{e}{c}\vec{\dot{x}}\cdot\vec{A}+e\phi\\&=\dfrac{1}{2}m\dot{x}^2+e\phi\end{align} But from the expression of the canonical momentum we found earlier, you can rewrite $\vec{\dot{x}}$ as: $$\vec{\dot{x}}=\dfrac{1}{m}\left(\vec{p}-\dfrac{e}{c}\vec{A}\right)$$ Such that: $$\dot{x}^2=\dfrac{1}{m^2}\left|\vec{p}-\dfrac{e}{c}\vec{A}\right|^2$$ Plugging this result into $H$: $$H=\dfrac{1}{2m}\left|\vec{p}-\dfrac{e}{c}\vec{A}\right|^2+e\phi$$ Make the transition to quantum mechanics by promoting the classical momentum $\vec{p}$ to the operator $\hat{p}=-i\hbar\nabla$ and you're done.

$\endgroup$
  • $\begingroup$ Nota: $-i=\frac{1}{i}$. $\endgroup$ – Demosthene Mar 26 '15 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.