Lagrangian density of Schroedinger's Equation and problem with momentum

Question

This question is certainly related to the one here from several years ago, but is not answered there.

The Lagrangian density of the Schrödinger's Eqn can be expressed as

$$ \mathscr{L} = \frac{i\hbar}{2}(\psi^*\dot{\psi}-\dot{\psi}^*\psi)-\frac{\hbar^2}{2m}\nabla\psi\cdot\nabla\psi^* -V(x)\psi^*\psi . $$

And, as far as I can tell, when you put this into the usual Lagrangian equation,

$$ \frac{\partial}{\partial t}\frac{\partial\mathscr{L}}{\partial\dot{\psi}^*} + \partial_i \frac{\partial\mathscr{L}}{\partial \psi^*_{,i}}-\frac{\partial\mathscr{L}}{\partial \psi^*}=0$$ you do indeed get the Schrödinger's eqn for $\psi$.

The problem is that now I want to find $\pi$, the momentum conjugate to $\psi$, which I read everywhere is $i\hbar\psi^*$. But try as I might, when I try to make the calculation $\pi = \partial\mathscr{L}/\partial \dot{\psi}$, I get a factor of 1/2 in the result: $\partial\mathscr{L}/\partial \dot{\psi}=i\hbar\psi^*/2$. Needless to say, this messes up my efforts to construct the Hamiltonian and to try to obtain the S.E. from that. What am I missing?

To add to my confusion, can add $i\hbar(\psi^*\dot{\psi}+\dot{\psi}^*\psi)/2$ to the Lagrangian here, because the Lagrangian operator operating on it gives zero. This gives the Lagrangian

$$ \mathscr{L} = {i\hbar}\psi^*\dot{\psi}-\frac{\hbar^2}{2m}\nabla\psi\cdot\nabla\psi^* -V(x)\psi^*\psi. $$

This also gives the correct S.E. and the correct momentum $\pi$. So is this somehow the "correct" Lagrangian? If so, how are we to know that a priori, and not just after we see that it give the correct momentum? (And I'm sure the first form also gives the correct momentum if I could just see how.)

UPDATE: So, apparently, my calculation of $\pi_\psi$ is correct. (It happens about twice a year for me.) And there is also $\pi_{\psi^*}$ to think about, which I can also get. This leads to the Hamiltonian

$$ \mathscr{H} = -\frac{\hbar^2}{2m}\nabla\psi\nabla\psi^* + V\psi^*\psi $$

But now I'm in a quandary. Namely, how exactly should I express $\mathscr{H}$? I mean, I want a Hamiltonian that will give the S.E. through the usual equations

$$ \dot{x} = \frac{\delta\mathscr{H}}{\delta p} $$ etc.

AFAICT, leaving $\mathscr{H}$ in the above form won't do it. There are other ways to write it. For instance the potential term could be written as $V\pi_\psi\psi$. (Forgetting some multiplying constants. Sorry, I'm in a rush.) And that would help, but not completely. And then there are $\psi$ and $\psi^*$ in the derivative terms. Should I replace one or both of those with $\pi_\psi$ or $\pi_{\psi^*}$? Showing my ignorance, would that even make sense?

So how do I express the Hamiltonian properly and, more to the point, how do I know the proper expression a priori? Surely there's a better prescription than just fiddling with it until I get the right S.E.?

Answer 1

The Lagrangian density of the Schrödinger's Eqn can be expressed as

$$ \mathscr{L} = \frac{i\hbar}{2}(\psi^*\dot{\psi}-\dot{\psi}^*\psi)-\frac{\hbar^2}{2m}\nabla\psi\cdot\nabla\psi^* -V(x)\psi^*\psi \tag{1}$$

...The problem is that now I want to find $\pi$, the momentum conjugate to $\psi$, which I read everywhere is $i\hbar\psi^*$.

Well, not everywhere. For example, not presently in this post, or here, or here.

But try as I might, when I try to make the calculation $\pi = \partial\mathscr{L}/\partial \dot{\psi}$, I get a factor of 1/2 in the result: $\partial\mathscr{L}/\partial \dot{\psi}=i\hbar\psi^*/2$.

Yes. Your calculation is correct.

Needless to say, this messes up my efforts to construct the Hamiltonian and to try to obtain the S.E. from that. What am I missing?

Remember, given the form of the Lagrangian that you have presented at the top of your post (the one that I labeled as Eq. (1)), you have two conjugate momenta that you need to consider: $$ \pi_\psi = \frac{i\hbar}{2}\psi^* $$ and $$ \pi_{\psi^*} = \frac{i\hbar}{2}\psi $$

So your expression for $\mathcal{H}$ should read: $$ \mathcal{H} = \pi_\psi \dot\psi + \pi_{\psi^*}\dot\psi^* - \mathcal{L} $$ $$ =\frac{\hbar^2}{2m}|\nabla\psi|^2 + V|\psi|^2\;. $$

... $$ \mathscr{L} = {i\hbar}\psi^*\dot{\psi}-\frac{\hbar^2}{2m}\nabla\psi\cdot\nabla\psi^* -V(x)\psi^*\psi\tag{2} $$ ...So is this somehow the "correct" Lagrangian?

All Lagrangians that give the correct equation of motion are correct. But you can't just switch back and forth willy-nilly. For example, if you use the Lagrangian of Eq. (1) above, you have two conjugate momenta to consider, but if you use the Lagrangian of Eq. (2) above you only have one conjugate momentum to consider (the other is identically zero for Eq. (2), since there is no $\dot\psi^*$ in Eq. (2)).

---

Update

UPDATE: So, apparently, my calculation of $\pi_\psi$ is correct. (It happens about twice a year for me.) And there is also $\pi_{\psi^*}$ to think about, which I can also get. This leads to the Hamiltonian

$$ \mathscr{H} = -\frac{\hbar^2}{2m}\nabla\psi\nabla\psi^* + V\psi^*\psi $$

But now I'm in a quandary. Namely, how exactly should I express $\mathscr{H}$?

First of all, you have a minus sign error. The correct expresion for the Hamiltonian density is $$ \mathcal{H} = +\frac{\hbar^2}{2m}\nabla\psi\nabla\psi^* + V\psi^*\psi $$

Now, following this reference, which is following Henley and Thirring, the proper expression for the Hamiltonian is made by the following (somewhat odd, but ultimately rather symmetric) replacements of certain terms by their conjugate momenta, to end up with: $$ \mathcal{H} = \frac{i\hbar}{2m} (\vec\nabla\psi^*\cdot\vec\nabla\pi^*-\vec\nabla\psi\cdot\vec\nabla\pi) -\frac{i}{\hbar}V(x)(\psi\pi - \psi^*\pi^*)\;, $$ which generates the correct equations of motions via the usual prescription: $$ \frac{\partial \mathcal{H}}{\partial \pi}\to \frac{i\hbar}{2m}\nabla^2\psi - \frac{i}{\hbar}V\psi = \dot\psi $$