Schrödinger picture formulation of a velocity-dependent potential of the form $V(x,\dot{x}) = a + bx + cx^{2} + d\dot{x} + ex\dot{x}$

Question

In Shankar Chapter 8, there is a section at the end of the chapter on the path integral formulation for a potential of the form $$V(x,\dot{x}) = a + bx + cx^{2} + d\dot{x} + ex\dot{x}.$$ I follow the arguments given related to the propagator, but I am wondering how this problem would be approached in the Schrödinger picture?

What would the form of the Hamiltonian look like when the potential has $x$ and $\dot{x}$ dependence in the same term? I think it is clear that it is some sort of harmonic oscillator, but I am confused on how to deal with the velocity-dependence in $V(x,\dot{x})$

I understand it is likely a more difficult problem to approach in this way, but I am curious on how these different formulations connect.

Answer 1

General strategy:

1. Start with the Lagrangian function $L(x, \dot{x})$.

2. Compute the canonical momentum $p= \frac{\partial L(x, \dot{x})}{\partial \dot{x}}$.

3. Express $\dot{x}$ in terms of $p$ and $x$: $\dot{x}=\dot{x}(p,x)$.

4. Write down the Hamiltonian $H= p \dot{x}-L$ as a function of $x$ and $p$ using the expression for $\dot{x}$ obtained in the previous step: $H(x,p)= p \dot{x}(p,x)-L(x, \dot{x}(p,x))$.

In your special case, you have $L(x,\dot{x})= m \dot{x}^2/2-a-bx-cx^2-d\dot{x}-e x \dot{x}$. The canonical momentum is thus given by $p= m \dot{x}-d-ex$. Writing $\dot{x}$ as a function of $p$ and $x$ gives $\dot{x}= (p+d+ex)/m$. Finding now the Hamiltonian $H(x,p)$ is left as an easy exercise.

Remark: The term $$-a-d\dot{x}-ex \dot{x}=\frac{d}{dt}(-at-dx-e x^2/2),$$ present in the Lagrangian, is a total derivative. As such, it does not affect the equation of motion (show this explicitly by using the Euler-Lagrange equation). In other words, your Lagrangian $L(x, \dot{x})$ is physically equivalent to the simpler Lagrangian $L^\prime(x, \dot{x}) =m \dot{x}^2/2-bx-cx^2$. How is this fact expressed in the Hamilton formalism?