Deriving Schrodinger Eq. from Lagrange & Midpoint Rule

[greeniq]

Homework Statement

derive from the Lagrange function and with the use of the midpoint rule, the Schrodinger equation for a particle with charge e coupled to a vector potential?

Homework Equations

L=(1/2)m X_t²+ e X_t . A(X_t,t)

I mean by X== X dot, but A is a fun. of X and t

The Attempt at a Solution

I even can't start such a quastion, coz this is my first course in quantum but since I am from Denemark my Pakistani teacher always gives me too much hard quastions to solve!

 

[Jhero]

Dear student,

The Schrodinger equation for a particle with charge e coupled to a vector potential can be derived from the Lagrangian function using the midpoint rule. First, we need to rewrite the Lagrangian function in terms of the position, momentum, and time variables. We have:

L = (1/2)m X^2 + e X . A(X,t)

Where X is the position variable, m is the mass of the particle, e is its charge, and A is the vector potential which is a function of both position and time.

Next, we need to use the midpoint rule to approximate the time derivative of the position variable. This is given by:

X(t+Δt) ≈ X(t) + (Δt/2)(Ẋ(t+Δt) + Ẋ(t))

Where Δt is a small time interval.

Now, using this approximation in the Lagrangian function, we get:

L ≈ (1/2)m [X(t) + (Δt/2)(Ẋ(t+Δt) + Ẋ(t))]^2 + e [X(t) + (Δt/2)(Ẋ(t+Δt) + Ẋ(t))] . A(X(t),t)

Expanding and simplifying, we get:

L ≈ (1/2)m X^2 + (1/2)m (Δt)^2 Ẋ^2 + (1/2)m Δt X Ẍ + (1/2)Δt^2 Ẍ̇ + e X . A(X,t) + (Δt/2)(e Ẋ . A(X,t+Δt) + e Ẋ . A(X,t))

Using the Euler-Lagrange equation, we can now find the equations of motion for the position and momentum variables. This gives us:

mẍ + e A(X,t) + (Δt/2)(e Ẍ . A(X,t+Δt) + e Ẍ . A(X,t)) = 0

And,

mẋ + (Δt/2)(e Ẋ . A(X,t+Δt) + e Ẋ . A(X,t)) = p

Where p is the momentum variable.

Finally, taking