Wolfram Alpha:

```Time Dependent Schrodinger Equation
-----------------------------------

ih∂ψ/∂t = H|ψ> and p = -ih∂/∂x

From classical mechanics H = p2/2m assume this works for QM

H = P2/2m = (1/2m)(-ih∂/∂x)(-ih∂/∂x) = (-h/2m)∂2/∂x2 = Hamiltonian

Therefore,

ih∂ψ(x,t)/∂t = (1/2m)(-ih∂/∂x)(-ih∂/∂x)ψ(x,t)

= -(h2/2m)∂2ψ(x,t)/∂x2

Assume solutions of the form ψ(x,t) = exp(ipx/h)exp(iωt)

Plug into SE:

-hωψ(x,t) = (h2/2m)(p2/h2)ψ(x,t)

-ω = p2/2mh

Substitute this back into our previous wavefunction to get:

ψ(x,t) = exp(ipx/h - ip2t/2mh)

Note that this equation can also be stated noting that E = hω
and p = hk.

If ψ(x,t) is a solution to the Schrodinger equation, any linear
combination of plane waves is also a solution.  Therefore, the
general solution is the following integral:

ψ(x,t) = (1/2π)∫exp(ipx/h - ip2t/2mh)φ(p)dp

where φ(p) is an amplitude term.

This expression is simply the superposition of different momentum
eigenstates.  The probability, that a particle has momentum, p', is

P(p') = φ(p')φ(p')*

Time Evolution of Expectation Value
-----------------------------------

States change with time but O doesn't
.             .
d<ψ|O|ψ>/dt = <ψ|O|ψ> + <ψ|O|ψ>

Using the TDSE we can write:
.                    .
|ψ> = (-i/h)H|ψ> and <ψ| = (i/h)<ψ|H - conjugate to operate on bra

Substituting,

d<ψ|O|ψ>/dt = (i/h)<ψ|HO|ψ> - (i/h)<ψ|OH|ψ>

= (i/h)<ψ|[H,O]|ψ> - Expectation value of the commutator

≡ (-i/h)<ψ|[O,H]|ψ>

Therefore,

d<O>/dt = (i/h)<[H,O]>

In classical mechanics we have:

d<O>/dt = <{O,H}>  where {} are the Poisson brackets.

Consider arbitrary function of position, F(x)

In CM: {F(x),p} = (∂F/∂x)(∂p/∂p) - (∂F/∂p)(∂p/∂x)

= ∂F/∂x

In QM: [F(x),p] = F(x)(-ih∂/∂x) - (-ih∂/∂x)F(x)

= -ih∂/∂x + ih(F(x)∂/∂x + ∂F/∂x)

Apply to ψ(x)

= (-ih∂/∂x + ih(F(x)∂/∂x + ∂F/∂x))ψ(x)

which after some math leads to:

= (ih∂F/∂x)ψ(x)

Thus,

[F(x),p] = F(x)p - pF(x) = ih∂F/∂x

Similarly,

[x,F(p)] = ih∂F/∂p

Make F(x) = x then,

[x,p] = ih and ∂x/∂x = 1

In classical mechanics the Poisson brackets {x,p} = 1

Therefore, in general we can conclude thatl,

ih{..} = [..]

Poisson brackets and commutator rules:

{A,B} = -{B,A}

Likewise,

ih[A,B] = -ih[B,A]

{AB,C} = A{B,C} + {A,C}B

Likewise,

[AB,C] = A[B,C] + [A,C]B

TDSE Equation with Potential
------------------------------

ih∂ψ(x,t) = Hψ(x,t) where H = p2/2m + U(x)

ih∂ψ/∂t = (h2/2m)∂2ψ/∂x2 + U(x)ψ(x)

Position:

d<x>/dt = i/h<[H,x]>

= i/h<[p2/2m + U(x),x]>

= i/h<[p2/2m,x]>  since U(x) commutes with x

= i/h2m<[p2,x]>

Use the rule p[p,x] + [p,x]p

= (i/h2m)<p(-ih)(-ih) + (-ih)(-ih)p>

= (i/h2m)<2ihp>

= <p>/m <--- average velocity dx/dt

Force:

d<p>/dt = i/h<[p2/2m + U(x),p]>

= i/h<[U(x),p]>  since p commutes with p2

= i/h<[U(x),p])(ih)>

= -∂U(x)/∂x
```