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Time Dependent Schrodinger Equation
-----------------------------------
ih∂ψ/∂t = H|ψ> and p = -ih∂/∂x
From classical mechanics H = p^{2}/2m assume this works for QM
H = P^{2}/2m = (1/2m)(-ih∂/∂x)(-ih∂/∂x) = (-h/2m)∂^{2}/∂x^{2} = Hamiltonian
Therefore,
ih∂ψ(x,t)/∂t = (1/2m)(-ih∂/∂x)(-ih∂/∂x)ψ(x,t)
= -(h^{2}/2m)∂^{2}ψ(x,t)/∂x^{2}
Assume solutions of the form ψ(x,t) = exp(ipx/h)exp(iωt)
Plug into SE:
-hωψ(x,t) = (h^{2}/2m)(p^{2}/h^{2})ψ(x,t)
This leads to:
-ω = p^{2}/2mh
Substitute this back into our previous wavefunction to get:
ψ(x,t) = exp(ipx/h - ip^{2}t/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 - ip^{2}t/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 = p^{2}/2m + U(x)
ih∂ψ/∂t = (h^{2}/2m)∂^{2}ψ/∂x^{2} + U(x)ψ(x)
Position:
d<x>/dt = i/h<[H,x]>
= i/h<[p^{2}/2m + U(x),x]>
= i/h<[p^{2}/2m,x]> since U(x) commutes with x
= i/h2m<[p^{2},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<[p^{2}/2m + U(x),p]>
= i/h<[U(x),p]> since p commutes with p^{2}
= i/h<[U(x),p])(ih)>
= -∂U(x)/∂x