working on exam 1, still

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caes 2016-02-23 19:41:02 -05:00
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Compatible Observables occur when commutator is equal to 0.
[Â,B̂] = ÂB̂ - B̂Â
Constant of Motion
------------------
if d/dt  = 0 and [Ĥ,Â] = 0, where Ĥ is the Hamiltonian operator.
then  is a constant of motion
Thm: Two operators such that [Â,B̂] always have common Eigenstates

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Computed Derivative (pic) d/dt <Â> (pic)
Discovered/Introduced Commutator
Cpmpatible Observables occur when commutator is equal to 0.
Computed Derivative d/dt <Â>:
[Ĥ,Â] = ĤÂ - ÂĤ
d <Â> = d 〈Ψ(t)|Â|Ψ(t)〉
dt dt
⎛d 〈Ψ(t)⎞ Â|Ψ(t)〉 + 〈Ψ(t)|Â ⎛d Ψ(t)〉⎞ + 〈Ψ(t)|d Â|Ψ(t)〉
⎝dt ⎠ ⎝dt ⎠ dt
ι 〈Ψ(t)|Ĥ Â|Ψ(t)〉 + 〈Ψ(t)|Â Ĥ -ι|Ψ(t)〉 + 〈Ψ(t)|d Â|Ψ(t)〉
ħ ħ dt
ι 〈Ψ(t)|Ĥ Â - Â Ĥ|Ψ(t)〉 + 〈Ψ(t)|d Â|Ψ(t)〉
ħ dt
Constant of Motion
------------------
if d/dt  = 0 and [Ĥ,Â] = 0,
then  is a constant of motion
Thm: Two operators such that [Â,B̂] always have common Eigenstates

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Show that ͟d͟〈͟p͟〉͟ = -͟d͟V͟(͟x͟)͟〵 when a particle is subjected to a potential 〈V(x)〉.
dt 〵 dx
dt 〵dx
The time derivative of the expectation value of the momentum is a known quantity, from
Time Dependence of Expectation Value of General Momentum Operator:
d〈p〉 = 1.
dt ιħ
The problem is therefore reduced to finding whether -/dV(x)\ reduces to 1.
\dx / ιħ
-/dV(x)\ = -〈Ψ| d V(x) |Ψ〉.
\dx / dx
Viewing the expression in this form reveals a relationship between the space derivative and the operators V(x) and |Ψ〉. The chain rule allows this derivative to be computed.
-〈Ψ| d V(x) |Ψ〉 = -〈Ψ| ⎛d V(x)|Ψ> + d |Ψ> V(x)⎞.
dx ⎝dx dx ⎠

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a)
🔋