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othocaes 2016-02-24 10:26:14 -05:00
commit 1df3870588
5 changed files with 942 additions and 89 deletions

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@ -0,0 +1,10 @@
Ĥ = p̂²
2mₑ
|Ψ(t)〉 = exp( -ι Ĥ t) ∫ dp |p〉 〈p|Ψ〉
ħ -∞
〈x|Ψ(t)〉 = exp( -ι Ĥ t ) dp 〈x|p〉 〈p|Ψ〉
ħ

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@ -30,4 +30,10 @@ Similarly, S𝓍 (S𝓍 + ħ)(S𝓍 - ħ) ≐
⎛ 0 1 0 ⎞ ⎛ 1 1 0 ⎞ ⎛ -1 1 0 ⎞ ⎛ 1 1 1 ⎞ ⎛ -1 1 0 ⎞ ⎛ 0 1 0 ⎞
ħ³ ⎜ 1 0 1 ⎟ ⎜ 1 1 1 ⎟ ⎜ 1 -1 1 ⎟ = ħ³ ⎜ 1 2 1 ⎟ ⎜ 1 -1 1 ⎟ = ħ³ ⎜ 1 0 1 ⎟.
⎝ 0 1 0 ⎠ ⎝ 0 1 1 ⎠ ⎝ 0 1 -1 ⎠ ⎝ 1 1 1 ⎠ ⎝ 0 1 -1 ⎠ ⎝ 0 1 0 ⎠
⎝ 0 1 0 ⎠ ⎝ 0 1 1 ⎠ ⎝ 0 1 -1 ⎠ ⎝ 1 1 1 ⎠ ⎝ 0 1 -1 ⎠ ⎝ 0 1 0 ⎠
This expression results in the non-zero matrix
⎛ 0 1 0 ⎞
ħ³ ⎜ 1 0 1 ⎟.
⎝ 0 1 0 ⎠

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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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@ -4,3 +4,5 @@
a) The eigenstates of the Hamiltonian can be determined by
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