Quantenmechanikvorlesung von Prof. Dr. E. Schöll, PhD
Der Artikel Dynamik des 2- Zustands- Systems basiert auf der Vorlesungsmitschrift von Franz- Josef Schmitt des 4.Kapitels (Abschnitt 2) der Quantenmechanikvorlesung von Prof. Dr. E. Schöll, PhD .
Die potenzielle Energie des magnetischen Moments des Elektronen- Spins
μ
¯
{\displaystyle {\bar {\mu }}}
B
¯
=
B
e
¯
3
{\displaystyle {\bar {B}}=B{{\bar {e}}_{3}}}
V
=
−
μ
¯
^
⋅
B
¯
{\displaystyle V=-{\hat {\bar {\mu }}}\cdot {\bar {B}}}
μ
¯
^
=
+
g
e
2
m
0
S
¯
^
=
+
e
ℏ
2
m
0
σ
¯
^
{\displaystyle {\hat {\bar {\mu }}}=+g{\frac {e}{2{{m}_{0}}}}{\hat {\bar {S}}}=+{\frac {e\hbar }{2{{m}_{0}}}}{\hat {\bar {\sigma }}}}
Somit:
V
^
=
−
e
ℏ
2
m
0
σ
¯
^
⋅
B
¯
=
−
e
ℏ
B
2
m
0
σ
¯
^
3
=
ℏ
ω
l
σ
¯
^
3
{\displaystyle {\hat {V}}=-{\frac {e\hbar }{2{{m}_{0}}}}{\hat {\bar {\sigma }}}\cdot {\bar {B}}=-{\frac {e\hbar B}{2{{m}_{0}}}}{{\hat {\bar {\sigma }}}_{3}}=\hbar {{\omega }_{l}}{{\hat {\bar {\sigma }}}_{3}}}
Mit der Larmor-Frequenz
ω
l
:=
|
e
|
B
2
m
0
{\displaystyle {{\omega }_{l}}:={\frac {|e|B}{2{{m}_{0}}}}}
Wenn der Spin an keine weitere Variable ankoppelt, so ist
H
^
=
V
^
{\displaystyle {\hat {H}}={\hat {V}}}
σ
¯
^
∘
=
i
ℏ
[
H
^
,
σ
¯
^
]
=
i
ω
l
[
σ
¯
^
3
,
σ
¯
^
]
{\displaystyle {{\hat {\bar {\sigma }}}^{\circ }}={\frac {i}{\hbar }}\left[{\hat {H}},{\hat {\bar {\sigma }}}\right]=i{{\omega }_{l}}\left[{{\hat {\bar {\sigma }}}_{3}},{{\hat {\bar {\sigma }}}_{}}\right]}
Berechnung der Erwartungswerte mit
[
σ
¯
^
j
,
σ
¯
^
k
]
=
2
i
ε
j
k
l
σ
¯
^
l
{\displaystyle \left[{{\hat {\bar {\sigma }}}_{j}},{{\hat {\bar {\sigma }}}_{k}}\right]=2i{{\varepsilon }_{jkl}}{{\hat {\bar {\sigma }}}_{l}}}
d
d
t
⟨
σ
¯
^
1
⟩
=
i
ℏ
⟨
[
H
,
σ
¯
^
1
]
⟩
=
i
ω
l
⟨
[
σ
¯
^
3
,
σ
¯
^
1
]
⟩
=
−
2
ω
l
⟨
σ
¯
^
2
⟩
{\displaystyle {\frac {d}{dt}}\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle ={\frac {i}{\hbar }}\left\langle \left[H,{{\hat {\bar {\sigma }}}_{1}}\right]\right\rangle =i{{\omega }_{l}}\left\langle \left[{{\hat {\bar {\sigma }}}_{3}},{{\hat {\bar {\sigma }}}_{1}}\right]\right\rangle =-2{{\omega }_{l}}\left\langle {{\hat {\bar {\sigma }}}_{2}}\right\rangle }
d
d
t
⟨
σ
¯
^
1
⟩
=
−
2
ω
l
⟨
σ
¯
^
2
⟩
d
d
t
⟨
σ
¯
^
2
⟩
=
2
ω
l
⟨
σ
¯
^
1
⟩
d
d
t
⟨
σ
¯
^
3
⟩
=
0
{\displaystyle {\begin{aligned}&{\frac {d}{dt}}\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle =-2{{\omega }_{l}}\left\langle {{\hat {\bar {\sigma }}}_{2}}\right\rangle \\&{\frac {d}{dt}}\left\langle {{\hat {\bar {\sigma }}}_{2}}\right\rangle =2{{\omega }_{l}}\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle \\&{\frac {d}{dt}}\left\langle {{\hat {\bar {\sigma }}}_{3}}\right\rangle =0\\\end{aligned}}}
Dies läßt sich reduzieren:
d
2
d
t
2
⟨
σ
¯
^
1
⟩
+
(
2
ω
l
)
2
⟨
σ
¯
^
1
⟩
=
0
{\displaystyle {\frac {{d}^{2}}{d{{t}^{2}}}}\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle +{{\left(2{{\omega }_{l}}\right)}^{2}}\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle =0}
Die Dynamik der Spins bildet also einen Oszillator in der x-y- Ebene.
Die zeitliche Unabhängigkeit der Spin3- Komponente liegt dabei alleine an der Wahl des Koordinatensystems, bzw. der Basis! Wir haben diese gerade so gewählt, dass die 3- Komponente zeitlich unabhängig wird.
Die Lösung der Diffgleichung liefert:
⟨
σ
¯
^
1
⟩
t
=
⟨
σ
¯
^
2
⟩
0
sin
(
2
ω
l
t
)
+
⟨
σ
¯
^
1
⟩
0
cos
(
2
ω
l
t
)
⟨
σ
¯
^
2
⟩
t
=
⟨
σ
¯
^
2
⟩
0
cos
(
2
ω
l
t
)
−
⟨
σ
¯
^
1
⟩
0
sin
(
2
ω
l
t
)
⟨
σ
¯
^
3
⟩
t
=
⟨
σ
¯
^
3
⟩
0
{\displaystyle {\begin{aligned}&{{\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle }_{t}}={{\left\langle {{\hat {\bar {\sigma }}}_{2}}\right\rangle }_{0}}\sin \left(2{{\omega }_{l}}t\right)+{{\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle }_{0}}\cos \left(2{{\omega }_{l}}t\right)\\&{{\left\langle {{\hat {\bar {\sigma }}}_{2}}\right\rangle }_{t}}={{\left\langle {{\hat {\bar {\sigma }}}_{2}}\right\rangle }_{0}}\cos \left(2{{\omega }_{l}}t\right)-{{\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle }_{0}}\sin \left(2{{\omega }_{l}}t\right)\\&{{\left\langle {{\hat {\bar {\sigma }}}_{3}}\right\rangle }_{t}}={{\left\langle {{\hat {\bar {\sigma }}}_{3}}\right\rangle }_{0}}\\\end{aligned}}}
klassischer Kreisel Die Anfangsbedingungen können ebenfalls durch Wahl des Koordinatensystems (feste x-y- Ebene) beeinflusst werden.
Wähle:
o.B. d.A.:
⟨
σ
¯
^
2
⟩
0
=
0
{\displaystyle {{\left\langle {{\hat {\bar {\sigma }}}_{2}}\right\rangle }_{0}}=0}
Wir können uns den Betrag des Erwartungswertes des gesamten Spinvektors ansehen und es zeigt sich :
|
⟨
σ
¯
^
⟩
t
|
2
=
⟨
σ
¯
^
1
⟩
t
2
+
⟨
σ
¯
^
2
⟩
t
2
+
⟨
σ
¯
^
3
⟩
t
2
=
⟨
σ
¯
^
1
⟩
0
2
[
cos
2
(
2
ω
l
t
)
+
sin
2
(
2
ω
l
t
)
]
+
⟨
σ
¯
^
3
⟩
0
2
=
⟨
σ
¯
^
1
⟩
0
2
+
⟨
σ
¯
^
3
⟩
0
2
{\displaystyle {{\left|{{\left\langle {{\hat {\bar {\sigma }}}_{}}\right\rangle }_{t}}\right|}^{2}}={{\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle }_{t}}^{2}+{{\left\langle {{\hat {\bar {\sigma }}}_{2}}\right\rangle }_{t}}^{2}+{{\left\langle {{\hat {\bar {\sigma }}}_{3}}\right\rangle }_{t}}^{2}={{\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle }_{0}}^{2}\left[{{\cos }^{2}}\left(2{{\omega }_{l}}t\right)+{{\sin }^{2}}\left(2{{\omega }_{l}}t\right)\right]+{{\left\langle {{\hat {\bar {\sigma }}}_{3}}\right\rangle }_{0}}^{2}={{\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle }_{0}}^{2}+{{\left\langle {{\hat {\bar {\sigma }}}_{3}}\right\rangle }_{0}}^{2}}
Mit anderen Worten:
|
⟨
σ
¯
^
⟩
t
|
2
=
|
⟨
σ
¯
^
⟩
0
|
2
=
c
o
n
s
t
{\displaystyle {{\left|{{\left\langle {{\hat {\bar {\sigma }}}_{}}\right\rangle }_{t}}\right|}^{2}}={{\left|{{\left\langle {{\hat {\bar {\sigma }}}_{}}\right\rangle }_{0}}\right|}^{2}}=const}
Der Erwartungswert des Spins präzediert also mit der Frequenz
2
ω
l
{\displaystyle 2{{\omega }_{l}}}
Schrödingergleichung für die Spinzustände
ℏ
ω
l
σ
¯
^
3
|
a
(
t
)
⟩
=
i
ℏ
∂
∂
t
|
a
(
t
)
⟩
{\displaystyle \hbar {{\omega }_{l}}{{\hat {\bar {\sigma }}}_{3}}\left|a(t)\right\rangle =i\hbar {\frac {\partial }{\partial t}}\left|a(t)\right\rangle }
(Schrödingergleichung für Spinzustände)
Achtung! Nur Spin- Hamiltonian!
Dabei muss der Zustand
|
a
(
t
)
⟩
{\displaystyle \left|a(t)\right\rangle }
|
a
(
t
)
⟩
=
a
1
(
t
)
|
↑
⟩
+
a
2
(
t
)
|
↓
⟩
{\displaystyle \left|a(t)\right\rangle ={{a}_{1}}(t)\left|\uparrow \right\rangle +{{a}_{2}}(t)\left|\downarrow \right\rangle }
Matrix- Darstellung:
ℏ
ω
l
(
1
0
0
−
1
)
(
a
1
(
t
)
a
2
(
t
)
)
=
i
ℏ
∂
∂
t
(
a
1
(
t
)
a
2
(
t
)
)
⇔
−
i
ω
l
a
1
=
a
˙
1
i
ω
l
a
2
=
a
˙
2
{\displaystyle \hbar {{\omega }_{l}}\left({\begin{matrix}1&0\\0&-1\\\end{matrix}}\right)\left({\begin{matrix}{{a}_{1}}(t)\\{{a}_{2}}(t)\\\end{matrix}}\right)=i\hbar {\frac {\partial }{\partial t}}\left({\begin{matrix}{{a}_{1}}(t)\\{{a}_{2}}(t)\\\end{matrix}}\right)\Leftrightarrow {\begin{matrix}-i{{\omega }_{l}}{{a}_{1}}={{\dot {a}}_{1}}\\i{{\omega }_{l}}{{a}_{2}}={{\dot {a}}_{2}}\\\end{matrix}}}
Die Lösung lautet:
a
1
(
t
)
=
a
10
e
−
i
ω
l
t
a
2
(
t
)
=
a
20
e
i
ω
l
t
{\displaystyle {\begin{aligned}&{{a}_{1}}(t)={{a}_{10}}{{e}^{-i{{\omega }_{l}}t}}\\&{{a}_{2}}(t)={{a}_{20}}{{e}^{i{{\omega }_{l}}t}}\\\end{aligned}}}
|
a
(
t
)
⟩
=
a
10
e
−
i
ω
l
t
|
↑
⟩
+
a
20
e
i
ω
l
t
|
↓
⟩
{\displaystyle \left|a(t)\right\rangle ={{a}_{10}}{{e}^{-i{{\omega }_{l}}t}}\left|\uparrow \right\rangle +{{a}_{20}}{{e}^{i{{\omega }_{l}}t}}\left|\downarrow \right\rangle }
Nebenbemerkung: Hieraus gewinnt man
⟨
σ
¯
^
j
⟩
t
{\displaystyle {{\left\langle {{\hat {\bar {\sigma }}}_{j}}\right\rangle }_{t}}}
![{\displaystyle {{\hat {\bar {\sigma }}}^{\circ }}={\frac {i}{\hbar }}\left[{\hat {H}},{\hat {\bar {\sigma }}}\right]=i{{\omega }_{l}}\left[{{\hat {\bar {\sigma }}}_{3}},{{\hat {\bar {\sigma }}}_{}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c5b89fe06eed7a803be68a6ad72da20b84ec8de)
![{\displaystyle \left[{{\hat {\bar {\sigma }}}_{j}},{{\hat {\bar {\sigma }}}_{k}}\right]=2i{{\varepsilon }_{jkl}}{{\hat {\bar {\sigma }}}_{l}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5cffe4f32b910f52197f129f52da99142b74c9b)
![{\displaystyle {\frac {d}{dt}}\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle ={\frac {i}{\hbar }}\left\langle \left[H,{{\hat {\bar {\sigma }}}_{1}}\right]\right\rangle =i{{\omega }_{l}}\left\langle \left[{{\hat {\bar {\sigma }}}_{3}},{{\hat {\bar {\sigma }}}_{1}}\right]\right\rangle =-2{{\omega }_{l}}\left\langle {{\hat {\bar {\sigma }}}_{2}}\right\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5bf655898dba3a6dcfd60856fe46c33cac042d3)
![{\displaystyle {{\left|{{\left\langle {{\hat {\bar {\sigma }}}_{}}\right\rangle }_{t}}\right|}^{2}}={{\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle }_{t}}^{2}+{{\left\langle {{\hat {\bar {\sigma }}}_{2}}\right\rangle }_{t}}^{2}+{{\left\langle {{\hat {\bar {\sigma }}}_{3}}\right\rangle }_{t}}^{2}={{\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle }_{0}}^{2}\left[{{\cos }^{2}}\left(2{{\omega }_{l}}t\right)+{{\sin }^{2}}\left(2{{\omega }_{l}}t\right)\right]+{{\left\langle {{\hat {\bar {\sigma }}}_{3}}\right\rangle }_{0}}^{2}={{\left\langle {{\hat {\bar {\sigma }}}_{1}}\right\rangle }_{0}}^{2}+{{\left\langle {{\hat {\bar {\sigma }}}_{3}}\right\rangle }_{0}}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b82ac71557b22e816dca46112a50e23bfbc5817)
