Definition Transition dipole moment
1 definition
1.1 single charged particle
1.2 multiple charged particles
1.3 in terms of momentum
definition
a single charged particle
for transition single charged particle changes state
|
ψ
a
⟩
{\displaystyle |\psi _{a}\rangle }
|
ψ
b
⟩
{\displaystyle |\psi _{b}\rangle }
, transition dipole moment
(t.d.m.)
{\displaystyle {\text{(t.d.m.)}}}
is
(
t.d.m.
a
→
b
)
=
⟨
ψ
b
|
(
q
r
)
|
ψ
a
⟩
=
q
∫
ψ
b
∗
(
r
)
r
ψ
a
(
r
)
d
3
r
{\displaystyle ({\text{t.d.m. }}a\rightarrow b)=\langle \psi _{b}|(q\mathbf {r} )|\psi _{a}\rangle =q\int \psi _{b}^{*}(\mathbf {r} )\,\mathbf {r} \,\psi _{a}(\mathbf {r} )\,d^{3}\mathbf {r} }
where q particle s charge, r position, , integral on space (
∫
d
3
r
{\displaystyle \int d^{3}\mathbf {r} }
shorthand
∭
d
x
d
y
d
z
{\displaystyle \iiint dx\,dy\,dz}
). transition dipole moment vector; example x-component is
(
x-component of t.d.m.
a
→
b
)
=
⟨
ψ
b
|
(
q
x
)
|
ψ
a
⟩
=
q
∫
ψ
b
∗
(
r
)
x
ψ
a
(
r
)
d
3
r
{\displaystyle ({\text{x-component of t.d.m. }}a\rightarrow b)=\langle \psi _{b}|(qx)|\psi _{a}\rangle =q\int \psi _{b}^{*}(\mathbf {r} )\,x\,\psi _{a}(\mathbf {r} )\,d^{3}\mathbf {r} }
in other words, transition dipole moment off-diagonal matrix element of position operator, multiplied particle s charge.
multiple charged particles
when transition involves more 1 charged particle, transition dipole moment defined in analogous way electric dipole moment: sum of positions, weighted charge. if ith particle has charge qi , position operator ri, transition dipole moment is:
(
t.d.m.
a
→
b
)
=
⟨
ψ
b
|
(
q
1
r
1
+
q
2
r
2
+
⋯
)
|
ψ
a
⟩
=
{\displaystyle ({\text{t.d.m. }}a\rightarrow b)=\langle \psi _{b}|(q_{1}\mathbf {r} _{1}+q_{2}\mathbf {r} _{2}+\cdots )|\psi _{a}\rangle =}
=
∫
ψ
b
∗
(
r
1
,
r
2
,
…
)
(
q
1
r
1
+
q
2
r
2
+
⋯
)
ψ
a
(
r
1
,
r
2
,
…
)
d
3
r
1
d
3
r
2
⋯
{\displaystyle =\int \psi _{b}^{*}(\mathbf {r} _{1},\mathbf {r} _{2},\ldots )\,(q_{1}\mathbf {r} _{1}+q_{2}\mathbf {r} _{2}+\cdots )\,\psi _{a}(\mathbf {r} _{1},\mathbf {r} _{2},\ldots )\,d^{3}\mathbf {r} _{1}\,d^{3}\mathbf {r} _{2}\cdots }
in terms of momentum
for single, nonrelativistic particle of mass m, in 0 magnetic field, transition dipole moment can alternatively written in terms of momentum operator, using relationship
⟨
ψ
a
|
r
|
ψ
b
⟩
=
i
ℏ
(
e
b
−
e
a
)
m
⟨
ψ
a
|
p
|
ψ
b
⟩
{\displaystyle \langle \psi _{a}|\mathbf {r} |\psi _{b}\rangle ={\frac {i\hbar }{(e_{b}-e_{a})m}}\langle \psi _{a}|\mathbf {p} |\psi _{b}\rangle }
this relationship can proven starting commutation relation between position x , hamiltonian h:
[
h
,
x
]
=
[
p
2
2
m
+
v
(
x
,
y
,
z
)
,
x
]
=
[
p
2
2
m
,
x
]
=
1
2
m
(
p
x
[
p
x
,
x
]
+
[
p
x
,
x
]
p
x
)
=
−
i
ℏ
p
x
/
m
{\displaystyle [h,x]=\left[{\frac {p^{2}}{2m}}+v(x,y,z),x\right]=\left[{\frac {p^{2}}{2m}},x\right]={\frac {1}{2m}}(p_{x}[p_{x},x]+[p_{x},x]p_{x})=-i\hbar p_{x}/m}
then
⟨
ψ
a
|
(
h
x
−
x
h
)
|
ψ
b
⟩
=
−
i
ℏ
m
⟨
ψ
a
|
p
x
|
ψ
b
⟩
{\displaystyle \langle \psi _{a}|(hx-xh)|\psi _{b}\rangle ={\frac {-i\hbar }{m}}\langle \psi _{a}|p_{x}|\psi _{b}\rangle }
however, assuming ψa , ψb energy eigenstates energy ea , eb, can write
⟨
ψ
a
|
(
h
x
−
x
h
)
|
ψ
b
⟩
=
(
⟨
ψ
a
|
h
)
x
|
ψ
b
⟩
−
⟨
ψ
a
|
x
(
h
|
ψ
b
⟩
)
=
(
e
a
−
e
b
)
⟨
ψ
a
|
x
|
ψ
b
⟩
{\displaystyle \langle \psi _{a}|(hx-xh)|\psi _{b}\rangle =(\langle \psi _{a}|h)x|\psi _{b}\rangle -\langle \psi _{a}|x(h|\psi _{b}\rangle )=(e_{a}-e_{b})\langle \psi _{a}|x|\psi _{b}\rangle }
similar relations hold y , z, give relationship above.
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