Kvantkeemia on keemia haru, milles probleemide lahendamiseks rakendatakse kvantmehaanikat.
Kvantmehaanika sai alguse Max Plancki aastal 1900 tähelepanekust, et musta keha kiirguse spektri seletamiseks peaks valgus tekkima ja neelduma diskreetsete "energiaportsjonitena" – kvantidena, mis erinevad teineteisest Plancki konstandi
h
{\displaystyle h}
E
=
h
ν
{\displaystyle E=h\nu }
ν
{\displaystyle \nu }
p
{\displaystyle p}
p
=
h
k
{\displaystyle p=hk}
k
{\displaystyle k}
Osake 1D karbis mudel
muuda
2
L
x
=
n
x
λ
{\displaystyle 2L_{x}=n_{x}\lambda }
p
=
h
λ
=
h
2
n
x
L
x
{\displaystyle p={\frac {h}{\lambda }}={\frac {h}{2}}{\frac {n_{x}}{L_{x}}}}
E
K
=
m
v
2
2
=
p
2
2
m
{\displaystyle E_{\mathrm {K} }={\frac {mv^{2}}{2}}={\frac {p^{2}}{2m}}}
E
K
=
h
2
8
m
(
n
x
2
L
x
2
)
{\displaystyle E_{\mathrm {K} }={\frac {h^{2}}{8m}}\left({\frac {n_{x}^{2}}{L_{x}^{2}}}\right)}
Osake 2D karbis mudel
muuda
E
K
=
h
2
8
m
(
n
x
2
L
x
2
+
n
y
2
L
y
2
)
{\displaystyle E_{\mathrm {K} }={\frac {h^{2}}{8m}}\left({\frac {n_{x}^{2}}{L_{x}^{2}}}+{\frac {n_{y}^{2}}{L_{y}^{2}}}\right)}
Osake 3D karbis mudel
muuda
E
K
=
h
2
8
m
(
n
x
2
L
x
2
+
n
y
2
L
y
2
+
n
z
2
L
z
2
)
{\displaystyle E_{\mathrm {K} }={\frac {h^{2}}{8m}}\left({\frac {n_{x}^{2}}{L_{x}^{2}}}+{\frac {n_{y}^{2}}{L_{y}^{2}}}+{\frac {n_{z}^{2}}{L_{z}^{2}}}\right)}
2
π
R
=
l
λ
{\displaystyle 2\pi R=l\lambda }
p
=
h
λ
=
h
2
π
l
R
{\displaystyle p={\frac {h}{\lambda }}={\frac {h}{2\pi }}{\frac {l}{R}}}
E
K
=
m
v
2
2
=
p
2
2
m
{\displaystyle E_{\mathrm {K} }={\frac {mv^{2}}{2}}={\frac {p^{2}}{2m}}}
E
K
=
h
2
8
π
2
m
(
l
2
R
2
)
{\displaystyle E_{\mathrm {K} }={\frac {h^{2}}{8\pi ^{2}m}}\left({\frac {l^{2}}{R^{2}}}\right)}
muuda
E
K
=
h
2
8
π
2
m
(
l
(
l
+
1
)
R
2
)
{\displaystyle E_{\mathrm {K} }={\frac {h^{2}}{8\pi ^{2}m}}\left({\frac {l(l+1)}{R^{2}}}\right)}
The virial theorem relates the expectation kinetic energy of a quantum system to the potential. Let's consider a quantum system in a stationary state, which does not have to be the group state. Let's assume that there is a single particle with position
q
{\displaystyle q}
V
(
q
)
{\displaystyle V(q)}
E
K
{\displaystyle E_{\mathrm {K} }}
V
{\displaystyle V}
2
E
K
=
⟨
q
⋅
∂
V
∂
q
⟩
=
n
E
P
{\displaystyle 2E_{\mathrm {K} }={\Big \langle }q\cdot {\frac {\partial V}{\partial q}}{\Big \rangle }=nE_{\mathrm {P} }}
Harmooniline kvantostsillaator
muuda
For harminic osciallator
V
=
1
2
k
x
2
{\displaystyle V={\frac {1}{2}}kx^{2}}
2
E
K
=
⟨
1
2
x
⋅
∂
k
x
2
∂
x
⟩
=
2
V
{\displaystyle 2E_{\mathrm {K} }={\Big \langle }{\frac {1}{2}}x\cdot {\frac {\partial kx^{2}}{\partial x}}{\Big \rangle }=2V}
E
T
=
E
K
+
E
P
=
2
E
K
=
2
E
P
{\displaystyle E_{\mathrm {T} }=E_{\mathrm {K} }+E_{\mathrm {P} }=2E_{\mathrm {K} }=2E_{\mathrm {P} }}
Kineetiline energia avaldub kui
E
K
=
⟨
p
2
⟩
/
2
μ
{\displaystyle E_{\mathrm {K} }={\langle p^{2}\rangle }/{2\mu }}
⟨
p
2
⟩
−
⟨
p
⟩
2
=
⟨
Δ
p
⟩
2
{\displaystyle {\langle p^{2}\rangle }-{\langle p\rangle }^{2}={\langle \Delta p\rangle }^{2}}
⟨
p
⟩
=
0
{\displaystyle {\langle p\rangle }=0}
⟨
p
2
⟩
=
⟨
Δ
p
⟩
2
{\displaystyle {\langle p^{2}\rangle }={\langle \Delta p\rangle }^{2}}
E
K
=
⟨
Δ
p
⟩
2
2
μ
{\displaystyle E_{\mathrm {K} }={\frac {{\langle \Delta p\rangle }^{2}}{2\mu }}}
Potentsiaalne energia avaldub kui
E
P
=
k
⟨
x
2
⟩
{\displaystyle E_{\mathrm {P} }=k\langle x^{2}\rangle }
⟨
x
2
⟩
−
⟨
x
⟩
2
=
⟨
Δ
x
⟩
2
{\displaystyle {\langle x^{2}\rangle }-{\langle x\rangle }^{2}={\langle \Delta x\rangle }^{2}}
⟨
x
⟩
=
0
{\displaystyle {\langle x\rangle }=0}
⟨
x
2
⟩
=
⟨
Δ
x
⟩
2
{\displaystyle {\langle x^{2}\rangle }={\langle \Delta x\rangle }^{2}}
E
P
=
k
⟨
Δ
x
⟩
2
2
{\displaystyle E_{\mathrm {P} }={\frac {k{\langle \Delta x\rangle }^{2}}{2}}}
Väljundame
E
T
{\displaystyle E_{\mathrm {T} }}
2
E
K
⋅
2
E
P
{\displaystyle {\sqrt {2E_{\mathrm {K} }\cdot 2E_{\mathrm {P} }}}}
E
T
=
⟨
Δ
x
⟩
2
/
μ
⋅
2
k
⟨
Δ
x
⟩
2
=
⟨
Δ
p
⟩
⟨
Δ
x
⟩
k
μ
{\displaystyle E_{\mathrm {T} }={\sqrt {{\langle \Delta x\rangle }^{2}/{\mu }\cdot 2k{\langle \Delta x\rangle }^{2}}}={\langle \Delta p\rangle }{\langle \Delta x\rangle }{\sqrt {\frac {k}{\mu }}}}
Vastavalt Heisenbergi määramatuse printsiibile,
⟨
Δ
p
⟩
⟨
Δ
x
⟩
≥
h
4
π
{\displaystyle {\langle \Delta p\rangle }{\langle \Delta x\rangle }\geq {\frac {h}{4\pi }}}
E
T
≥
k
μ
h
4
π
{\displaystyle E_{\mathrm {T} }\geq {\sqrt {\frac {k}{\mu }}}{\frac {h}{4\pi }}}
E
T
≥
1
2
h
ν
{\displaystyle E_{\mathrm {T} }\geq {\frac {1}{2}}h\nu }
ν
=
1
2
π
k
μ
{\displaystyle \nu ={\frac {1}{2\pi }}{\sqrt {\frac {k}{\mu }}}}
Harmoonilise kvantostsillaatori nullenergia võrdub
E
0
=
1
2
h
ν
{\displaystyle E_{0}={\frac {1}{2}}h\nu }
h
ν
{\displaystyle h\nu }
E
n
=
h
ν
(
n
+
1
2
)
{\displaystyle E_{n}=h\nu \left(n+{\frac {1}{2}}\right)}
Vesinikuaatomi Bohri mudel
muuda
Kineetiline energia avaldub kui
E
K
=
p
2
/
2
μ
{\displaystyle E_{\mathrm {K} }={p^{2}}/{2\mu }}
p
=
h
λ
{\displaystyle p={\frac {h}{\lambda }}}
2
π
r
=
n
λ
{\displaystyle 2\pi r=n\lambda }
E
K
=
h
2
n
2
8
π
2
r
2
μ
{\displaystyle E_{\mathrm {K} }={\frac {h^{2}n^{2}}{8\pi ^{2}r^{2}\mu }}}
Potentsiaalne energia avaldub kui
E
P
=
−
Z
e
2
4
π
ϵ
0
r
{\displaystyle E_{\mathrm {P} }=-{\frac {Ze^{2}}{4\pi \epsilon _{0}r}}}
Vesinikuaatomi puhul Coulomb'i potentsiaal on
V
=
−
Z
e
2
4
π
ϵ
0
r
{\displaystyle V=-{\frac {Ze^{2}}{4\pi \epsilon _{0}r}}}
2
E
K
=
⟨
−
r
⋅
∂
∂
r
Z
e
2
4
π
ϵ
0
r
⟩
=
−
V
{\displaystyle 2E_{\mathrm {K} }={\Big \langle }-r\cdot {\frac {\partial }{\partial r}}{\frac {Ze^{2}}{4\pi \epsilon _{0}r}}{\Big \rangle }=-V}
2
h
2
n
2
8
π
2
r
2
μ
=
Z
e
2
4
π
ϵ
0
r
{\displaystyle 2{\frac {h^{2}n^{2}}{8\pi ^{2}r^{2}\mu }}={\frac {Ze^{2}}{4\pi \epsilon _{0}r}}}
Saame avaldada
r
{\displaystyle r}
r
=
ϵ
0
h
2
Z
e
2
μ
n
2
{\displaystyle r={\frac {\epsilon _{0}h^{2}}{Ze^{2}\mu }}n^{2}}
Aatomi koguenergia on
E
T
=
E
K
+
E
P
=
−
E
K
=
1
2
E
P
{\displaystyle E_{\mathrm {T} }=E_{\mathrm {K} }+E_{\mathrm {P} }=-E_{\mathrm {K} }={\frac {1}{2}}E_{\mathrm {P} }}
E
n
=
−
Z
2
e
4
μ
8
π
ϵ
0
2
h
2
1
n
2
{\displaystyle E_{n}=-{\frac {Z^{2}e^{4}\mu }{8\pi \epsilon _{0}^{2}h^{2}}}{\frac {1}{n^{2}}}}
