Relation between energy and frequency of a quanta of radiation
edit
E
=
h
f
{\displaystyle E=hf\;}
E
=
ℏ
ω
{\displaystyle E=\hbar \omega }
p
=
ℏ
k
{\displaystyle \mathbf {p} =\hbar \mathbf {k} \;}
Energy:
E
{\displaystyle E}
Frequency:
f
{\displaystyle f}
Angular Frequency:
ω
=
2
π
f
{\displaystyle \omega =2\pi f}
Wavenumber:
k
=
2
π
/
λ
{\displaystyle k=2\pi /\lambda }
Plank's Constant:
h
{\displaystyle h}
De Broglie Hypothesis
edit
p
=
h
/
λ
{\displaystyle p=h/\lambda \;}
Wavelength:
λ
{\displaystyle \lambda }
Momentum:
p
{\displaystyle p}
Phase of a Plane Wave Expressed as a Complex Phase Factor
edit
ψ
≈
e
i
(
k
⋅
x
−
ω
t
)
{\displaystyle \psi \approx e^{i(\mathbf {k} \cdot \mathbf {x} -\omega t)}}
Time-Dependent Schrodinger Equation
edit
i
ℏ
∂
∂
t
Ψ
=
−
ℏ
2
2
m
∇
2
Ψ
+
V
Ψ
{\displaystyle i\hbar {\frac {\partial }{\partial t}}\Psi =-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\Psi +V\Psi \;}
i
ℏ
d
d
t
|
ψ
(
t
)
⟩
=
H
(
t
)
|
ψ
(
t
)
⟩
{\displaystyle \mathrm {i} \hbar {\frac {d}{dt}}\left|\psi \left(t\right)\right\rangle =H(t)\left|\psi \left(t\right)\right\rangle }
Ket:
|
ψ
(
t
)
⟩
{\displaystyle |\psi (t)\rangle }
Reduced Planck's Constant:
ℏ
{\displaystyle \hbar }
Hamiltonian:
H
(
t
)
{\displaystyle H(t)}
The Hamiltonian describes the total energy of the system. Partial Derivative:
∂
/
∂
t
{\displaystyle \partial /\partial t}
Mass:
m
{\displaystyle m}
Potential:
V
{\displaystyle V}
Begin with a step from the time-independent derivation
1
Ψ
d
2
Ψ
d
x
2
=
1
c
2
ζ
d
2
ζ
d
t
2
{\displaystyle {\frac {1}{\Psi }}{\frac {d^{2}\Psi }{dx^{2}}}={\frac {1}{c^{2}\zeta }}{\frac {d^{2}\zeta }{dt^{2}}}}
Set each side equal to a constant,
−
κ
2
{\displaystyle -\kappa ^{2}}
−
κ
2
=
1
c
2
ζ
d
2
ζ
d
t
2
{\displaystyle -\kappa ^{2}={\frac {1}{c^{2}\zeta }}{\frac {d^{2}\zeta }{dt^{2}}}}
Multiply by
c
2
{\displaystyle c^{2}}
−
β
2
=
1
ζ
d
2
ζ
d
t
2
{\displaystyle -\beta ^{2}={\frac {1}{\zeta }}{\frac {d^{2}\zeta }{dt^{2}}}}
The solution is similar to what was found previously
ζ
(
t
)
=
N
e
±
i
β
t
{\displaystyle \zeta (t)=Ne^{\pm i\beta t}}
The amplitude at a point
t
{\displaystyle t}
t
+
τ
{\displaystyle t+\tau }
N
e
±
i
β
t
=
N
e
±
i
β
(
t
+
τ
)
{\displaystyle Ne^{\pm i\beta t}=Ne^{\pm i\beta (t+\tau )}}
The following equation must be true:
β
τ
=
2
π
{\displaystyle \beta \tau =2\pi \;}
Rewrite
β
{\displaystyle \beta }
β
=
2
π
v
{\displaystyle \beta =2\pi v\;}
Enter the equation into the expression of
ζ
{\displaystyle \zeta }
ζ
(
t
)
=
N
e
±
2
π
i
v
t
{\displaystyle \zeta (t)=Ne^{\pm 2\pi ivt}}
ζ
(
t
)
=
N
e
−
i
E
t
/
ℏ
{\displaystyle \zeta (t)=Ne^{-iEt/\hbar }}
The time-dependent Schrodinger equation is a product of two 'sub-functions'
Ψ
(
x
,
t
)
=
ψ
(
x
)
ζ
(
t
)
{\displaystyle \Psi (x,t)=\psi (x)\zeta (t)\;}
Ψ
(
x
,
t
)
=
ψ
e
−
i
E
t
/
ℏ
{\displaystyle \Psi (x,t)=\psi e^{-iEt/\hbar }}
To extract
E
{\displaystyle E}
∂
Ψ
∂
t
=
−
i
E
ℏ
ψ
e
−
i
E
t
/
ℏ
{\displaystyle {\frac {\partial \Psi }{\partial t}}={\frac {-iE}{\hbar }}\psi e^{-iEt/\hbar }}
∂
Ψ
∂
t
=
E
i
ℏ
ψ
e
−
i
E
t
/
ℏ
{\displaystyle {\frac {\partial \Psi }{\partial t}}={\frac {E}{i\hbar }}\psi e^{-iEt/\hbar }}
Rearrange:
i
ℏ
∂
Ψ
∂
t
=
E
Ψ
{\displaystyle i\hbar {\frac {\partial \Psi }{\partial t}}=E\Psi }
H
^
Ψ
=
E
Ψ
{\displaystyle {\hat {H}}\Psi =E\Psi }
Time-Independent Schrodinger Equation
edit
H
Ψ
=
E
Ψ
{\displaystyle H\Psi =E\Psi \;}
−
ℏ
2
2
m
d
2
ψ
(
x
)
d
x
2
+
U
(
x
)
ψ
(
x
)
=
E
ψ
(
x
)
{\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi (x)}{dx^{2}}}+U(x)\psi (x)=E\psi (x)}
[
−
ℏ
2
2
m
∇
2
+
U
(
r
)
]
ψ
(
r
)
=
E
ψ
(
r
)
{\displaystyle \left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+U(\mathbf {r} )\right]\psi (\mathbf {r} )=E\psi (\mathbf {r} )}
−
ℏ
2
2
m
∇
2
ψ
+
(
U
−
E
)
ψ
=
0
{\displaystyle -{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi +(U-E)\psi =0}
H
|
ψ
n
⟩
=
E
n
|
ψ
n
⟩
.
{\displaystyle H\left|\psi _{n}\right\rangle =E_{n}\left|\psi _{n}\right\rangle .}
Del Operator:
∇
{\displaystyle \nabla }
The Schrodinger Equation is based on two formulas:
The classical wave function derived from the Newton's Second Law
The de Broglie wave expression Formula of a classical wave:
d
2
z
d
x
2
=
1
c
2
d
2
z
d
t
2
{\displaystyle {\frac {d^{2}z}{dx^{2}}}={\frac {1}{c^{2}}}{\frac {d^{2}z}{dt^{2}}}}
Separate the function into two variables:
z
(
x
,
t
)
=
Ψ
(
x
)
ζ
(
t
)
{\displaystyle z(x,t)=\Psi (x)\zeta (t)\;}
Insert the function into the wave equation:
ζ
d
2
Ψ
d
x
2
=
Ψ
c
2
d
2
ζ
d
t
2
{\displaystyle \zeta {\frac {d^{2}\Psi }{dx^{2}}}={\frac {\Psi }{c^{2}}}{\frac {d^{2}\zeta }{dt^{2}}}}
Rearrange to separate
Ψ
{\displaystyle \Psi }
ζ
{\displaystyle \zeta }
1
Ψ
d
2
Ψ
d
x
2
=
1
c
2
ζ
d
2
ζ
d
t
2
{\displaystyle {\frac {1}{\Psi }}{\frac {d^{2}\Psi }{dx^{2}}}={\frac {1}{c^{2}\zeta }}{\frac {d^{2}\zeta }{dt^{2}}}}
Set each side equal to an arbitrary constant,
−
κ
2
{\displaystyle -\kappa ^{2}}
1
Ψ
d
2
Ψ
d
x
2
=
−
κ
2
{\displaystyle {\frac {1}{\Psi }}{\frac {d^{2}\Psi }{dx^{2}}}=-\kappa ^{2}}
d
2
Ψ
d
x
2
=
−
κ
2
Ψ
{\displaystyle {\frac {d^{2}\Psi }{dx^{2}}}=-\kappa ^{2}\Psi }
Solve this equation
Ψ
(
x
)
=
N
e
±
i
κ
x
{\displaystyle \Psi (x)=Ne^{\pm i\kappa x}}
The amplitude at one point needs to be equal to the amplitude at another point:
N
e
±
i
κ
x
=
N
e
±
i
κ
(
x
+
λ
)
{\displaystyle Ne^{\pm i\kappa x}=Ne^{\pm i\kappa (x+\lambda )}}
The following condition must be true:
κ
λ
=
2
π
{\displaystyle \kappa \lambda =2\pi \;}
Incorporate the de Broglie wave expression
h
m
v
=
λ
{\displaystyle {\frac {h}{mv}}=\lambda }
κ
=
2
π
m
v
h
{\displaystyle \kappa ={\frac {2\pi mv}{h}}}
Use the symbol
ℏ
{\displaystyle \hbar }
ℏ
=
h
2
π
{\displaystyle \hbar ={\frac {h}{2\pi }}}
d
2
Ψ
d
x
2
=
−
m
2
v
2
ℏ
2
Ψ
{\displaystyle {\frac {d^{2}\Psi }{dx^{2}}}=-{\frac {m^{2}v^{2}}{\hbar ^{2}}}\Psi }
−
ℏ
2
m
2
v
2
d
2
Ψ
d
x
2
=
Ψ
{\displaystyle {\frac {-\hbar ^{2}}{m^{2}v^{2}}}{\frac {d^{2}\Psi }{dx^{2}}}=\Psi }
Use the expression of kinetic energy,
E
k
i
n
e
t
i
c
=
1
2
m
v
2
{\displaystyle E_{kinetic}={\frac {1}{2}}mv^{2}}
−
ℏ
2
2
m
d
2
Ψ
d
x
2
=
E
Ψ
{\displaystyle {\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}\Psi }{dx^{2}}}=E\Psi }
Modify the equation by adding a potential energy term and the Laplacian operator
−
ℏ
2
2
m
∇
2
Ψ
+
V
Ψ
=
E
Ψ
{\displaystyle {\frac {-\hbar ^{2}}{2m}}\nabla ^{2}\Psi +V\Psi =E\Psi }
H
^
Ψ
=
E
Ψ
{\displaystyle {\hat {H}}\Psi =E\Psi }
Non-Relativistic Schrodinger Wave Equation
edit
In non-relativistic quantum mechanics, the Hamiltonian of a particle can be expressed as the sum of two operators, one corresponding to kinetic energy and the other to potential energy. The Hamiltonian of a particle with no electric charge and no spin in this case is:
H
ψ
(
r
,
t
)
=
(
T
+
V
)
ψ
(
r
,
t
)
{\displaystyle H\psi \left(\mathbf {r} ,t\right)=\left(T+V\right)\psi \left(\mathbf {r} ,t\right)}
H
ψ
(
r
,
t
)
=
[
−
ℏ
2
2
m
∇
2
+
V
(
r
)
]
ψ
(
r
,
t
)
{\displaystyle H\psi \left(\mathbf {r} ,t\right)=\left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V\left(\mathbf {r} \right)\right]\psi \left(\mathbf {r} ,t\right)}
H
ψ
(
r
,
t
)
=
i
ℏ
∂
ψ
∂
t
(
r
,
t
)
{\displaystyle H\psi \left(\mathbf {r} ,t\right)=\mathrm {i} \hbar {\frac {\partial \psi }{\partial t}}\left(\mathbf {r} ,t\right)}
kinetic energy operator:
T
=
p
2
2
m
{\displaystyle T={\frac {p^{2}}{2m}}}
mass of the particle:
m
{\displaystyle m\;}
momentum operator:
p
=
−
i
ℏ
∇
{\displaystyle \mathbf {p} =-\mathrm {i} \hbar \nabla }
potential energy operator:
V
=
V
(
r
)
{\displaystyle V=V\left(\mathbf {r} \right)}
real scalar function of the position operator
r
{\displaystyle \mathbf {r} }
V
{\displaystyle V}
Gradient operator:
∇
{\displaystyle \nabla }
Laplace operator:
∇
2
{\displaystyle \nabla ^{2}}
![{\displaystyle \left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+U(\mathbf {r} )\right]\psi (\mathbf {r} )=E\psi (\mathbf {r} )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bdf858301e2a61ba8e87afde1f2929f2371f8ff6)
![{\displaystyle H\psi \left(\mathbf {r} ,t\right)=\left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V\left(\mathbf {r} \right)\right]\psi \left(\mathbf {r} ,t\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/418bcff45edc843cc06873169625ac29b8482388)
