Equations for the Physics 264L Second Midterm
Professor Greenside
Monday, November 9, 2015
E = ℏω,
iℏ ∂t Ψ(t, x) =
p = ℏk,
−ℏ2 2
∂ Ψ + V (x)Ψ,
2m x
∫
⟨ Ψ1 | Ψ2 ⟩ =
∞
−∞
ω
,
k
vϕ =
vg =
dω
.
dk
−ℏ2 2
∂ ψ(x) + V (x)ψ = Eψ.
2m x
Ψ∗1 (x, t)Ψ2 (x, t) dx.
(1)
(2)
(3)
For functions that are square-integrable (decay to zero at ±∞ sufficiently quickly), integration by parts let’s
you move a derivative onto everything else in an integrand, provided you reverse the sign:
∫ ∞
∫ ∞
dh(x)
d[f (x)g(x)]
f (x)g(x)
dx = −
h(x) dx
(4)
dx
dx
−∞
−∞
This can also be written in terms of dot products as
⟨ f (x)g(x) |
√
ψn,L =
dh
d[f (x)g(x)]
⟩ = −⟨
| h(x) dx.
dx
dx
( πnx )
2
sin
,
L
L
En =
E > V0
ψ(x) = c1 eikx + c2 e−ikx ,
V0 > E
ψ(x) = c1 eκx + c2 e−κx ,
π 2 ℏ2 2
n ,
2mL2
(5)
n ≥ 1.
2m(E − V0 )
ℏ2
2m(V
0 − E)
κ2 =
.
ℏ2
k2 =
If
Ψ(t = 0, x) = c1 ψ1 (x) + c2 ψ2 (x) + c3 ψ2 (x) + · · · =
∞
∑
cn ψn (x),
(6)
(7)
(8)
(9)
n=1
then
Ψ(t, x) = c1 ψ1 (x)e−i(E1 /ℏ)t + c2 ψ2 (x)e−i(E2 /ℏ)t + c3 ψ2 (x)e−i(E3 /ℏ)t + · · ·
∞
∑
=
cn ψn (x)e−i(En /ℏ)t ,
where cn = ⟨ ψn | Ψ ⟩ since ⟨ψm |ψn ⟩ = δmn .
(10)
(11)
n=1
∫
⟨xk ⟩ =
∞
−∞
[
]
Ψ∗ (t, x) xk Ψ(t, x) dx,
∫
⟨p⟩ =
1
(
)
d⟨x⟩
Ψ∗ (t, x) −iℏ∂x Ψ dx = m
dt
−∞
∞
(12)