設想一個二階齊次線性微分方程式
ϵ
2
d
2
y
d
x
2
=
Q
(
x
)
y
{\displaystyle \epsilon ^{2}{\frac {d^{2}y}{dx^{2}}}=Q(x)y\,\!}
其中，
Q
(
x
)
≠
0
{\displaystyle Q(x)\neq 0\,\!}
猜想解答的形式為
y
(
x
)
=
exp
[
1
δ
∑
n
=
0
∞
δ
n
S
n
(
x
)
]
{\displaystyle y(x)=\exp \left[{\frac {1}{\delta }}\sum _{n=0}^{\infty }\delta ^{n}S_{n}(x)\right]\,\!}
將猜想代入微分方程式，可以得到
ϵ
2
[
1
δ
2
(
∑
n
=
0
∞
δ
n
S
n
′
)
2
+
1
δ
∑
n
=
0
∞
δ
n
S
n
″
]
=
Q
(
x
)
{\displaystyle \epsilon ^{2}\left[{\frac {1}{\delta ^{2}}}\left(\sum _{n=0}^{\infty }\delta ^{n}S_{n}'\right)^{2}+{\frac {1}{\delta }}\sum _{n=0}^{\infty }\delta ^{n}S_{n}''\right]=Q(x)\,\!}
取
δ
→
0
{\displaystyle \delta \rightarrow 0\,\!}
ϵ
2
δ
2
S
0
′
2
∼
Q
(
x
)
{\displaystyle {\frac {\epsilon ^{2}}{\delta ^{2}}}S_{0}'^{2}\sim Q(x)\,\!}
我們可以察覺，
δ
{\displaystyle \delta \,\!}
ϵ
{\displaystyle \epsilon \,\!}
δ
=
ϵ
{\displaystyle \delta =\epsilon \,\!}
ϵ
{\displaystyle \epsilon \,\!}
ϵ
0
:
S
0
′
2
=
Q
(
x
)
{\displaystyle \epsilon ^{0}:\qquad S_{0}'^{2}=Q(x)\,\!}
我們立刻認出這是程函方程 。解答為
S
0
(
x
)
=
±
∫
x
0
x
Q
(
t
)
d
t
{\displaystyle S_{0}(x)=\pm \int _{x_{0}}^{x}{\sqrt {Q(t)}}\,dt\,\!}
檢查
ϵ
{\displaystyle \epsilon \,\!}
ϵ
1
:
2
S
0
′
S
1
′
+
S
0
″
=
0
{\displaystyle \epsilon ^{1}:\qquad 2S_{0}'S_{1}'+S_{0}''=0\,\!}
這是一個一維傳輸方程式 。解答為
S
1
(
x
)
=
−
1
4
ln
(
Q
(
x
)
)
+
k
1
{\displaystyle S_{1}(x)=-{\frac {1}{4}}\ln \left(Q(x)\right)+k_{1}\,\!}
其中，
k
1
{\displaystyle k_{1}\,\!}
我們現在有一對近似解（因為
S
0
{\displaystyle S_{0}\,\!}
y
(
x
)
≈
c
1
Q
−
1
4
(
x
)
exp
[
1
ϵ
∫
x
0
x
Q
(
t
)
d
t
]
+
c
2
Q
−
1
4
(
x
)
exp
[
−
1
ϵ
∫
x
0
x
Q
(
t
)
d
t
]
{\displaystyle y(x)\approx c_{1}Q^{-{\frac {1}{4}}}(x)\exp \left[{\frac {1}{\epsilon }}\int _{x_{0}}^{x}{\sqrt {Q(t)}}dt\right]+c_{2}Q^{-{\frac {1}{4}}}(x)\exp \left[-{\frac {1}{\epsilon }}\int _{x_{0}}^{x}{\sqrt {Q(t)}}dt\right]\,\!}
檢查
ϵ
{\displaystyle \epsilon \,\!}
n
>
2
{\displaystyle n>2\,\!}
2
S
0
′
S
n
′
+
S
n
−
1
″
+
∑
j
=
1
n
−
1
S
j
′
S
n
−
j
′
=
0
{\displaystyle 2S_{0}'S_{n}'+S''_{n-1}+\sum _{j=1}^{n-1}S'_{j}S'_{n-j}=0\,\!}
解析一個量子系統的薛丁格方程式，WKB近似涉及以下步驟：
將波函數 重寫為一個指數函數 ，
將這指數函數代入薛丁格方程式 ，
展開指數函數的參數為約化普朗克常數 的冪級數 ，
匹配約化普朗克常數同次冪的項目，會得到一組方程式，
解析這些方程式，就會得到波函數的近似。 一維不含時薛丁格方程式 為
−
ℏ
2
2
m
d
2
d
x
2
ψ
(
x
)
+
V
(
x
)
ψ
(
x
)
=
E
ψ
(
x
)
{\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\psi (x)+V(x)\psi (x)=E\psi (x)\,\!}
其中，
ℏ
{\displaystyle \hbar \,\!}
約化普朗克常數 ，
m
{\displaystyle m\,\!}
x
{\displaystyle x\,\!}
V
(
x
)
{\displaystyle V(x)\,\!}
位勢 ，
E
{\displaystyle E\,\!}
ψ
{\displaystyle \psi \,\!}
稍加編排，重寫為
ℏ
2
d
2
d
x
2
ψ
(
x
)
=
2
m
(
V
(
x
)
−
E
)
ψ
(
x
)
{\displaystyle \hbar ^{2}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\psi (x)=2m\left(V(x)-E\right)\psi (x)\,\!}
(1) 假設波函數的形式為另外一個函數
ϕ
{\displaystyle \phi \,\!}
ϕ
{\displaystyle \phi \,\!}
作用量 有很密切的關係）：
ψ
(
x
)
=
e
ϕ
(
x
)
/
ℏ
{\displaystyle \psi (x)=e^{\phi (x)/\hbar }\,\!}
代入方程式(1)，
ℏ
ϕ
″
(
x
)
+
[
ϕ
′
(
x
)
]
2
=
2
m
(
V
(
x
)
−
E
)
{\displaystyle \hbar \phi ''(x)+\left[\phi '(x)\right]^{2}=2m\left(V(x)-E\right)\,\!}
(2) 其中，
ϕ
′
{\displaystyle \phi '\,\!}
ϕ
{\displaystyle \phi \,\!}
x
{\displaystyle x\,\!}
ϕ
′
{\displaystyle \phi '\,\!}
A
(
x
)
{\displaystyle A(x)\,\!}
B
(
x
)
{\displaystyle B(x)\,\!}
ϕ
′
(
x
)
=
A
(
x
)
+
i
B
(
x
)
{\displaystyle \phi '(x)=A(x)+iB(x)\,\!}
注意到波函數的波幅是
exp
[
∫
x
A
(
x
′
)
d
x
′
/
ℏ
]
{\displaystyle \exp \left[\int ^{x}A(x')dx'/\hbar \right]\,\!}
∫
x
B
(
x
′
)
d
x
′
/
ℏ
{\displaystyle \int ^{x}B(x')dx'/\hbar \,\!}
ϕ
′
{\displaystyle \phi '\,\!}
ℏ
A
′
(
x
)
+
A
(
x
)
2
−
B
(
x
)
2
=
2
m
(
V
(
x
)
−
E
)
{\displaystyle \hbar A'(x)+A(x)^{2}-B(x)^{2}=2m\left(V(x)-E\right)\,\!}
(3)
ℏ
B
′
(
x
)
+
2
A
(
x
)
B
(
x
)
=
0
{\displaystyle \hbar B'(x)+2A(x)B(x)=0\,\!}
(4)
將
A
(
x
)
{\displaystyle A(x)\,\!}
B
(
x
)
{\displaystyle B(x)\,\!}
ℏ
{\displaystyle \hbar \,\!}
冪級數 ：
A
(
x
)
=
∑
n
=
0
∞
ℏ
n
A
n
(
x
)
{\displaystyle A(x)=\sum _{n=0}^{\infty }\hbar ^{n}A_{n}(x)\,\!}
B
(
x
)
=
∑
n
=
0
∞
ℏ
n
B
n
(
x
)
{\displaystyle B(x)=\sum _{n=0}^{\infty }\hbar ^{n}B_{n}(x)\,\!}
將兩個冪級數代入方程式(3)與(4)。
ℏ
{\displaystyle \hbar \,\!}
A
0
(
x
)
2
−
B
0
(
x
)
2
=
2
m
(
V
(
x
)
−
E
)
{\displaystyle A_{0}(x)^{2}-B_{0}(x)^{2}=2m\left(V(x)-E\right)\,\!}
A
0
(
x
)
B
0
(
x
)
=
0
{\displaystyle A_{0}(x)B_{0}(x)=0\,\!}
假若波幅變化地足夠慢於相位（
A
0
(
x
)
≪
B
0
(
x
)
{\displaystyle A_{0}(x)\ll B_{0}(x)\,\!}
A
0
(
x
)
=
0
{\displaystyle A_{0}(x)=0\,\!}
B
0
(
x
)
=
±
2
m
(
E
−
V
(
x
)
)
{\displaystyle B_{0}(x)=\pm {\sqrt {2m\left(E-V(x)\right)}}\,\!}
只有當
E
≥
V
(
x
)
{\displaystyle E\geq V(x)\,\!}
更精確一點，
ℏ
{\displaystyle \hbar \,\!}
A
0
′
+
2
A
0
A
1
−
2
B
0
B
1
=
−
2
B
0
B
1
=
0
{\displaystyle A_{0}'+2A_{0}A_{1}-2B_{0}B_{1}=-2B_{0}B_{1}=0\,\!}
B
0
′
+
2
A
0
B
1
+
2
B
0
A
1
=
B
0
′
+
2
B
0
A
1
=
0
{\displaystyle B_{0}'+2A_{0}B_{1}+2B_{0}A_{1}=B_{0}'+2B_{0}A_{1}=0\,\!}
所以，
B
1
=
0
{\displaystyle B_{1}=0\,\!}
A
1
=
−
B
0
′
2
B
0
=
d
d
x
l
n
B
0
−
1
/
2
{\displaystyle A_{1}=-{\frac {B_{0}'}{2B_{0}}}={\frac {d}{dx}}lnB_{0}^{-1/2}\,\!}
波函數的波幅是
exp
[
∫
x
A
(
x
′
)
d
x
′
/
ℏ
]
=
1
B
0
{\displaystyle \exp \left[\int ^{x}A(x')dx'/\hbar \right]={\frac {1}{\sqrt {B_{0}}}}\,\!}
定義動量
p
(
x
)
=
2
m
(
E
−
V
(
x
)
)
{\displaystyle p(x)={\sqrt {2m\left(E-V(x)\right)}}\,\!}
ψ
(
x
)
≈
C
±
p
(
x
)
e
±
i
∫
x
0
x
p
(
x
′
)
d
x
′
/
ℏ
{\displaystyle \psi (x)\approx {\cfrac {C_{\pm }}{\sqrt {p(x)}}}e^{\pm i\int _{x_{0}}^{x}p(x')\mathrm {d} x'/\hbar }\,\!}
(5) 其中，
C
+
{\displaystyle C_{+}\,\!}
C
−
{\displaystyle C_{-}\,\!}
x
0
{\displaystyle x_{0}\,\!}
換到另一方面，假若相位變化地足夠慢於波幅（
B
0
(
x
)
≪
A
0
(
x
)
{\displaystyle B_{0}(x)\ll A_{0}(x)\,\!}
A
0
(
x
)
=
±
2
m
(
V
(
x
)
−
E
)
{\displaystyle A_{0}(x)=\pm {\sqrt {2m\left(V(x)-E\right)}}\,\!}
B
0
(
x
)
=
0
{\displaystyle B_{0}(x)=0\,\!}
只有當
V
(
x
)
≥
E
{\displaystyle V(x)\geq E\,\!}
量子穿隧效應 。類似地計算，可以求得波函數的近似為
ψ
(
x
)
≈
C
±
p
(
x
)
e
±
∫
x
0
x
p
(
x
′
)
d
x
′
/
ℏ
{\displaystyle \psi (x)\approx {\frac {C_{\pm }}{\sqrt {p(x)}}}e^{\pm \int _{x_{0}}^{x}p(x')\mathrm {d} x'/\hbar }\,\!}
(6) 其中，
p
(
x
)
=
2
m
(
V
(
x
)
−
E
)
{\displaystyle p(x)={\sqrt {2m\left(V(x)-E\right)}}\,\!}
顯而易見地，我們可以從分母觀察出來，在經典轉向點
E
=
V
(
x
)
{\displaystyle E=V(x)\,\!}
x
1
<
x
<
x
2
{\displaystyle x_{1}<x<x_{2}\,\!}
E
>
V
(
x
)
{\displaystyle E>V(x)\,\!}
x
<
x
1
{\displaystyle x<x_{1}\,\!}
x
2
<
x
{\displaystyle x_{2}<x\,\!}
x
2
{\displaystyle x_{2}\,\!}
2
m
ℏ
2
(
V
(
x
)
−
E
)
{\displaystyle {\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)\,\!}
2
m
ℏ
2
(
V
(
x
)
−
E
)
=
U
1
(
x
−
x
2
)
+
U
2
(
x
−
x
2
)
2
+
⋯
{\displaystyle {\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)=U_{1}(x-x_{2})+U_{2}(x-x_{2})^{2}+\cdots \,\!}
其中，
U
1
,
U
2
,
⋯
{\displaystyle U_{1},\,U_{2},\,\cdots \,\!}
取至一階，方程式(1)變為
d
2
d
x
2
ψ
(
x
)
=
U
1
(
x
−
x
2
)
ψ
(
x
)
{\displaystyle {\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\psi (x)=U_{1}(x-x_{2})\psi (x)\,\!}
這微分方程式稱為艾里方程式 ，其解為著名的艾里函數 ：
ψ
(
x
)
=
C
2
A
Ai
(
U
1
3
(
x
−
x
2
)
)
+
C
2
B
Bi
(
U
1
3
(
x
−
x
2
)
)
{\displaystyle \psi (x)=C_{2A}{\textrm {Ai}}\left({\sqrt[{3}]{U_{1}}}(x-x_{2})\right)+C_{2B}{\textrm {Bi}}\left({\sqrt[{3}]{U_{1}}}(x-x_{2})\right)\,\!}
匹配艾里函數和在
x
<
x
2
{\displaystyle x<x_{2}\,\!}
x
2
<
x
{\displaystyle x_{2}<x\,\!}
x
2
{\displaystyle x_{2}\,\!}
連接公式 （connection formula ）[ 1]
ψ
(
x
)
=
{
2
C
2
p
(
x
)
sin
(
1
ℏ
∫
x
x
2
p
(
x
′
)
d
x
′
+
π
4
)
if
x
<
x
2
C
2
|
p
(
x
)
|
exp
(
−
∫
x
2
x
|
p
(
x
′
)
|
d
x
′
/
ℏ
)
if
x
2
<
x
{\displaystyle \psi (x)={\begin{cases}{\cfrac {2C_{2}}{\sqrt {p(x)}}}\sin \left({\cfrac {1}{\hbar }}\int _{x}^{x_{2}}p(x')dx'+{\cfrac {\pi }{4}}\right)&{\mbox{if }}x<x_{2}\\{\cfrac {C_{2}}{\sqrt {|p(x)|}}}\exp \left(-\int _{x_{2}}^{x}|p(x')|dx'/{\hbar }\right)&{\mbox{if }}x_{2}<x\end{cases}}\,\!}
。 類似地，也可以得到在
x
1
{\displaystyle x_{1}\,\!}
ψ
(
x
)
=
{
C
1
|
p
(
x
)
|
exp
(
−
∫
x
x
1
|
p
(
x
′
)
|
d
x
′
/
ℏ
)
if
x
<
x
1
2
C
1
p
(
x
)
sin
(
1
ℏ
∫
x
1
x
p
(
x
′
)
d
x
′
+
π
4
)
if
x
1
<
x
{\displaystyle \psi (x)={\begin{cases}{\cfrac {C_{1}}{\sqrt {|p(x)|}}}\exp \left(-\int _{x}^{x_{1}}|p(x')|dx'/{\hbar }\right)&{\mbox{if }}x<x_{1}\\{\cfrac {2C_{1}}{\sqrt {p(x)}}}\sin \left({\cfrac {1}{\hbar }}\int _{x_{1}}^{x}p(x')dx'+{\cfrac {\pi }{4}}\right)&{\mbox{if }}x_{1}<x\end{cases}}\,\!}
。
在經典運動允許區域
x
1
<
x
<
x
2
{\displaystyle x_{1}<x<x_{2}\,\!}
θ
1
=
−
1
ℏ
∫
x
1
x
p
(
x
′
)
d
x
′
−
π
4
{\displaystyle \theta _{1}=-{\frac {1}{\hbar }}\int _{x_{1}}^{x}p(x')dx'-{\frac {\pi }{4}}\,\!}
θ
2
=
1
ℏ
∫
x
x
2
p
(
x
′
)
d
x
′
+
π
4
{\displaystyle \theta _{2}=~{\frac {1}{\hbar }}\int _{x}^{x_{2}}p(x')dx'+{\frac {\pi }{4}}\,\!}
α
=
∫
x
1
x
2
p
(
x
)
d
x
/
ℏ
{\displaystyle \alpha =\int _{x_{1}}^{x_{2}}p(x)dx/\hbar \,\!}
那麼，
α
=
θ
2
−
θ
1
−
π
/
2
{\displaystyle \alpha =\theta _{2}-\theta _{1}-\pi /2\,\!}
−
C
1
sin
θ
1
=
C
2
sin
θ
2
=
C
2
sin
(
θ
1
+
α
+
π
/
2
)
{\displaystyle -C_{1}\sin \theta _{1}=C_{2}\sin \theta _{2}=C_{2}\sin(\theta _{1}+\alpha +\pi /2)\,\!}
立刻，我們可以認定
|
C
1
|
=
|
C
2
|
{\displaystyle |C_{1}|=|C_{2}|\,\!}
C
1
=
C
2
{\displaystyle C_{1}=C_{2}\,\!}
α
+
π
/
2
=
(
2
m
−
1
)
π
,
m
=
1
,
2
,
3
,
…
{\displaystyle \alpha +\pi /2=(2m-1)\pi ,\qquad m=1,\,2,\,3,\,\dots \,\!}
所以，
α
=
(
2
m
−
3
/
2
)
π
,
m
=
1
,
2
,
3
,
…
{\displaystyle \alpha =(2m-3/2)\pi ,\qquad m=1,\,2,\,3,\,\dots \,\!}
假若
C
1
=
−
C
2
{\displaystyle C_{1}=-C_{2}\,\!}
α
+
π
/
2
=
2
m
π
,
m
=
1
,
2
,
3
,
…
{\displaystyle \alpha +\pi /2=2m\pi ,\qquad m=1,\,2,\,3,\,\dots \,\!}
所以，
α
=
(
2
m
−
1
/
2
)
π
,
m
=
1
,
2
,
3
,
…
{\displaystyle \alpha =(2m-1/2)\pi ,\qquad m=1,\,2,\,3,\,\dots \,\!}
總結，量子系統必須滿足量子化守則：
∫
x
1
x
2
p
(
x
)
d
x
=
(
n
−
1
/
2
)
π
ℏ
,
n
=
1
,
2
,
3
,
…
{\displaystyle \int _{x_{1}}^{x_{2}}p(x)dx=(n-1/2)\pi \hbar ,\qquad n=1,\,2,\,3,\,\dots \,\!}
考慮一個量子諧振子 系統，一個質量為
m
{\displaystyle m\,\!}
V
(
x
)
=
1
2
m
ω
2
x
2
{\displaystyle V(x)={\frac {1}{2}}m\omega ^{2}x^{2}\,\!}
ω
{\displaystyle \omega \,\!}
本徵能級
E
n
{\displaystyle E_{n}\,\!}
能量為
E
{\displaystyle E\,\!}
x
t
{\displaystyle x_{t}\,\!}
E
=
1
2
m
ω
2
x
t
2
{\displaystyle E={\frac {1}{2}}m\omega ^{2}x_{t}^{2}\,\!}
所以，
x
t
=
±
2
E
m
ω
2
{\displaystyle x_{t}=\pm {\sqrt {\frac {2E}{m\omega ^{2}}}}\,\!}
粒子的動量為
p
(
x
)
=
2
m
(
E
−
1
2
m
ω
2
x
2
)
{\displaystyle p(x)={\sqrt {2m\left(E-{\frac {1}{2}}m\omega ^{2}x^{2}\right)}}\,\!}
將這些變量代入量子化守則：
∫
−
2
E
/
m
ω
2
2
E
/
m
ω
2
2
m
(
E
−
1
2
m
ω
2
x
2
)
d
x
=
(
n
−
1
/
2
)
π
ℏ
,
n
=
1
,
2
,
3
,
…
{\displaystyle \int _{-2E/m\omega ^{2}}^{2E/m\omega ^{2}}\,{\sqrt {2m\left(E-{\frac {1}{2}}m\omega ^{2}x^{2}\right)}}\,dx=(n-1/2)\pi \hbar ,\qquad n=1,\,2,\,3,\,\dots \,\!}
經過一番運算，可以得到本徵能量
E
n
=
(
n
−
1
/
2
)
ω
ℏ
,
n
=
1
,
2
,
3
,
…
{\displaystyle E_{n}=(n-1/2)\omega \hbar ,\qquad n=1,\,2,\,3,\,\dots \,\!}
藉由以上之計算，發現近似解與精確解完全一樣。
![{\displaystyle y(x)\sim \exp \left[{\frac {1}{\delta }}\sum _{n=0}^{\infty }\delta ^{n}S_{n}(x)\right]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16955980a418738921d7fcb12e804bf3d0b30c10)
![{\displaystyle y(x)=\exp \left[{\frac {1}{\delta }}\sum _{n=0}^{\infty }\delta ^{n}S_{n}(x)\right]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d7f19608f7f038075b346d014da71a5c043d350)
![{\displaystyle \epsilon ^{2}\left[{\frac {1}{\delta ^{2}}}\left(\sum _{n=0}^{\infty }\delta ^{n}S_{n}'\right)^{2}+{\frac {1}{\delta }}\sum _{n=0}^{\infty }\delta ^{n}S_{n}''\right]=Q(x)\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2990b480e61e94fb7509374040c0669703d354ba)
![{\displaystyle y(x)\approx c_{1}Q^{-{\frac {1}{4}}}(x)\exp \left[{\frac {1}{\epsilon }}\int _{x_{0}}^{x}{\sqrt {Q(t)}}dt\right]+c_{2}Q^{-{\frac {1}{4}}}(x)\exp \left[-{\frac {1}{\epsilon }}\int _{x_{0}}^{x}{\sqrt {Q(t)}}dt\right]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0283901fe6502a928ea16f053dde91d07a6cc1cd)
![{\displaystyle \hbar \phi ''(x)+\left[\phi '(x)\right]^{2}=2m\left(V(x)-E\right)\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57761c223b34e50bcd0d1aa7f5dcb40d938ed087)
![{\displaystyle \exp \left[\int ^{x}A(x')dx'/\hbar \right]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9621eca139247216bfbdc9b971564ec082e74437)
![{\displaystyle \exp \left[\int ^{x}A(x')dx'/\hbar \right]={\frac {1}{\sqrt {B_{0}}}}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6f90fa64e9c4aeef934dd780512a3325d7a4042)
![{\displaystyle \psi (x)=C_{2A}{\textrm {Ai}}\left({\sqrt[{3}]{U_{1}}}(x-x_{2})\right)+C_{2B}{\textrm {Bi}}\left({\sqrt[{3}]{U_{1}}}(x-x_{2})\right)\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e9832e88590ccf2150c89024e3f53c88722ab43)
