1 引言
在《量子力學》這本書中，其中一節的內容就是關於「一維諧振子」的定態 Schr?dinger 方程。內容如下﹕
一維諧振子的勢場
V
(
x
)
=
1
2
m
ω
2
x
2
V(x) = \frac{1}{2}m\omega^2x^2
V(x)=21?mω2x2
其定態波函數
ψ
(
x
)
\psi(x)
ψ(x) 滿足
?
?
2
2
m
d
2
ψ
d
x
2
+
1
2
m
ω
2
x
2
ψ
=
E
ψ
-\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2}+\frac{1}{2}m\omega^2x^2\psi=E\psi
?2m?2?dx2d2ψ?+21?mω2x2ψ=Eψ
解方程後得
ψ
n
(
x
)
=
N
n
exp
?
(
?
α
2
x
2
2
)
H
n
(
α
x
)
\psi_n(x)=N_n\exp(-\frac{\alpha^2x^2}{2})H_n(\alpha x)
ψn?(x)=Nn?exp(?2α2x2?)Hn?(αx)
N
n
N_n
Nn? 是歸一化因子
α
π
1
2
?
2
n
?
n
!
\sqrt{\frac{\alpha}{\pi^{\frac{1}{2}}\cdot 2^n\cdot n!}}
π21??2n?n!α?
?
α
\alpha
α 是位移無量綱化因子
m
ω
?
\sqrt{\frac{m\omega}{\hbar}}
?mω?
?
H
n
(
x
)
H_n(x)
Hn?(x) 是
n
n
n 階 Hermite 多項式
(
?
1
)
n
exp
?
(
x
2
)
d
n
d
x
n
exp
?
(
?
x
2
)
(-1)^n\exp(x^2)\frac{d^n}{dx^n}\exp(-x^2)
(?1)nexp(x2)dxndn?exp(?x2)
E
n
E_n
En? 是粒子能級分佈
(
n
+
1
2
)
?
ω
(n+\frac{1}{2})\hbar\omega
(n+21?)?ω。
我基於一維諧振子的理論，延伸到二維諧振子，逐步延伸 n 維諧振子，然後由二維諧振子延伸到二維「正定勢場」下的波函數，最後擴展到 n 維空間「正定勢場」下的波函數。
2 二維諧振子
二維諧振子下的勢場
V
(
x
,
y
)
=
1
2
m
ω
2
(
x
2
+
y
2
)
V(x,y)=\frac{1}{2}m\omega^2(x^2+y^2)
V(x,y)=21?mω2(x2+y2)
其定態波函數
ψ
(
x
,
y
)
\psi(x,y)
ψ(x,y) 滿足
?
?
2
2
m
?
2
ψ
(
x
,
y
)
+
1
2
m
ω
2
(
x
2
+
y
2
)
ψ
(
x
,
y
)
=
E
ψ
(
x
,
y
)
-\frac{\hbar^2}{2m}\nabla^2\psi(x,y)+\frac{1}{2}m\omega^2(x^2+y^2)\psi(x,y)=E\psi(x,y)
?2m?2??2ψ(x,y)+21?mω2(x2+y2)ψ(x,y)=Eψ(x,y)
定義算子
H
x
^
=
?
?
2
2
m
?
2
?
x
2
+
1
2
m
ω
2
x
2
\hat{H_x}=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac{1}{2}m\omega^2x^2
Hx?^?=?2m?2??x2?2?+21?mω2x2
H
y
^
=
?
?
2
2
m
?
2
?
y
2
+
1
2
m
ω
2
y
2
\hat{H_y}=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial y^2}+\frac{1}{2}m\omega^2y^2
Hy?^?=?2m?2??y2?2?+21?mω2y2
利用分離變量法，設
ψ
(
x
,
y
)
=
f
(
x
)
g
(
y
)
\psi(x,y)=f(x)g(y)
ψ(x,y)=f(x)g(y)，所以有
H
^
ψ
=
g
(
y
)
H
x
^
f
(
x
)
+
f
(
x
)
H
y
^
g
(
y
)
=
E
f
(
x
)
g
(
y
)
f
(
x
)
H
y
^
g
(
y
)
=
(
E
f
(
x
)
?
H
x
^
f
(
x
)
)
g
(
y
)
H
y
^
g
(
y
)
=
E
y
g
(
y
)
?且?
H
x
^
f
(
x
)
)
=
(
E
?
E
y
)
f
(
x
)
=
E
x
f
(
x
)
\hat{H}\psi=g(y)\hat{H_x}f(x)+f(x)\hat{H_y}g(y)=Ef(x)g(y)\\[1em] f(x)\hat{H_y}g(y)=(Ef(x)-\hat{H_x}f(x))g(y)\\[1em] \hat{H_y}g(y)=E_yg(y)\ \text{且}\ \hat{H_x}f(x))=(E-E_y)f(x)=E_xf(x)
H^ψ=g(y)Hx?^?f(x)+f(x)Hy?^?g(y)=Ef(x)g(y)f(x)Hy?^?g(y)=(Ef(x)?Hx?^?f(x))g(y)Hy?^?g(y)=Ey?g(y)?且?Hx?^?f(x))=(E?Ey?)f(x)=Ex?f(x)
由一維諧振子的結論可得，
H
x
^
\hat{H_x}
Hx?^? 和
H
y
^
\hat{H_y}
Hy?^? 的本征值都是
{
E
1
,
n
}
n
≥
0
\{E_{1,n}\}_{n\geq0}
{E1,n?}n≥0?，對應的本征函數是
{
ψ
n
(
x
)
}
n
≥
0
\{\psi_n(x)\}_{n\geq0}
{ψn?(x)}n≥0? 和
{
ψ
n
(
y
)
}
n
≥
0
\{\psi_n(y)\}_{n\geq0}
{ψn?(y)}n≥0?。
現在
f
f
f 和
g
g
g 分別是
H
x
^
\hat{H_x}
Hx?^? 和
H
y
^
\hat{H_y}
Hy?^? 的本征函數，所以
f
∈
{
ψ
n
(
x
)
}
n
≥
0
f \in \{\psi_n(x)\}_{n\geq0}
f∈{ψn?(x)}n≥0? 和
g
∈
{
ψ
n
(
y
)
}
n
≥
0
g \in \{\psi_n(y)\}_{n\geq0}
g∈{ψn?(y)}n≥0?，即
ψ
m
+
n
(
x
,
y
)
=
ψ
m
(
x
)
ψ
n
(
y
)
\psi_{m+n}(x,y)=\psi_m(x)\psi_n(y)
ψm+n?(x,y)=ψm?(x)ψn?(y)。與此同時，對應的能級是
E
2
,
N
=
E
m
n
=
E
1
,
m
+
E
1
,
n
=
(
m
+
n
+
1
)
?
ω
E_{2,N}=E_{mn}=E_{1,m}+E_{1,n}=(m+n+1)\hbar\omega
E2,N?=Emn?=E1,m?+E1,n?=(m+n+1)?ω，簡並度是
(
N
+
1
1
)
=
N
+
1
\left(\begin{matrix}N+1\\1\end{matrix}\right)=N+1
(N+11?)=N+1。
3
n
n
n 維諧振子
以下用數學歸納法來證明如下命題﹕
當
n
n
n 維諧振子的勢場為
V
(
r
n
?
)
=
V
(
x
1
,
x
2
,
?
?
,
x
n
)
=
1
2
m
ω
2
r
n
?
2
V(\vec{r_n}) = V(x_1,x_2,\cdots,x_n) = \frac{1}{2}m\omega^2\vec{r_n}^2
V(rn?
?)=V(x1?,x2?,?,xn?)=21?mω2rn?
?2，
本征波函數族為
{
∏
k
=
1
n
ψ
N
k
(
x
k
)
∣
∑
k
=
1
n
N
k
=
N
}
\{\prod\limits_{k=1}^{n}\psi_{N_k}(x_k)|\sum\limits_{k=1}^{n}N_k=N\}
{k=1∏n?ψNk??(xk?)∣k=1∑n?Nk?=N}，其中
ψ
k
\psi_k
ψk? 是一維諧振子各能級
E
k
E_k
Ek? 對應的定態波函數。與此同時，對應的本征能級是
E
n
,
N
=
(
∑
k
=
1
n
N
k
+
n
2
)
?
ω
=
(
N
+
n
2
)
?
ω
E_{n,N}=(\sum\limits_{k=1}^nN_k+\frac{n}{2})\hbar\omega=(N+\frac{n}{2})\hbar\omega
En,N?=(k=1∑n?Nk?+2n?)?ω=(N+2n?)?ω，簡並度是
(
N
+
n
?
1
n
?
1
)
\left(\begin{matrix}N+n-1\\n-1\end{matrix}\right)
(N+n?1n?1?)
(1.) 當
n
=
1
n = 1
n=1 時，命題等價於一維諧振子結論
(2.) 假設
n
=
t
?
1
n = t - 1
n=t?1 時命題成立，那麼當
n
=
t
n = t
n=t 時，設
ψ
(
r
t
?
)
=
ψ
(
r
t
?
1
→
,
x
t
)
=
f
(
r
t
?
1
→
)
g
(
x
t
)
\psi(\vec{r_t})=\psi(\overrightarrow{r_{t-1}},x_t) = f(\overrightarrow{r_{t-1}})g(x_t)
ψ(rt?
?)=ψ(rt?1?
?,xt?)=f(rt?1?
?)g(xt?)。
記
?
r
t
?
1
→
2
=
∑
i
=
1
t
?
1
?
2
?
x
i
2
H
r
t
?
1
→
^
=
?
?
2
2
m
?
r
t
?
1
→
2
+
1
2
m
ω
2
r
t
1
→
2
H
x
t
^
=
?
?
2
2
m
?
2
?
x
t
2
+
1
2
m
ω
2
x
t
2
\nabla_{\overrightarrow{r_{t-1}}}^2=\sum_{i=1}^{t-1}\frac{\partial^2}{\partial x_i^2}\\[1em] \hat{H_{\overrightarrow{r_{t-1}}}}=-\frac{\hbar^2}{2m}\nabla_{\overrightarrow{r_{t-1}}}^2+\frac{1}{2}m\omega^2\overrightarrow{r_{t_1}}^2\\[1em] \hat{H_{x_t}}=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_t^2}+\frac{1}{2}m\omega^2x_t^2
?rt?1?
?2?=i=1∑t?1??xi2??2?Hrt?1?
??^?=?2m?2??rt?1?
?2?+21?mω2rt1??
?2Hxt??^?=?2m?2??xt2??2?+21?mω2xt2?
所以
H
^
=
?
?
2
2
m
?
r
t
→
2
+
1
2
m
ω
2
r
t
→
2
=
(
?
?
2
2
m
?
r
t
?
1
→
2
+
1
2
m
ω
2
r
t
1
→
2
)
+
(
?
?
2
2
m
?
2
?
x
t
2
+
1
2
m
ω
2
x
t
2
)
=
H
r
t
?
1
→
^
+
H
x
t
^
\hat{H}=-\frac{\hbar^2}{2m}\nabla_{\overrightarrow{r_t}}^2+\frac{1}{2}m\omega^2\overrightarrow{r_t}^2=\left(-\frac{\hbar^2}{2m}\nabla_{\overrightarrow{r_{t-1}}}^2+\frac{1}{2}m\omega^2\overrightarrow{r_{t_1}}^2\right)+\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_t^2}+\frac{1} {2}m\omega^2x_t^2\right)=\hat{H_{\overrightarrow{r_{t-1}}}}+\hat{H_{x_t}}
H^=?2m?2??rt?
?2?+21?mω2rt?
?2=(?2m?2??rt?1?
?2?+21?mω2rt1??
?2)+(?2m?2??xt2??2?+21?mω2xt2?)=Hrt?1?
??^?+Hxt??^?
利用分離變量法，設
ψ
(
r
t
→
)
=
f
(
r
t
?
1
→
)
g
(
r
t
)
\psi(\overrightarrow{r_t})=f(\overrightarrow{r_{t-1}})g(r_t)
ψ(rt?
?)=f(rt?1?
?)g(rt?)，類似二維諧振子般，可以得到如下關係式
H
x
t
^
g
(
y
)
=
E
x
t
g
(
x
t
)
?且?
H
r
t
?
1
→
^
f
(
r
t
?
1
→
)
)
=
(
E
?
E
x
t
)
f
(
r
t
?
1
→
)
=
E
r
t
?
1
→
f
(
r
t
?
1
→
)
\hat{H_{x_t}}g(y)=E_{x_t}g(x_t)\ \text{且}\ \hat{H_{\overrightarrow{r_{t-1}}}}f(\overrightarrow{r_{t-1}}))=(E-E_{x_t})f(\overrightarrow{r_{t-1}})=E_{\overrightarrow{r_{t-1}}}f(\overrightarrow{r_{t-1}})
Hxt??^?g(y)=Ext??g(xt?)?且?Hrt?1?
??^?f(rt?1?
?))=(E?Ext??)f(rt?1?
?)=Ert?1?
??f(rt?1?
?)
由 (1.) 知，
H
x
t
^
\hat{H_{x_t}}
Hxt??^? 的本征值是
{
E
1
,
N
}
N
≥
0
=
{
(
N
+
1
2
)
?
ω
}
N
≥
0
\{E_{1,N}\}_{N\geq0}=\{(N+\frac{1}{2})\hbar\omega\}_{N\geq0}
{E1,N?}N≥0?={(N+21?)?ω}N≥0?，對應的本征函數是
{
ψ
N
(
x
t
)
}
N
≥
0
\{\psi_N(x_t)\}_{N\geq0}
{ψN?(xt?)}N≥0?。
由歸納假設知，
H
r
t
?
1
→
^
\hat{H_{\overrightarrow{r_{t-1}}}}
Hrt?1?
??^? 的本征值是
{
E
t
?
1
,
N
}
N
≥
0
=
{
(
N
+
t
?
1
2
)
?
ω
}
n
≥
0
\{E_{t-1,N}\}_{N\geq0}=\{(N+\frac{t-1}{2})\hbar\omega\}_{n\geq0}
{Et?1,N?}N≥0?={(N+2t?1?)?ω}n≥0?，每一個本征值
E
t
?
1
,
N
E_{t-1,N}
Et?1,N? 對應的本征函數族是
{
∏
k
=
1
t
?
1
ψ
N
k
(
x
k
)
∣
∑
k
=
1
t
?
1
N
k
=
N
}
\{\prod\limits_{k=1}^{t-1}\psi_{N_k}(x_k)|\sum\limits_{k=1}^{t-1}N_k=N\}
{k=1∏t?1?ψNk??(xk?)∣k=1∑t?1?Nk?=N}
因此，
H
^
\hat{H}
H^ 的本征值便是
{
E
t
,
N
}
N
≥
0
=
{
E
t
?
1
,
N
+
E
1
,
0
}
N
≥
0
=
{
(
N
+
t
?
1
2
+
0
+
1
2
)
?
ω
}
N
≥
0
=
{
(
N
+
t
2
)
}
N
≥
0
\{E_{t,N}\}_{N\geq0}=\{E_{t-1,N}+E_{1,0}\}_{N\geq0}=\{(N+\frac{t-1}{2}+0+\frac{1}{2})\hbar\omega\}_{N\geq0}=\{(N+\frac{t}{2})\}_{N\geq0}
{Et,N?}N≥0?={Et?1,N?+E1,0?}N≥0?={(N+2t?1?+0+21?)?ω}N≥0?={(N+2t?)}N≥0?，對應的本征值函數族是
{
∏
k
=
1
t
ψ
N
k
(
x
k
)
∣
∑
k
=
1
t
N
k
=
N
}
\{\prod\limits_{k=1}^{t}\psi_{N_k}(x_k)|\sum\limits_{k=1}^{t}N_k=N\}
{k=1∏t?ψNk??(xk?)∣k=1∑t?Nk?=N}
結合(1.) 和 (2.)，根據數學歸納原理可知，原命題對於任意正整數
n
n
n 皆成立。
4 二維「正定勢場」下的定態 Schr?dinger 方程
我先不引入「正定勢場」的概念，而是用另一個特殊勢場作為引入，求解其定態 Schr?dinger 方程。
V
(
x
,
y
)
=
1
2
m
ω
2
(
x
2
+
x
y
+
y
2
)
V(x,y)=\frac{1}{2}m\omega^2(x^2+xy+y^2)
V(x,y)=21?mω2(x2+xy+y2)
考慮座標變換(讀者可以先自行思考)
(
u
v
)
=
(
1
2
1
2
1
2
?
1
2
)
(
x
y
)
\left( \begin{matrix} u\\ v \end{matrix} \right)= \left( \begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{matrix} \right) \left( \begin{matrix} x\\ y \end{matrix} \right)
(uv?)=(2
?1?2
?1??2
?1??2
?1??)(xy?)
用
u
u
u、
v
v
v 表示
?
2
\nabla^2
?2﹕
?
?
x
=
?
?
u
?
u
?
x
+
?
?
v
?
v
?
x
=
1
2
(
?
?
u
+
?
?
v
)
?
2
?
x
2
=
1
2
(
?
?
u
+
?
?
v
)
?
1
2
(
?
?
u
+
?
?
v
)
=
1
2
(
?
2
?
u
2
+
?
2
?
v
2
+
?
2
?
u
?
v
+
?
2
?
v
?
u
)
\frac{\partial}{\partial x}=\frac{\partial}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial}{\partial v}\frac{\partial v}{\partial x}=\frac{1}{\sqrt{2}}(\frac{\partial}{\partial u}+\frac{\partial}{\partial v})\\[1em] \frac{\partial^2}{\partial x^2}=\frac{1}{\sqrt{2}}(\frac{\partial}{\partial u}+\frac{\partial}{\partial v})\cdot\frac{1}{\sqrt{2}}(\frac{\partial}{\partial u}+\frac{\partial}{\partial v})=\frac{1}{2}(\frac{\partial^2}{\partial u^2}+\frac{\partial^2}{\partial v^2}+\frac{\partial^2}{\partial u\partial v}+\frac{\partial^2}{\partial v\partial u})
?x??=?u???x?u?+?v???x?v?=2
?1?(?u??+?v??)?x2?2?=2
?1?(?u??+?v??)?2
?1?(?u??+?v??)=21?(?u2?2?+?v2?2?+?u?v?2?+?v?u?2?)
同理可得
?
2
?
y
2
=
1
2
(
?
2
?
u
2
+
?
2
?
v
2
?
?
2
?
u
?
v
?
?
2
?
v
?
u
)
?
2
=
?
2
?
x
2
+
?
2
?
y
2
=
?
2
?
u
2
+
?
2
?
v
2
\frac{\partial^2}{\partial y^2}=\frac{1}{2}(\frac{\partial^2}{\partial u^2}+\frac{\partial^2}{\partial v^2}-\frac{\partial^2}{\partial u\partial v}-\frac{\partial^2}{\partial v\partial u})\\[1em] \nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}=\frac{\partial^2}{\partial u^2}+\frac{\partial^2}{\partial v^2}
?y2?2?=21?(?u2?2?+?v2?2???u?v?2???v?u?2?)?2=?x2?2?+?y2?2?=?u2?2?+?v2?2?
簡記
?
x
y
2
=
?
u
v
2
\nabla^2_{xy}=\nabla^2_{uv}
?xy2?=?uv2?
用
u
u
u、
v
v
v 改寫勢場﹕
V
=
1
2
m
ω
2
(
3
2
u
2
+
1
2
v
2
)
=
1
2
m
(
3
2
ω
)
2
u
2
+
1
2
m
(
1
2
ω
)
2
v
2
V=\frac{1}{2}m\omega^2(\frac{3}{2}u^2+\frac{1}{2}v^2)=\frac{1}{2}m(\sqrt{\frac{3}{2}}\omega)^2u^2+\frac{1}{2}m(\sqrt{\frac{1}{2}}\omega)^2v^2
V=21?mω2(23?u2+21?v2)=21?m(23?
?ω)2u2+21?m(21?
?ω)2v2
記
H
x
,
ω
^
=
?
?
2
2
m
?
2
?
x
2
+
1
2
m
ω
2
x
2
ψ
ω
,
n
(
x
)
=
m
ω
?
π
1
2
?
2
n
?
n
!
exp
?
(
?
m
ω
?
x
2
2
)
H
n
(
m
ω
?
x
)
\hat{H_{x,\omega}}=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac{1}{2}m\omega^2x^2\\[1em] \psi_{\omega,n}(x)=\sqrt{\frac{\sqrt{\frac{m\omega}{\hbar}}}{\pi^{\frac{1}{2}}\cdot 2^n\cdot n!}}\exp(-\frac{\frac{m\omega}{\hbar}x^2}{2})H_n(\sqrt{\frac{m\omega}{\hbar}} x)
Hx,ω?^?=?2m?2??x2?2?+21?mω2x2ψω,n?(x)=π21??2n?n!?mω?
??
?exp(?2?mω?x2?)Hn?(?mω?
?x)
所以在這個勢場下的哈密頓算符是
H
^
=
H
u
,
3
2
ω
^
+
H
v
,
1
2
ω
^
\hat{H}=\hat{H_{u,\sqrt{\frac{3}{2}}\omega}}+\hat{H_{v,\sqrt{\frac{1}{2}}\omega}}
H^=Hu,23?
?ω?^?+Hv,21?
?ω?^?
如二維諧振子般利用分離變量法，可得本征波函數
ψ
(
x
,
y
)
=
ψ
3
2
ω
,
m
(
u
)
ψ
1
2
ω
,
n
(
v
)
=
ψ
3
2
ω
,
m
(
x
+
y
2
)
ψ
1
2
ω
,
n
(
x
?
y
2
)
\psi(x,y)=\psi_{\sqrt{\frac{3}{2}}\omega,m}(u)\psi_{\sqrt{\frac{1}{2}}\omega,n}(v)=\psi_{\sqrt{\frac{3}{2}}\omega,m}(\frac{x+y}{\sqrt{2}})\psi_{\sqrt{\frac{1}{2}}\omega,n}(\frac{x-y}{\sqrt{2}})
ψ(x,y)=ψ23?
?ω,m?(u)ψ21?
?ω,n?(v)=ψ23?
?ω,m?(2
?x+y?)ψ21?
?ω,n?(2
?x?y?)
對應的本征能級是
E
m
,
n
=
[
(
m
+
1
2
)
3
2
+
(
n
+
1
2
)
1
2
]
?
ω
E_{m,n}=\left[(m+\frac{1}{2})\sqrt{\frac{3}{2}}+(n+\frac{1}{2})\sqrt{\frac{1}{2}}\right]\hbar\omega
Em,n?=[(m+21?)23?
?+(n+21?)21?
?]?ω
顯然地，這個能級是非簡並的。
忽略能級簡並度的因素，我們去擴展這個問題﹕
如果勢函數
V
(
x
,
y
)
V(x,y)
V(x,y) 可以寫成
V
(
x
,
y
)
=
1
2
m
ω
2
(
x
y
)
(
a
b
b
c
)
(
x
y
)
V(x,y)=\frac{1}{2}m\omega^2\left(\begin{matrix}x&y\end{matrix}\right)\left(\begin{matrix}a&b\\b&c\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)
V(x,y)=21?mω2(x?y?)(ab?bc?)(xy?)
且當
(
x
,
y
)
≠
(
0
,
0
)
(x,y) \neq (0,0)
(x,y)?=(0,0) 時
V
(
x
,
y
)
>
0
V(x,y) > 0
V(x,y)>0
那麼定態波函數
ψ
(
x
,
y
)
\psi(x,y)
ψ(x,y) 是否可以寫成
ψ
λ
1
ω
,
m
(
k
1
x
+
b
1
y
)
?
ψ
λ
2
ω
,
n
(
k
2
x
+
b
2
y
)
\psi_{\lambda_1\omega,m}(k_1x+b_1y)\cdot\psi_{\lambda_2\omega,n}(k_2x+b_2y)
ψλ1?ω,m?(k1?x+b1?y)?ψλ2?ω,n?(k2?x+b2?y)的形式呢？(其中，
k
1
,
b
1
,
k
2
,
b
2
,
λ
1
,
λ
2
k_1,b_1,k_2,b_2,\lambda_1,\lambda_2
k1?,b1?,k2?,b2?,λ1?,λ2? 都是常數)
注意到矩陣
(
a
b
b
c
)
\left(\begin{matrix}a&b\\b&c\end{matrix}\right)
(ab?bc?) 是實對稱正定矩陣 (若讀者對正定矩陣不瞭解，可以參考 MIT 的 Introduction to Linear Algebra 書籍)，所以可以進行正交對角化。即存在一個正交矩陣
Q
=
(
q
11
q
12
q
21
g
22
)
Q=\left(\begin{matrix}q_{11}&q_{12}\\q_{21}&g_{22}\end{matrix}\right)
Q=(q11?q21??q12?g22??) 和一個對角矩陣
Λ
=
d
i
a
g
(
λ
1
2
,
λ
2
2
)
\Lambda=diag(\lambda_1^2,\lambda_2^2)
Λ=diag(λ12?,λ22?)，滿足
A
=
Q
T
Λ
Q
V
(
x
,
y
)
=
1
2
m
ω
2
(
x
y
)
Q
T
Λ
Q
(
x
y
)
=
1
2
m
ω
2
q
?
T
Λ
q
?
=
1
2
m
(
λ
1
ω
)
2
u
2
+
1
2
m
(
λ
2
ω
)
2
v
2
A=Q^T\Lambda Q\\[1em] V(x,y)=\frac{1}{2}m\omega^2\left(\begin{matrix}x&y\end{matrix}\right)Q^T\Lambda Q\left(\begin{matrix}x\\y\end{matrix}\right)=\frac{1}{2}m\omega^2\vec{q}^T\Lambda \vec{q}=\frac{1}{2}m(\lambda_1\omega)^2u^2+\frac{1}{2}m(\lambda_2\omega)^2v^2
A=QTΛQV(x,y)=21?mω2(x?y?)QTΛQ(xy?)=21?mω2q
?TΛq
?=21?m(λ1?ω)2u2+21?m(λ2?ω)2v2
此時，
u
=
q
11
x
+
q
12
y
u=q_{11}x+q_{12}y
u=q11?x+q12?y、
v
=
q
21
x
+
q
22
y
v=q_{21}x+q_{22}y
v=q21?x+q22?y
下面證明﹕
?
2
?
x
2
+
?
2
?
y
2
=
?
2
?
u
2
+
?
2
?
v
2
\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}= \frac{\partial^2}{\partial u^2}+ \frac{\partial^2}{\partial v^2}
?x2?2?+?y2?2?=?u2?2?+?v2?2?
證﹕
?
?
x
=
?
?
u
?
u
?
x
+
?
?
v
?
v
?
x
=
q
11
?
?
u
+
q
21
?
?
v
?
2
?
x
2
=
?
?
x
?
?
?
x
=
(
q
11
?
?
u
+
q
21
?
?
v
)
(
q
11
?
?
u
+
q
21
?
?
v
)
=
q
11
2
?
2
?
u
2
+
q
21
2
?
2
?
u
2
+
q
11
q
21
(
?
2
?
u
?
v
+
?
2
?
v
?
u
)
\frac{\partial}{\partial x} = \frac{\partial}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial}{\partial v}\frac{\partial v}{\partial x} = q_{11}\frac{\partial}{\partial u}+q_{21}\frac{\partial}{\partial v}\\[1em] \frac{\partial^2}{\partial x^2}=\frac{\partial}{\partial x} \cdot \frac{\partial}{\partial x}=(q_{11}\frac{\partial}{\partial u}+q_{21}\frac{\partial}{\partial v})(q_{11}\frac{\partial}{\partial u}+q_{21}\frac{\partial}{\partial v})=q_{11}^2\frac{\partial^2}{\partial u^2}+q_{21}^2\frac{\partial^2}{\partial u^2}+q_{11}q_{21}(\frac{\partial^2}{\partial u\partial v}+\frac{\partial^2}{\partial v\partial u})
?x??=?u???x?u?+?v???x?v?=q11??u??+q21??v???x2?2?=?x????x??=(q11??u??+q21??v??)(q11??u??+q21??v??)=q112??u2?2?+q212??u2?2?+q11?q21?(?u?v?2?+?v?u?2?)
同理可得
?
2
?
y
2
=
q
12
2
?
2
?
u
2
+
q
22
2
?
2
?
u
2
+
q
12
q
22
(
?
2
?
u
?
v
+
?
2
?
v
?
u
)
\frac{\partial^2}{\partial y^2}=q_{12}^2\frac{\partial^2}{\partial u^2}+q_{22}^2\frac{\partial^2}{\partial u^2}+q_{12}q_{22}(\frac{\partial^2}{\partial u\partial v}+\frac{\partial^2}{\partial v\partial u})
?y2?2?=q122??u2?2?+q222??u2?2?+q12?q22?(?u?v?2?+?v?u?2?)
利用正交矩陣的定義
q
11
2
+
q
12
2
=
q
21
2
+
q
22
2
=
1
q
11
q
21
+
q
12
q
22
=
0
q_{11}^2+q_{12}^2=q_{21}^2+q_{22}^2=1\\[1em] q_{11}q_{21}+q_{12}q_{22}=0
q112?+q122?=q212?+q222?=1q11?q21?+q12?q22?=0
可證得
?
2
?
x
2
+
?
2
?
y
2
=
?
2
?
u
2
+
?
2
?
v
2
\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}= \frac{\partial^2}{\partial u^2}+ \frac{\partial^2}{\partial v^2}
?x2?2?+?y2?2?=?u2?2?+?v2?2?
利用上方結論，可以得出算符量
T
^
=
T
u
^
+
T
v
^
V
^
=
V
λ
1
ω
^
+
V
λ
2
ω
^
H
^
=
H
λ
1
ω
,
u
^
+
H
λ
2
ω
,
v
^
\hat{T}=\hat{T_u}+\hat{T_v}\\[1em] \hat{V}=\hat{V_{\lambda_1\omega}}+\hat{V_{\lambda_2\omega}}\\[1em] \hat{H}=\hat{H_{\lambda_1\omega,u}}+\hat{H_{\lambda_2\omega,v}}
T^=Tu?^?+Tv?^?V^=Vλ1?ω?^?+Vλ2?ω?^?H^=Hλ1?ω,u?^?+Hλ2?ω,v?^?
因此，本征波函數為
ψ
(
x
,
y
)
=
ψ
λ
1
ω
,
m
(
u
)
ψ
λ
2
ω
,
n
(
v
)
=
ψ
λ
1
ω
,
m
(
q
11
x
+
q
12
y
)
ψ
λ
2
ω
,
n
(
q
21
x
+
q
22
y
)
\psi(x,y)=\psi_{\lambda_1\omega,m}(u)\psi_{\lambda_2\omega,n}(v)=\psi_{\lambda_1\omega,m}(q_{11}x+q_{12}y)\psi_{\lambda_2\omega,n}(q_{21}x+q_{22}y)
ψ(x,y)=ψλ1?ω,m?(u)ψλ2?ω,n?(v)=ψλ1?ω,m?(q11?x+q12?y)ψλ2?ω,n?(q21?x+q22?y)
對應的本征能級為
E
m
,
n
=
[
(
m
+
1
2
)
λ
1
+
(
n
+
1
2
)
λ
2
]
?
ω
=
(
m
+
1
2
n
+
1
2
)
Λ
(
1
1
)
?
ω
E_{m,n}=\left[(m+\frac{1}{2})\lambda_1+(n+\frac{1}{2})\lambda_2\right]\hbar\omega=\left(\begin{matrix}m+\frac{1}{2}&n+\frac{1}{2}\end{matrix}\right)\sqrt{\Lambda}\left(\begin{matrix}1\\1\end{matrix}\right)\hbar\omega
Em,n?=[(m+21?)λ1?+(n+21?)λ2?]?ω=(m+21??n+21??)Λ
?(11?)?ω
5 正定勢場的定義
定義﹕ 對於任意向量
r
?
=
(
x
1
,
x
2
,
?
?
,
x
n
)
T
≠
0
?
T
\vec{r}=(x_1,x_2,\cdots,x_n)^T\neq\vec{0}^T
r
=(x1?,x2?,?,xn?)T?=0
T，勢場
V
(
r
?
)
=
1
2
m
ω
2
r
?
T
A
r
?
>
0
V(\vec{r})=\frac{1}{2}m\omega^2\vec{r}^TA\vec{r}>0
V(r
)=21?mω2r
TAr
>0
其中
A
A
A 是一個實數矩陣，那麼勢場
V
V
V 就是正定勢場。
一般而言，
A
A
A 會被寫成實對稱矩陣 (若非，則
A
A
A 可用
A
+
A
T
2
\frac{A+A^T}{2}
2A+AT? 代替)，下一節的矩陣
A
A
A 都會默認是實對稱矩陣。根據線性代數知識可知，
A
A
A 就是正定矩陣。
6
n
n
n 維「正定勢場」下的定態 Schr?dinger 方程
試着把二維情況推廣到
n
n
n 維空間中，結論又是否會成立呢？
考慮
r
?
=
(
x
1
,
x
2
,
?
?
,
x
n
)
T
≠
0
?
T
V
(
r
?
)
=
1
2
m
ω
2
r
?
T
A
n
r
?
\vec{r}=(x_1,x_2,\cdots,x_n)^T\neq\vec{0}^T\\[1em] V(\vec{r})=\frac{1}{2}m\omega^2\vec{r}^TA_n\vec{r}
r
=(x1?,x2?,?,xn?)T?=0
TV(r
)=21?mω2r
TAn?r
V
V
V 是一個正定勢場，尋找其本征波函數
ψ
(
r
?
)
\psi(\vec{r})
ψ(r
)。
首先，對
A
n
A_n
An? 進行正交對角化，同時改寫勢能表達式﹕
A
n
=
Q
T
Λ
Q
Λ
=
d
i
a
g
(
λ
1
2
,
λ
2
2
,
?
?
,
λ
n
2
)
Q
=
(
q
i
j
)
n
×
n
=
(
q
1
?
T
q
2
?
T
?
q
n
?
T
)
V
=
1
2
m
ω
2
r
?
T
Q
T
Λ
Q
r
?
=
1
2
m
ω
2
u
?
T
Λ
u
?
=
∑
i
=
1
n
1
2
m
(
λ
i
ω
)
2
u
i
2
(
u
1
,
u
2
,
?
?
,
u
n
)
T
=
u
?
=
Q
r
?
A_n = Q^T\Lambda Q\\ \Lambda=diag(\lambda_1^2,\lambda_2^2,\cdots,\lambda_n^2)\\[1em] Q=(q_{ij})_{n\times n}=\left(\begin{matrix}\vec{q_1}^T\\\vec{q_2}^T\\\vdots\\\vec{q_n}^T\end{matrix}\right)\\[1em] V = \frac{1}{2}m\omega^2\vec{r}^TQ^T\Lambda Q\vec{r}=\frac{1}{2}m\omega^2\vec{u}^T\Lambda\vec{u}=\sum_{i=1}^n\frac{1}{2}m(\lambda_i \omega)^2u_i^2\\(u_1,u_2,\cdots,u_n)^T=\vec{u}=Q\vec{r}
An?=QTΛQΛ=diag(λ12?,λ22?,?,λn2?)Q=(qij?)n×n?=??????q1?
?Tq2?
?T?qn?
?T???????V=21?mω2r
TQTΛQr
=21?mω2u
TΛu
=i=1∑n?21?m(λi?ω)2ui2?(u1?,u2?,?,un?)T=u
=Qr
然後，我們證明一個結論﹕
若存在一個正交矩陣
Q
=
{
q
i
j
}
n
×
n
Q=\{q_{ij}\}_{n\times n}
Q={qij?}n×n?，滿足
u
?
=
Q
r
?
=
(
q
1
?
T
q
2
?
T
?
q
n
?
T
)
r
?
\vec{u}=Q\vec{r}=\left(\begin{matrix}\vec{q_1}^T\\\vec{q_2}^T\\\vdots\\\vec{q_n}^T\end{matrix}\right)\vec{r}
u
=Qr
=??????q1?
?Tq2?
?T?qn?
?T???????r
，那麼
?
r
?
2
=
?
u
?
2
?
(
∑
i
=
1
n
?
2
?
x
i
2
=
∑
i
=
1
n
?
2
?
u
i
2
)
\nabla^2_{\vec{r}}=\nabla^2_{\vec{u}}\ (\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}=\sum_{i=1}^n\frac{\partial^2}{\partial u_i^2})
?r
2?=?u
2??(i=1∑n??xi2??2?=i=1∑n??ui2??2?)
證﹕
?
?
x
i
=
∑
k
=
1
n
?
?
u
k
?
u
k
?
x
i
=
∑
k
=
1
n
q
k
i
?
?
u
k
?
2
?
x
i
2
=
(
∑
k
=
1
n
q
k
i
?
?
u
k
)
(
∑
m
=
1
n
q
m
i
?
?
u
m
)
=
∑
k
=
1
n
∑
m
=
1
n
q
k
i
q
m
i
?
2
?
u
k
?
u
m
?
r
?
2
=
∑
i
=
1
n
?
2
?
x
i
2
=
∑
i
=
1
n
∑
k
=
1
n
∑
m
=
1
n
q
k
i
q
m
i
?
2
?
u
k
?
u
m
=
∑
k
=
1
n
∑
m
=
1
n
[
?
2
?
u
k
?
u
m
(
∑
i
=
1
n
q
k
i
q
m
i
)
]
=
∑
k
=
1
n
∑
m
=
1
n
[
?
2
?
u
k
?
u
m
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\frac{\partial}{\partial x_i}=\sum_{k=1}^n\frac{\partial}{\partial u_k}\frac{\partial u_k}{\partial x_i}=\sum_{k=1}^nq_{ki}\frac{\partial}{\partial u_k}\\[1em] \frac{\partial^2}{\partial x_i^2} = (\sum_{k=1}^nq_{ki}\frac{\partial}{\partial u_k})(\sum_{m=1}^nq_{mi}\frac{\partial}{\partial u_m})=\sum_{k=1}^n\sum_{m=1}^nq_{ki}q_{mi}\frac{\partial^2}{\partial u_k\partial u_m}\\[1em] \nabla^2_{\vec{r}}=\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}=\sum_{i=1}^n\sum_{k=1}^n\sum_{m=1}^nq_{ki}q_{mi}\frac{\partial^2}{\partial u_k\partial u_m}=\sum_{k=1}^n\sum_{m=1}^n\left[\frac{\partial^2}{\partial u_k\partial u_m}\left(\sum_{i=1}^nq_{ki}q_{mi}\right)\right]=\sum_{k=1}^n\sum_{m=1}^n\left[\frac{\partial^2}{\partial u_k\partial u_m}\left(\vec{q_k}^T\vec{q_m}\right)\right]=\sum_{k=1}^n\sum_{m=1}^n\left[\frac{\partial^2}{\partial u_k\partial u_m}\delta_{km}\right]=\sum_{k=1}^n\frac{\partial^2}{\partial u_k^2}=\nabla^2_{\vec{u}}\\[1em]
?xi???=k=1∑n??uk????xi??uk??=k=1∑n?qki??uk????xi2??2?=(k=1∑n?qki??uk???)(m=1∑n?qmi??um???)=k=1∑n?m=1∑n?qki?qmi??uk??um??2??r
2?=i=1∑n??xi2??2?=i=1∑n?k=1∑n?m=1∑n?qki?qmi??uk??um??2?=k=1∑n?m=1∑n?[?uk??um??2?(i=1∑n?qki?qmi?)]=k=1∑n?m=1∑n?[?uk??um??2?(qk?
?Tqm?
?)]=k=1∑n?m=1∑n?[?uk??um??2?δkm?]=k=1∑n??uk2??2?=?u
2?
因此，對應的算符量是
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\hat{T_i}=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial u_i^2}\\[1em] \hat{V_{\lambda_i\omega,i}}=\frac{1}{2}m(\lambda_i \omega)^2u_i^2\\[1em] \hat{H_{\lambda_i\omega,i}}=\hat{T_i}+\hat{V_{\lambda\omega,i}}\\[1em] \hat{H}=\sum_{i=1}^n\hat{H_{\lambda_i\omega,i}}
Ti?^?=?2m?2??ui2??2?Vλi?ω,i?^?=21?m(λi?ω)2ui2?Hλi?ω,i?^?=Ti?^?+Vλω,i?^?H^=i=1∑n?Hλi?ω,i?^?
如第 3 節的「
n
n
n 維諧振子」般利用數學歸納法可得，本征波函數
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\psi(\vec{r})=\prod_{i=1}^{n}\psi_{\lambda_i\omega,m_i}(\vec{q_i}^T \vec{r})
ψ(r
)=i=1∏n?ψλi?ω,mi??(qi?
?Tr
)
對應的本征能級
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\vec{m}=(m_i)^T_{n\times 1}
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=(mi?)n×1T? 是
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E_{\vec{m}}=\hbar\omega\sum_{i=1}^n(m_i+\frac{1}{2})\lambda_i=(\vec{m}^T+(\frac{1}{2})_{1\times n})\sqrt{\Lambda}(1)_{n\times 1}\hbar\omega
Em
?=?ωi=1∑n?(mi?+21?)λi?=(m
T+(21?)1×n?)Λ
?(1)n×1??ω
7 資料來源
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