edit
If the potential V does not depend on time, then the Schrödinger equation has solutions that are products of a time-independent function
ψ
(
r
)
{\displaystyle \psi (\mathbf {r} )}
e
−
(
i
/
ℏ
)
E
t
{\displaystyle e^{-(i/\hbar )\,E\,t}}
ψ
(
t
,
r
)
=
ψ
(
r
)
e
−
(
i
/
ℏ
)
E
t
.
{\displaystyle \psi (t,\mathbf {r} )=\psi (\mathbf {r} )\,e^{-(i/\hbar )\,E\,t}.}
Because the probability density
|
ψ
(
t
,
r
)
|
2
{\displaystyle |\psi (t,\mathbf {r} )|^{2}}
stationary .
Plug
ψ
(
r
)
e
−
(
i
/
ℏ
)
E
t
{\displaystyle \psi (\mathbf {r} )\,e^{-(i/\hbar )\,E\,t}}
i
ℏ
∂
ψ
∂
t
=
−
ℏ
2
2
m
∂
∂
r
⋅
∂
∂
r
ψ
+
V
ψ
{\displaystyle i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial }{\partial \mathbf {r} }}\cdot {\frac {\partial }{\partial \mathbf {r} }}\psi +V\psi }
to find that
ψ
(
r
)
{\displaystyle \psi (\mathbf {r} )}
E
ψ
(
r
)
=
−
ℏ
2
2
m
(
∂
2
∂
x
2
+
∂
2
∂
y
2
+
∂
2
∂
z
2
)
ψ
(
r
)
+
V
(
r
)
ψ
(
r
)
.
{\displaystyle E\psi (\mathbf {r} )=-{\hbar ^{2} \over 2m}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}\right)\psi (\mathbf {r} )+V(\mathbf {r} )\,\psi (\mathbf {r} ).}
