機率密度 ρ ( t , r ) = | ψ ( t , r ) | 2 {\displaystyle \rho (t,\mathbf {r} )=|\psi (t,\mathbf {r} )|^{2}} (在固定位置 r {\displaystyle \mathbf {r} } ) 對時間的變化率由下式給出
∂ ρ ∂ t = ψ ∗ ∂ ψ ∂ t + ψ ∂ ψ ∗ ∂ t . {\displaystyle {\frac {\partial \rho }{\partial t}}=\psi ^{*}{\frac {\partial \psi }{\partial t}}+\psi {\frac {\partial \psi ^{*}}{\partial t}}.}
藉助薛定諤方程及其複共軛,
i ℏ ∂ ψ ∂ t = 1 2 m ( ℏ i ∂ ∂ r − A ) ⋅ ( ℏ i ∂ ∂ r − A ) ψ + V ψ , {\displaystyle i\hbar {\frac {\partial \psi }{\partial t}}={\frac {1}{2m}}\left({\frac {\hbar }{i}}{\frac {\partial }{\partial \mathbf {r} }}-\mathbf {A} \right)\cdot \left({\frac {\hbar }{i}}{\frac {\partial }{\partial \mathbf {r} }}-\mathbf {A} \right)\psi +V\psi ,}
ℏ i ∂ ψ ∗ ∂ t = 1 2 m ( i ℏ ∂ ∂ r − A ) ⋅ ( i ℏ ∂ ∂ r − A ) ψ ∗ + V ψ ∗ , {\displaystyle {\hbar \over i}{\partial \psi ^{*} \over \partial t}={\frac {1}{2m}}\left(i\hbar {\frac {\partial }{\partial \mathbf {r} }}-\mathbf {A} \right)\cdot \left(i\hbar {\frac {\partial }{\partial \mathbf {r} }}-\mathbf {A} \right)\psi ^{*}+V\psi ^{*},}
得到
∂ ρ ∂ t = − i ℏ ψ ∗ [ 1 2 m ( ℏ i ∂ ∂ r − A ) ⋅ ( ℏ i ∂ ∂ r − A ) ψ + V ψ ] {\displaystyle {\frac {\partial \rho }{\partial t}}=-{\frac {i}{\hbar }}\psi ^{*}\left[{\frac {1}{2m}}\left({\frac {\hbar }{i}}{\frac {\partial }{\partial \mathbf {r} }}-\mathbf {A} \right)\cdot \left({\frac {\hbar }{i}}{\frac {\partial }{\partial \mathbf {r} }}-\mathbf {A} \right)\psi +V\psi \right]}
+ i ℏ ψ [ 1 2 m ( i ℏ ∂ ∂ r − A ) ⋅ ( i ℏ ∂ ∂ r − A ) ψ ∗ + V ψ ∗ ] . {\displaystyle +{\frac {i}{\hbar }}\psi \left[{\frac {1}{2m}}\left(i\hbar {\frac {\partial }{\partial \mathbf {r} }}-\mathbf {A} \right)\cdot \left(i\hbar {\frac {\partial }{\partial \mathbf {r} }}-\mathbf {A} \right)\psi ^{*}+V\psi ^{*}\right].}
包含 V {\displaystyle V} 的項相互抵消,因此剩下
∂ ρ ∂ t = − i 2 m ℏ [ ψ ∗ ( i ℏ ∂ ∂ r + A ) ⋅ ( i ℏ ∂ ∂ r + A ) ψ − ψ ( i ℏ ∂ ∂ r − A ) ⋅ ( i ℏ ∂ ∂ r − A ) ψ ∗ ] {\displaystyle {\frac {\partial \rho }{\partial t}}=-{\frac {i}{2m\hbar }}\left[\psi ^{*}\left(i\hbar {\frac {\partial }{\partial \mathbf {r} }}+\mathbf {A} \right)\cdot \left(i\hbar {\frac {\partial }{\partial \mathbf {r} }}+\mathbf {A} \right)\psi -\psi \left(i\hbar {\frac {\partial }{\partial \mathbf {r} }}-\mathbf {A} \right)\cdot \left(i\hbar {\frac {\partial }{\partial \mathbf {r} }}-\mathbf {A} \right)\psi ^{*}\right]}
= ⋯ = − ℏ 2 m i ( ∂ 2 ψ ∂ r 2 ψ ∗ − ψ ∂ 2 ψ ∗ ∂ r 2 ) + 1 m ( ψ ψ ∗ ∂ ∂ r ⋅ A + A ∂ ψ ∂ r ψ ∗ + A ψ ∂ ψ ∗ ∂ r ) . {\displaystyle =\dots =-{\frac {\hbar }{2mi}}\left({\frac {\partial ^{2}\psi }{\partial \mathbf {r} ^{2}}}\psi ^{*}-\psi {\frac {\partial ^{2}\psi ^{*}}{\partial \mathbf {r} ^{2}}}\right)+{\frac {1}{m}}\left(\psi \psi ^{*}{\frac {\partial }{\partial \mathbf {r} }}\cdot \mathbf {A} +\mathbf {A} {\frac {\partial \psi }{\partial \mathbf {r} }}\psi ^{*}+\mathbf {A} \psi {\frac {\partial \psi ^{*}}{\partial \mathbf {r} }}\right).}
接下來,我們計算 j = ℏ 2 m i ( ψ ∗ ∂ ψ ∂ r − ∂ ψ ∗ ∂ r ψ ) − 1 m A ψ ∗ ψ {\displaystyle \mathbf {j} ={\frac {\hbar }{2mi}}\left(\psi ^{*}{\frac {\partial \psi }{\partial \mathbf {r} }}-{\frac {\partial \psi ^{*}}{\partial \mathbf {r} }}\psi \right)-{\frac {1}{m}}\mathbf {A} \psi ^{*}\psi } 的散度。
∂ ∂ r ⋅ j = ℏ 2 m i ( ∂ 2 ψ ∂ r 2 ψ ∗ − ψ ∂ 2 ψ ∗ ∂ r 2 ) − 1 m ( ψ ψ ∗ ∂ ∂ r ⋅ A + A ∂ ψ ∂ r ψ ∗ + A ψ ∂ ψ ∗ ∂ r ) . {\displaystyle {\frac {\partial }{\partial \mathbf {r} }}\cdot \mathbf {j} ={\frac {\hbar }{2mi}}\left({\frac {\partial ^{2}\psi }{\partial \mathbf {r} ^{2}}}\psi ^{*}-\psi {\frac {\partial ^{2}\psi ^{*}}{\partial \mathbf {r} ^{2}}}\right)-{\frac {1}{m}}\left(\psi \psi ^{*}{\frac {\partial }{\partial \mathbf {r} }}\cdot \mathbf {A} +\mathbf {A} {\frac {\partial \psi }{\partial \mathbf {r} }}\psi ^{*}+\mathbf {A} \psi {\frac {\partial \psi ^{*}}{\partial \mathbf {r} }}\right).}
結果
∂ ρ ∂ t = − ∂ ∂ r ⋅ j . {\displaystyle {\frac {\partial \rho }{\partial t}}=-{\frac {\partial }{\partial \mathbf {r} }}\cdot \mathbf {j} .}
在空間區域 R {\displaystyle R} 上積分,邊界 ∂ R : {\displaystyle \partial R:} 不變。
∂ ∂ t ∫ R ρ d 3 r = − ∫ R ∂ ∂ r ⋅ j d 3 r . {\displaystyle {\partial \over \partial t}\int _{R}\rho \,d^{3}r=-\int _{R}{\partial \over \partial \mathbf {r} }\cdot \mathbf {j} \,d^{3}r.}
根據 高斯定律 , j {\displaystyle \mathbf {j} } 穿過 ∂ R {\displaystyle \partial R} 的外向通量等於 j {\displaystyle \mathbf {j} } 在 R : {\displaystyle R:} 上 散度 的積分:
∮ ∂ R j ⋅ d Σ = ∫ R ∂ ∂ r ⋅ j d 3 r . {\displaystyle \oint _{\partial R}\mathbf {j} \cdot d\Sigma =\int _{R}{\partial \over \partial \mathbf {r} }\cdot \mathbf {j} \,d^{3}r.}
因此,我們有
∂ ∂ t ∫ R ρ d 3 r = − ∮ ∂ R j ⋅ d Σ . {\displaystyle {\partial \over \partial t}\int _{R}\rho \,d^{3}r=-\oint _{\partial R}\mathbf {j} \cdot d\Sigma .}
如果 ρ {\displaystyle \rho } 是某種物質的連續密度(每單位體積的物質量),並且 j {\displaystyle \mathbf {j} } 是它的通量(每單位面積每單位時間的物質量),那麼在等式的左側,我們得到了物質在 R {\displaystyle R} 內增加的速率,在等式的右側,我們得到了物質透過 R {\displaystyle R} 表面進入的速率。因此,如果某些物質從 A 處移動到 B 處,它將穿過包含 A 或 B 的任何區域的邊界。這就是為什麼這個公式被稱為 連續性方程 。
然而,在量子世界中,不存在連續分佈和/或連續移動的物質。 ρ {\displaystyle \rho } 和 j , {\displaystyle \mathbf {j} ,} 分別僅僅在形式意義上是密度(每單位體積的某事物)和通量(每單位面積每單位時間的某事物)。如果 ψ {\displaystyle \psi } 是與一個粒子相關的波函式,那麼積分 ∫ R ρ d 3 r = ∫ R | ψ | 2 d 3 r {\displaystyle \int _{R}\rho \,d^{3}r=\int _{R}|\psi |^{2}\,d^{3}r} 給出了在 R {\displaystyle R} 內找到粒子的機率, _如果進行了適當的測量_,而這個公式告訴我們:如果找到粒子在 R {\displaystyle R} 內的機率(作為進行測量的時間的函式)增加,那麼在 R {\displaystyle R} 外找到粒子的機率(作為相同時間的函式)會以相同的值減少。(如果 ψ {\displaystyle \psi } 與具有 n {\displaystyle n} 個自由度的系統相關,並且 R {\displaystyle R} 是系統配置空間中的一個區域,那麼情況也是如此。)這有時被表達為“機率是(區域性)守恆的”。當你聽到這句話時,記住某件事在給定時間和地點發生的機率,並非位於那個地點,也並非存在於那個時間。