Probability Flow
정의 1 2
Wave Function $\psi (x, t)$ of Probability Current is defined as follows.
$$ j(x,t) := \frac{\hbar}{2m\i}\left( \psi^{\ast}\dfrac{\partial \psi}{\partial x} - \psi\frac{\partial \psi^{\ast}}{\partial x}\right) \tag{1} $$
Formula
The rate of change of the probability current is equal to the time derivative of the Probability Density. That is, the following equation holds.
$$ \dfrac{\partial \left| \psi(x, t) \right|^{2}}{\partial t} = - \dfrac{\partial j(x,t)}{\partial x} \tag{2} $$
Explanation
In fact, as you will see from the derivation process, it is more accurate to say that the equation $(2)$ holds because we defined it as $(1)$. If we denote the Probability Density of the Wave Function as $P$, then $(2)$ is as follows.
$$ \dfrac{\partial P}{\partial t} = - \dfrac{\partial j(x,t)}{\partial x} $$
If we denote the Probability of finding a particle between the interval $[a, b]$ as $P_{ab} = \displaystyle \int_{a}^{b}P$, we get the following.
$$ \dfrac{\partial P_{ab}}{\partial t} = j(a, t) - j(b, t) $$
Therefore, $j(x, t)$ can be interpreted as the quantity of probability passing through the point $x$ when the time is $t$. In other words, it is the Flux of the probability density, and it is the flux of the wave function. For this reason, we call $j$ Probability Current. The left-hand side represents the total change in probability within the interval $[a, b]$, while the right-hand side represents the sum of the changes in probability at both ends of the interval $[a, b]$. In simple terms, it is like saying that the change in the total number of people in a room is equal to the total number of people who entered or exited the room. A similar concept exists in electromagnetism called the Continuity Equation.
$$ \dfrac{\partial \rho}{\partial t} = -\nabla $$
The Continuity Equation signifies that the total change in charge quantity is equal to the sum of changes at the boundaries, implying that the charge is conserved. Similarly, $(2)$ signifies the conservation of probability. Naturally, when we integrate the probability density over the entire space, it must equal $1$, which means that if the value decreases in one place, it must increase in another.
Derivation
Let’s start from the Schrödinger Equation.
$$ \i \hbar \frac{\partial \psi (x,t)}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial ^2 \psi (x,t)}{\partial x^2} $$
Multiplying both sides by $\psi^{\ast}$, we get the following.
$$ \i \hbar \psi^{\ast}\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \psi^{\ast} \frac{\partial ^2 \psi}{\partial x^2} \tag{3} $$
Next, applying the Complex Conjugate to the Schrödinger equation and multiplying by $\psi$, we get the following.
$$ -\i \hbar \psi\frac{\partial \psi^{\ast}}{\partial t} = -\frac{\hbar^2}{2m} \psi \frac{\partial ^2 \psi^{\ast}}{\partial x^2} \tag{4} $$
Calculating $(3) - (4)$, we obtain the following.
$$ \begin{align*} \i\hbar \left[ \psi^{\ast} \frac{ \partial \psi }{\partial t} + \psi \frac{\partial \psi^{\ast}}{\partial t}\right] &= -\frac{ \hbar^2 }{2m}\left[\psi^{\ast} \frac{\partial^2 \psi}{\partial x^2} - \psi \frac{\partial^2 \psi^{\ast}}{\partial x^2} \right] \\ &= -\frac{ \hbar^2 }{2m} \left[ \frac{\partial}{\partial x} \left( \psi^{\ast}\frac{\partial \psi }{\partial x} - \psi \frac{\partial \psi^{\ast}}{\partial x }\right) \right] \end{align*} $$
Simplifying the left-hand side again, we get the following.
$$ \begin{align*} && \i\hbar \frac{\partial}{\partial t}\left( \psi^{\ast} \psi \right) &= -\frac{ \hbar^2 }{2m} \frac{\partial}{\partial x} \left( \psi^{\ast}\frac{\partial \psi }{\partial x} - \psi \frac{\partial \psi^{\ast}}{\partial x }\right) \\ \implies&& \frac{\partial}{\partial t}\left( \psi^{\ast} \psi \right) &= -{\color{blue}\frac{ \hbar }{2m\i} } \frac{\partial}{\partial x} {\color{blue}\left( \psi^{\ast}\frac{\partial \psi }{\partial x} - \psi \frac{\partial \psi^{\ast}}{\partial x }\right)} \end{align*} $$
Since $\psi^{\ast} \psi=|\psi|^{2} $, the left-hand side is the time derivative of the Probability Density. Denoting the blue part on the right-hand side as $j(x,t)$, we finally obtain the following.
$$ \frac{\partial }{\partial t} \left| \psi \right|^{2} = -\frac{\partial}{\partial x}j(x,t) $$
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