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Probability Flow 📂Quantum Mechanics

Probability Flow

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Wave Function ψ(x,t)\psi (x, t) of Probability Current is defined as follows.

j(x,t):=ℏ2mi(ψ∗∂ψ∂x−ψ∂ψ∗∂x)(1) 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.

∂∣ψ(x,t)∣2∂t=−∂j(x,t)∂x(2) \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)(2) holds because we defined it as (1)(1). If we denote the Probability Density of the Wave Function as PP, then (2)(2) is as follows.

∂P∂t=−∂j(x,t)∂x \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][a, b] as Pab=∫abPP_{ab} = \displaystyle \int_{a}^{b}P, we get the following.

∂Pab∂t=j(a,t)−j(b,t) \dfrac{\partial P_{ab}}{\partial t} = j(a, t) - j(b, t)

Therefore, j(x,t)j(x, t) can be interpreted as the quantity of probability passing through the point xx when the time is tt. 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 jj Probability Current. The left-hand side represents the total change in probability within the interval [a,b][a, b], while the right-hand side represents the sum of the changes in probability at both ends of the interval [a,b][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.

∂ρ∂t=−∇ \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)(2) signifies the conservation of probability. Naturally, when we integrate the probability density over the entire space, it must equal 11, which means that if the value decreases in one place, it must increase in another.

Derivation

Let’s start from the Schrödinger Equation.

iℏ∂ψ(x,t)∂t=−ℏ22m∂2ψ(x,t)∂x2 \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 ψ∗\psi^{\ast}, we get the following.

iℏψ∗∂ψ∂t=−ℏ22mψ∗∂2ψ∂x2(3) \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 ψ\psi, we get the following.

−iℏψ∂ψ∗∂t=−ℏ22mψ∂2ψ∗∂x2(4) -\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)(3) - (4), we obtain the following.

iℏ[ψ∗∂ψ∂t+ψ∂ψ∗∂t]=−ℏ22m[ψ∗∂2ψ∂x2−ψ∂2ψ∗∂x2]=−ℏ22m[∂∂x(ψ∗∂ψ∂x−ψ∂ψ∗∂x)] \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.

iℏ∂∂t(ψ∗ψ)=−ℏ22m∂∂x(ψ∗∂ψ∂x−ψ∂ψ∗∂x)  ⟹  ∂∂t(ψ∗ψ)=−ℏ2mi∂∂x(ψ∗∂ψ∂x−ψ∂ψ∗∂x) \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 ψ∗ψ=∣ψ∣2\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)j(x,t), we finally obtain the following.

∂∂t∣ψ∣2=−∂∂xj(x,t) \frac{\partial }{\partial t} \left| \psi \right|^{2} = -\frac{\partial}{\partial x}j(x,t)

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  1. Stephen Gasiorowicz, 양자물리학(Quantum Physics, 서강대학교 물리학과 공역) (3rd Edition, 2005), p41-42 ↩︎

  2. David J. Griffiths, 양자역학(Introduction to Quantum Mechanics, 권영준 역) (2nd Edition, 2006), p21-22 ↩︎