# Reversal of limits of integration in the derivation of probability current density

• hnicholls
In summary, the derivation of the probability current density involves changing the limits of integration from (b to a) to (a to b) and multiplying by a prefactor of -iħ/2m. This prefactor is necessary to satisfy the "continuity equation" and the Schrödinger equation and does not have a physical significance on its own.
hnicholls
In working out the derivation of the probability current density, I see (based on the definition of j(x,t)) that the limits of integration are changed from

d/dt∫(b to a) P(x.t) dx = iħ/2m[ψ*(x.t)∂/∂xψ(x.t) - ψ(x.t)∂/∂xψ*(x.t)](b to a)

to

d/dt∫(b to a) P(x.t) dx = -iħ/2m[ψ*(x.t)∂/∂xψ(x.t) - ψ(x.t)∂/∂xψ*(x.t)](a to b)

Thus, the prefactor becomes -iħ/2m as a result of reversing the limits of integration.

Is there a reason that the prefactor must be in terms of -i?

Thanks

The changed limits just changed the sign from + to -, the ##\frac{i\hbar}{2m}##-part is quantum mechanics.

I understand that the sign change is a result of reversing the limits and that the iħ/2m part is what quantizes the result, my question is why does the result need to me negative, rather than positive.

Thanks again.

Let me give you an alternative way to derive the current's expression, which maybe has more sense. What we are looking for is a function that satisfies the “continuity equation”:

$${{\partial }_{t}}\rho +{{\partial }_{x}}j=0$$

which comes from the requirement of local conservation of probability. In the above equation $\rho ={{\Psi }^{*}}\Psi$ is the probability density, so when you calculate ${\partial \rho }/{\partial t}\;$ using S.E. and its complex conjugate, you find:

$${{\partial }_{t}}\rho =-\frac{i\hbar }{2m}{{\partial }_{x}}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right)$$

So, when you compare this with the continuity equation, you have to set:

$$j=\frac{i\hbar }{2m}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right)$$

A minus sign does not mean that the result is negative, the integral itself can be negative as well. I don't know the context of that equation, but I think it is just a complex phase anyway. In other words, the sign (together with i) has no physical significance on its own.

cosmic dust said:
Let me give you an alternative way to derive the current's expression, which maybe has more sense. What we are looking for is a function that satisfies the “continuity equation”:

$${{\partial }_{t}}\rho +{{\partial }_{x}}j=0$$

which comes from the requirement of local conservation of probability. In the above equation $\rho ={{\Psi }^{*}}\Psi$ is the probability density, so when you calculate ${\partial \rho }/{\partial t}\;$ using S.E. and its complex conjugate, you find:

$${{\partial }_{t}}\rho =-\frac{i\hbar }{2m}{{\partial }_{x}}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right)$$

So, when you compare this with the continuity equation, you have to set:

$$j=\frac{i\hbar }{2m}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right)$$

But isn't j (x,t) defined as,

$$j=\frac{-i\hbar }{2m}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right)$$

The sign is uniquely defined by the Schrödinger equation, which reads (setting $\hbar=1$)
$$\mathrm{i} \partial_t \psi(t,x)=-\frac{\Delta}{2m} \psi(t,x)+ V(x) \psi(t,x).$$
Multiplying with $\psi^*$ leads to
$$\psi^*(t,x) \mathrm{i} \partial_t \psi(t,x)=\psi^*(t,x) \left [-\frac{\Delta}{2m} \psi(t,x)+ V(x) \psi(t,x) \right].$$
Then subtracting the conjugate complex of this equation and multiplying with (-i) leads to
$$\partial_t |\psi(t,x)|^2=\vec{\nabla} \cdot \frac{\mathrm{i}}{2m} [\psi^*(t,x) \vec{\nabla} \psi(t,x)-\psi(t,x) \vec{\nabla} \psi^*(t,x)].$$
Comparing this with the continuity equation leads to
$$\rho(t,x)=|\psi(t,x)|^2, \quad \vec{j}=-\frac{\mathrm{i}}{2m} [\psi^*(t,x) \vec{\nabla} \psi(t,x)-\psi(t,x) \vec{\nabla} \psi^*(t,x)],$$
as already given by cosmic dust.

So, proceeding with the assumption that ψ(t,x) satisfies the TDSE,

(−ℏ2/2m)∂x2ψ(t,x) = (iℏ)∂tψ(t,x)

with V(x)ψ(t,x) = 0

Dividing by iℏ

(−ℏ/2mi)∂x2ψ(t,x) = ∂tψ(t,x)

Multiply by -i

(−iℏ/2m)∂x2ψ(t,x) = ∂tψ(t,x)

Multiplying by ψ(t,x)∗

(−iℏ/2m)∂x2ψ(t,x)∗ψ(t,x) = ∂tψ(t,x)ψ(t,x)∗

P(t,x) = ψ(t,x)ψ(t,x)∗

So, (−iℏ/2m)∂x2ψ(t,x)∗ψ(t,x) = ∂tP(t,x)

and the left side of this equation can be rewritten as

(−iℏ/2m)∂x[ψ(t,x)∗∂xψ(t,x) - ψ(t,x)∂xψ(t,x)∗]

but, this is ∂xj(t,x) where j(t,x) is defined as

(−iℏ/2m)[ψ(t,x)∗∂xψ(t,x) - ψ(t,x)∂xψ(t,x)∗]

and ∂xj(t,x) = ∂tP(t,x) because the "continuity equation" requires that

∂xj(t,x) - ∂tP(t,x) = 0

So, the reversal of the limits of integration is necessary so that the TDSE and the "continuity equation" are both satisfied.

That seems right.

## 1. What is "Reversal of limits of integration" in the derivation of probability current density?

Reversal of limits of integration refers to the process of changing the order of integration in a mathematical equation. In the context of probability current density, this refers to switching the order of integration when calculating the current density from a probability density function.

## 2. Why is it necessary to reverse the limits of integration in the derivation of probability current density?

Reversing the limits of integration is necessary in order to correctly calculate the direction of the current flow. This is because the direction of the current is determined by the difference between the limits of integration.

## 3. How is the reversal of limits of integration done in the derivation of probability current density?

The reversal of limits of integration is done by simply swapping the limits of integration for the independent variable in the integration process. This is typically done in equations involving multiple variables, such as the probability current density equation.

## 4. What is the significance of the reversal of limits of integration in the derivation of probability current density?

The reversal of limits of integration is significant because it allows us to accurately determine the direction of current flow in a system. Without this reversal, the current density may be calculated incorrectly, leading to inaccurate results and predictions.

## 5. Can the reversal of limits of integration be applied to other mathematical equations?

Yes, the reversal of limits of integration can be applied to other mathematical equations that involve multiple variables and integration. It is a commonly used technique in calculus and is essential for accurately solving certain equations in physics and engineering.

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