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

[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

 

[mfb]

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

 

[hnicholls]

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.

 

[cosmic dust]

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:

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

 

[mfb]

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.

 

[hnicholls]

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,

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

 

[vanhees71]

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.

 

[hnicholls]

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.