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


by hnicholls
Tags: current, density, derivation, integration, limits, probability, reversal
hnicholls
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#1
Jan6-13, 01:47 AM
P: 25
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
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mfb
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#2
Jan6-13, 09:53 AM
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The changed limits just changed the sign from + to -, the ##\frac{i\hbar}{2m}##-part is quantum mechanics.
hnicholls
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#3
Jan6-13, 11:30 AM
P: 25
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
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#4
Jan6-13, 11:47 AM
P: 123

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


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”:

[tex]{{\partial }_{t}}\rho +{{\partial }_{x}}j=0[/tex]

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

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

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

[tex]j=\frac{i\hbar }{2m}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right) [/tex]
mfb
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#5
Jan6-13, 11:48 AM
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P: 10,853
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
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#6
Jan6-13, 12:54 PM
P: 25
Quote Quote by cosmic dust View Post
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”:

[tex]{{\partial }_{t}}\rho +{{\partial }_{x}}j=0[/tex]

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

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

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

[tex]j=\frac{i\hbar }{2m}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right) [/tex]
But isn't j (x,t) defined as,

[tex]j=\frac{-i\hbar }{2m}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right) [/tex]
vanhees71
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#7
Jan6-13, 03:06 PM
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P: 2,152
The sign is uniquely defined by the Schrödinger equation, which reads (setting [itex]\hbar=1[/itex])
[tex]\mathrm{i} \partial_t \psi(t,x)=-\frac{\Delta}{2m} \psi(t,x)+ V(x) \psi(t,x).[/tex]
Multiplying with [itex] \psi^*[/itex] leads to
[tex]\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].[/tex]
Then subtracting the conjugate complex of this equation and multiplying with (-i) leads to
[tex]\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)].[/tex]
Comparing this with the continuity equation leads to
[tex]\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)],[/tex]
as already given by cosmic dust.
hnicholls
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#8
Jan10-13, 08:28 PM
P: 25
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.


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