3.4: Ehrenfest's Theorem
( \newcommand{\kernel}{\mathrm{null}\,}\)
A simple way to calculate the expectation value of momentum is to evaluate the time derivative of ⟨x⟩, and then multiply by the mass m: that is,
⟨p⟩=md⟨x⟩dt=mddt∫∞−∞x|ψ|2dx=m∫∞−∞x∂|ψ|2∂tdx.
However, it is easily demonstrated that
∂|ψ|2∂t+∂j∂x=0
[this is just the differential form of Equation ([epc])], where j is the probability current defined in Equation ([eprobc]). Thus,
⟨p⟩=−m∫∞−∞x∂j∂xdx=m∫∞−∞jdx,
where we have integrated by parts. It follows from Equation ([eprobc]) that
⟨p⟩=−iℏ2∫∞−∞(ψ∗∂ψ∂x−∂ψ∗∂xψ)dx=−iℏ∫∞−∞ψ∗∂ψ∂xdx,
where we have again integrated by parts. Hence, the expectation value of the momentum can be written
It follows from the previous equation that d⟨p⟩dt=−iℏ∫∞−∞(∂ψ∗∂t∂ψ∂x+ψ∗∂2ψ∂t∂x)dx=∫∞−∞[(iℏ∂ψ∂t)∗∂ψ∂x+∂ψ∗∂x(iℏ∂ψ∂t)]dx,
Hence, according to Equations ([e4.34x]) and ([e3.41]),
Suppose that the potential V(x) is slowly varying. In this case, we can expand dV/dx as a Taylor series about ⟨x⟩. Keeping terms up to second order, we obtain
dV(x)dx=dV(⟨x⟩)d⟨x⟩+dV2(⟨x⟩)d⟨x⟩2(x−⟨x⟩)+12dV3(⟨x⟩)d⟨x⟩3(x−⟨x⟩)2.
Substitution of the previous expansion into Equation ([e3.43]) yields d⟨p⟩dt=−dV(⟨x⟩)d⟨x⟩−σ2x2dV3(⟨x⟩)d⟨x⟩3,
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)