Answer to Question #195988 in Quantum Mechanics for LUCKY

Question #195988

(a) Suppose, I have normalized the wave function at some point of time. The wave

function evolves with time according to time dependent Schrodinger equation. How do

I know that the wave function remains normalized after some time?

[Hint: Show that 𝑑

𝑑𝑑

βˆ«βˆ’βˆž

∞

|πœ“(π‘₯,𝑑)|

2𝑑π‘₯ = 0]

(b) Show that

π‘‘βŸ¨π‘βŸ©

𝑑𝑑

= βŸ¨βˆ’

πœ•π‘‰

πœ•π‘₯⟩

i.e. expectation values follow Newton’s law.


1
Expert's answer
2021-05-20T10:07:59-0400
  • As indicated, let us calculate "\\frac{d}{dt}\\int |\\psi(x,t)|^2 dx", using that we can differentiate under the integral sign :

"\\frac{d}{dt}\\int |\\psi(x,t)|^2dx = \\int(\\frac{\\partial}{\\partial t}|\\psi(x,t)|^2)dx"

Now we use that "|\\psi|^2=\\psi \\bar{\\psi}" :

"\\int \\frac{\\partial}{\\partial t}(\\psi\\bar{\\psi})dx =\\int ((\\frac{\\partial}{\\partial t}\\psi)\\bar{\\psi}+(\\frac{\\partial}{\\partial t}\\bar{\\psi})\\psi)dx"

Now we will use the fact that "\\psi" satisfies the Schrodinger equation

"\\hat{H}\\psi = i\\hbar\\partial_t \\psi"

And as hamoltonian is a real operator

"\\bar{\\hat{H}}=\\hat{H}" and thus

"\\hat H \\bar{\\psi}=-i\\hbar \\partial_t \\bar{\\psi}"

Using these expressions gives us

"\\int \\left((\\frac{-i}{\\hbar}\\hat H\\psi)\\bar{\\psi}+(\\frac{i}{\\hbar}\\hat H \\bar\\psi)\\psi\\right)dx"

But we know that as "\\hat H" is a hermitian operator, we have "\\int \\bar \\psi \\hat H \\psi dx = \\int\\psi\\hat H \\bar \\psi dx" and thus the integral is zero, "\\frac{d}{dt}\\int |\\psi(x,t)|^2dx=0", "\\int |\\psi(x,t)|^2dx=||\\psi||^2=const" and as initially it is 1, it is always normalized.

  • First let's write the expression for the expectation value of "p"

"<p>=\\int \\bar\\psi (-i\\hbar\\frac{\\partial}{\\partial x})\\psi dx"

Deriving with respect to time gives us

"-i\\hbar \\left( \\int(\\partial_t\\bar \\psi)(\\partial_x\\psi)dx+\\int(\\bar\\psi) (\\partial_t\\partial_x\\psi) dx \\right)"

As previously, we will use the Schrodinger equation to replace "\\partial \\psi" :

"-i\\hbar\\left( \\int\\frac{i}{\\hbar}(\\hat H \\bar{\\psi})(\\partial_x\\psi)dx+\\int(\\bar\\psi) \\frac{-i}{\\hbar}(\\partial_x\\hat H \\psi) \\right)"

Now, using that hamiltonian is a hermitian operator we can write it as

"\\int \\bar \\psi (\\hat H \\partial_x\\psi - \\partial_x \\hat H \\psi)dx"

Now using that "\\hat H = -\\frac{\\hbar^2}{2m}\\partial_{xx}+V" we find

"\\int \\bar\\psi (-\\frac{\\hbar^2}{2m}\\partial_{xxx}\\psi +V\\partial_x \\psi +\\frac{\\hbar^2}{2m}\\partial_{xxx}\\psi-(\\partial_xV)\\psi -V\\partial_x\\psi)dx"

"\\int \\bar\\psi (-\\partial_xV)\\psi dx=<-\\frac{\\partial}{\\partial x}V>"

"\\frac{d}{dt}<p>=<-\\frac{\\partial V}{\\partial x}>"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS