- #1
Smalde
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The problem is actually of an introductory leven in Quantum Mechanics. I am doing a course on atomic and molecular physics and they wanted us to practice again some of the basics.
I want to know where I went conceptually wrong because my answer doesn't give a total probability of one, which of course is wrong.
\hat{H}\Psi_1=E_1\Psi_1\quad and\quad\hat{H}\Psi_2=E_2\Psi_2
\phi_1=(\Psi_1-\Psi_2)/\sqrt2\quad and\quad\phi_2=(\Psi_1+\Psi_2)/\sqrt2
\hat{A}\phi_1=a_1\quad and\quad\hat{A}\phi_2=a_2
The question goes like this: Assuming you measure A^ and find a1, then how does the probability of finding the same result (a1) change over time?
Well, the only important equation is the time development for a wave function in an eigenstate of the Hamilton Operator.
\exp(-iE_jt/\hbar)
After measurement the wave function collapses so the state wave function becomes Φ1. Since we know that \phi_1=(\Psi_1-\Psi_2)/\sqrt2\ we know that the time-dependent wave function will be: \Psi=(\Psi_1\exp(-iE_jt/\hbar)-\Psi_2\exp(-iE_jt/\hbar))/\sqrt2\
Thus if we take <\Psi|\hat{A}|\Psi> it is
<(\Psi_1\exp(-iE_jt/\hbar)-\Psi_2\exp(-iE_jt/\hbar))/\sqrt2\ |\hat{A}|(\Psi_1\exp(-iE_jt/\hbar)-\Psi_2\exp(-iE_jt/\hbar))/\sqrt2\ >
= 1/2 (<\Psi_1|\hat{A}|\Psi_1> + <\Psi_2|\hat{A}|\Psi_2> - 2<\Psi_1|\hat{A}|\Psi_2>\cos((E1-E2)t/\hbar))
= 1/2 ( 1/2(a_1+a_2) + 1/2(a_2-a_1) - 2<\Psi_1|\hat{A}|\Psi_2>\cos((E1-E2)t/\hbar)))
= 1/2( a_2 - \cos((E_1-E_2)t/\hbar))(a_2-a_1))
= a_2/2 - \cos((E_1-E_2)t/\hbar))(a_2-a_1)/2
I don't think this can be true, since when the cosinus is 0 the result is a2/2, thus violating the fact that the sum of the factors should be one... And when one is to insert t=0 one becomes a1/2 and not a1 as expected.
Could someone tell me where I went conceptually wrong? Thanks in advance :D
I want to know where I went conceptually wrong because my answer doesn't give a total probability of one, which of course is wrong.
Homework Statement
\hat{H}\Psi_1=E_1\Psi_1\quad and\quad\hat{H}\Psi_2=E_2\Psi_2
\phi_1=(\Psi_1-\Psi_2)/\sqrt2\quad and\quad\phi_2=(\Psi_1+\Psi_2)/\sqrt2
\hat{A}\phi_1=a_1\quad and\quad\hat{A}\phi_2=a_2
The question goes like this: Assuming you measure A^ and find a1, then how does the probability of finding the same result (a1) change over time?
Homework Equations
Well, the only important equation is the time development for a wave function in an eigenstate of the Hamilton Operator.
\exp(-iE_jt/\hbar)
The Attempt at a Solution
After measurement the wave function collapses so the state wave function becomes Φ1. Since we know that \phi_1=(\Psi_1-\Psi_2)/\sqrt2\ we know that the time-dependent wave function will be: \Psi=(\Psi_1\exp(-iE_jt/\hbar)-\Psi_2\exp(-iE_jt/\hbar))/\sqrt2\
Thus if we take <\Psi|\hat{A}|\Psi> it is
<(\Psi_1\exp(-iE_jt/\hbar)-\Psi_2\exp(-iE_jt/\hbar))/\sqrt2\ |\hat{A}|(\Psi_1\exp(-iE_jt/\hbar)-\Psi_2\exp(-iE_jt/\hbar))/\sqrt2\ >
= 1/2 (<\Psi_1|\hat{A}|\Psi_1> + <\Psi_2|\hat{A}|\Psi_2> - 2<\Psi_1|\hat{A}|\Psi_2>\cos((E1-E2)t/\hbar))
= 1/2 ( 1/2(a_1+a_2) + 1/2(a_2-a_1) - 2<\Psi_1|\hat{A}|\Psi_2>\cos((E1-E2)t/\hbar)))
= 1/2( a_2 - \cos((E_1-E_2)t/\hbar))(a_2-a_1))
= a_2/2 - \cos((E_1-E_2)t/\hbar))(a_2-a_1)/2
I don't think this can be true, since when the cosinus is 0 the result is a2/2, thus violating the fact that the sum of the factors should be one... And when one is to insert t=0 one becomes a1/2 and not a1 as expected.
Could someone tell me where I went conceptually wrong? Thanks in advance :D