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titanandwire
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Homework Statement
Consider an experiment on a system that can be described using two basis functions. We begin in the ground state of a Hamiltonian H0 at a time t1, then rapidly change the hamiltonian to H1 at the time t1. At a later time tD>t1 you preform a measurement of an observable D.
[tex] H_0 =
\left(
\begin{array}{cc}
0 & -4\\
-4 & 6
\end{array}
\right),
\hspace{0.1in}
H_1 =
\left(
\begin{array}{cc}
1 & 0\\
0 & 3
\end{array}
\right),
\hspace{0.1in}
D =
\left(
\begin{array}{cc}
0 & 1\\
1 & 2
\end{array}
\right)
[/tex]
a ) What are the possible outcomes for the measurement of D? What is the state of the system after each measurement?
b ) If you preform many, many measurements what will be the average observed vaule of D as a function of t1 and tD (this is an exam task from 2004 from a MIT open course on quantum mechanics)
Homework Equations
The eigenvalue equation
[tex]
det(A - \lambda I ) = 0
[/tex]
Fouriers trick, for a state ψ represented in a particular orthogonal basis fnthe coefficients can be found by:
[tex]
c_n =\langle f_n | \psi \rangle
[/tex]
By fouriers trick the probability of getting a particular eigenvalue of an operator Q, qn, associated by the orthonormalized eigenfunction fn
Expectation value of an observable Q is given by
[tex]
\langle Q \rangle = \sum _n q_n |c_n|^2
[/tex]
The Attempt at a Solution
a)[/B]
Changing the hamiltonian rapidly results in an diabatic process which is characterized by the statement that the state doesn't change. Such that for t1 the system is in the ground state of H0, but expressed in the basis of H1. For the later time tD one can assume that the system has transitioned according to a tacked-on exponential time dependant term.
The eigenvalues of D is, from the eigenvalue equation. Which also are the possible measurements of D
[tex]
\lambda_{1, D} = -0.4 \\
\lambda_{2, D} = 2.4
[/tex]
The original hamiltonian has the normalized eigenvectors
[tex]
H_0 \frac{1}{\sqrt{5}} \left( \begin{array}{c} -2 \\ -1 \end{array}\right) = -2 \frac{1}{\sqrt{5}} \left( \begin{array}{c} -2 \\ -1 \end{array}\right) \\
H_0 \frac{1}{\sqrt{5}} \left( \begin{array}{c} -1 \\ 2 \end{array} \right)= 8 \frac{1}{\sqrt{5}} \left(\begin{array}{c} -1 \\ 2 \end{array}\right)
[/tex]
Such that the state with eigenvalue -2 is the groundstate of H0.
The new hamiltonian, H1 , then has eigenvectors that are the cartesian basis-vectors for ℝ2 (which we see quite trivially). With eigenvalues 1 and 3. At the time t1 before the experiment is started the system is then in the ground-state of H0. After the experiment is started and the new hamiltonian is in place the state of the system is then a linear combination of the basis-vectors of H1 and tacking on an exponential time dependence
[tex]
\frac{1}{\sqrt{5}} \left( \begin{array}{c} -2 \\ -1 \end{array} \right) = \frac{1}{\sqrt{5}} (-2e_1\cdot e^{-i(t- t_1)} - e_2 \cdot e^{-i3(t- t_1)})
[/tex]
Where e1 and e2 are the ordinairy cartesian basis vectors. At the time tD the system has gone to:
[tex]
|t_d \rangle= \frac{1}{\sqrt{5}} (-2e_1\cdot e^{-i(t_D- t_1)} - e_2 \cdot e^{-i3(t_D- t_1)})
[/tex]
b)
The expectation value of D at the time of measurement tD is given in the normal manner as
[tex]
\langle D \rangle =\langle t_d| D |t_d \rangle
[/tex]
Doing the matrix multiplication in the normal manner gives
[tex]
\langle D \rangle = \frac{2}{5}(e^{-6i(t_D - t_1)} + 2e^{-4i(t_D - t_1)})
[/tex]
EDIT1 : I realized I for some reason picked the excited state of H0. Corrected this!
EDIT2: Changed the statement "For the later time tD one can assume that the system has transitioned to the ground state of the changed hamiltonian H1." to "After the experiment is started and the new hamiltonian is in place the state of the system is then a linear combination of the basis-vectors of H1 and tacking on an exponential time dependence"
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