Rapidly changing Hamiltonian and an observable

[titanandwire]

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 t_D>t₁ you preform a measurement of an observable D.
$$H_0 = \left( \begin{array}{cc} 0 & -4\\ -4 & 6 \end{array} \right), \hspace{0.1in} H_1 = \left( \begin{array}{cc} 1 & 0\\ 0 & 3 \end{array} \right), \hspace{0.1in} D = \left( \begin{array}{cc} 0 & 1\\ 1 & 2 \end{array} \right)$$
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 t_D (this is an exam task from 2004 from a MIT open course on quantum mechanics)

Homework Equations

The eigenvalue equation
$$det(A - \lambda I ) = 0$$
Fouriers trick, for a state ψ represented in a particular orthogonal basis f_nthe coefficients can be found by:
$$c_n =\langle f_n | \psi \rangle$$
By fouriers trick the probability of getting a particular eigenvalue of an operator Q, q_n, associated by the orthonormalized eigenfunction f_n

Expectation value of an observable Q is given by

$$\langle Q \rangle = \sum _n q_n |c_n|^2$$

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 t_D 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
$$\lambda_{1, D} = -0.4 \\ \lambda_{2, D} = 2.4$$
The original hamiltonian has the normalized eigenvectors
$$H_0 \frac{1}{\sqrt{5}} \left( \begin{array}{c} -2 \\ -1 \end{array}\right) = -2 \frac{1}{\sqrt{5}} \left( \begin{array}{c} -2 \\ -1 \end{array}\right) \\ H_0 \frac{1}{\sqrt{5}} \left( \begin{array}{c} -1 \\ 2 \end{array} \right)= 8 \frac{1}{\sqrt{5}} \left(\begin{array}{c} -1 \\ 2 \end{array}\right)$$
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

$$\frac{1}{\sqrt{5}} \left( \begin{array}{c} -2 \\ -1 \end{array} \right) = \frac{1}{\sqrt{5}} (-2e_1\cdot e^{-i(t- t_1)} - e_2 \cdot e^{-i3(t- t_1)})$$

Where e₁ and e₂ are the ordinairy cartesian basis vectors. At the time t_D the system has gone to:

$$|t_d \rangle= \frac{1}{\sqrt{5}} (-2e_1\cdot e^{-i(t_D- t_1)} - e_2 \cdot e^{-i3(t_D- t_1)})$$
b)

The expectation value of D at the time of measurement t_D is given in the normal manner as

$$\langle D \rangle =\langle t_d| D |t_d \rangle$$

Doing the matrix multiplication in the normal manner gives

$$\langle D \rangle = \frac{2}{5}(e^{-6i(t_D - t_1)} + 2e^{-4i(t_D - t_1)})$$

EDIT1 : I realized I for some reason picked the excited state of H₀. Corrected this!
EDIT2: Changed the statement "For the later time t_D 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"

 

[blue_leaf77]

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.

Sounds a little off to me. In the beginning specified by the time marker $t_1$ the system's Hamiltonian is $H_0$ then at time $t_1$, which is the same instant as the first Hamiltonian, the system is changed to $H_1$. Are you sure it was not $t_0$ when the system was still in $H_0$?
What is the initial state? In your answer you assume that it starts from the ground state of $H_0$ but I think such information should be given in the question.

For the later time tD one can assume that the system has transitioned to the ground state of the changed hamiltonian H1.

It's unnecessary, there is no reason why the system de-excite to the ground level (we are not considering atoms where the upper level has a finite life time due to the interaction with vacuum photon).

 

[titanandwire]

Sounds a little off to me. In the beginning specified by the time marker $t_1$ the system's Hamiltonian is $H_0$ then at time $t_1$, which is the same instant as the first Hamiltonian, the system is changed to $H_1$. Are you sure it was not $t_0$ when the system was still in $H_0$?
What is the initial state? In your answer you assume that it starts from the ground state of $H_0$ but I think such information should be given in the question.

It's unnecessary, there is no reason why the system de-excite to the ground level (we are not considering atom where an upper level has a finite life time due to interaction with vacuum photon).

The initial state is the ground state of $H_0$. It's formulated such that it's supposed to be a near instantaneous change from $H_0$ to $H_1$, to reflect that it's a diabatic process I reckon they've chosen to use the same time parameter. It confused me a good bit too.

Yeah I that was an artifact of some earlier reasoning, edited it out just before you answered to reflect the new reasoning.

 

[blue_leaf77]

For a) you are also asked the final state corresponding to the possible outcomes of the measurement of $D$. It doesn't look like you have the answer to this part already.
For b), your step is correct, I just don't have the time to check your calculation. But I believe it's a trivial matrix multiplication problem that you yourself can spot any error if you do recheck.

 

[titanandwire]

For a) you are also asked the final state corresponding to the possible outcomes of the measurement of $D$. It doesn't look like you have the answer to this part already.
For b), your step is correct, I just don't have the time to check your calculation. But I believe it's a trivial matrix multiplication problem that you yourself can spot any error if you do recheck.

I probably misunderstood then. I figured the question " What is the state of the system after each measurement" meant measurement in the sense of "a change in the system" i.e. to find the ground state of $H_0$, then its representation in the basis of $H_1$ and finally tacking on the time dependence to the same state and setting $t = t_D$. If it instead means the state associated with the possible measurements of $D$ that would be to find the eigenvalues of the matrix representation of $D$, as far as I've understood.

The exam also has another question "At tD, you measure D, then the energy and then D again in rapid (i.e. essentially instantaneous) succession. What is the probability that the two measurements of D yield the same result?" After measuring D, you collapse the wavefunction into an eigenstate of D, $D_n$ with a particular measurement value, $d_n$. The probability of each of these states are given by their coefficients
$$P(D_n) = \frac{|d_n|^2}{\sum_n |d_n|^2}$$
This statement should hold for non-normalized states too. So the probability of getting a measurement twice is then the probability of the first, squared (to account for it being measured twice). Plus the probability of the second, also squared. Have I understood that correctly or am I way off?

 

[blue_leaf77]

If it instead means the state associated with the possible measurements of DD that would be to find the eigenvalues of the matrix representation of DD, as far as I've understood.

The possible states after measuring $D$ are its eigenvectors.

This statement should hold for non-normalized states too. So the probability of getting a measurement twice is then the probability of the first, squared (to account for it being measured twice). Plus the probability of the second, also squared.

If by first and second as underlined in the quoted statement above are the first and second eigenvectors of $H_1$, then yes I believe that should be the correct way.