Our first job is to find the state \(\left|{\psi(t)}\right\rangle \) at a time \(t\) later than zero. To find the time-dependent state \(|\psi(t)\rangle\), we follow these steps:
- Diagonalize \(\hat H\) (find the eigenvalues \(E_n\) and eigenvectors \(|E_n\rangle\)).
- Write \(|\psi(t=0)\rangle\) in terms of the energy eigenstates \(|E_n\rangle\). \[ |\psi(t=0)\rangle = \sum_n c_n |E_n\rangle . \]
- Multiply each eigenstate coefficient by \(e^{-iE_n t/\hbar}\) to get \(|\psi(t)\rangle\). \[ |\psi(t)\rangle = \sum_n c_n e^{-iE_n t/\hbar}\,|E_n\rangle . \]
The expectation value of the operator \(\hat{M}\) is given by \(\left\langle {\psi(t)}\right|\hat{M}\left|{\psi(t)}\right\rangle \). In matrix notation, both \(|\psi(t)\rangle\) and \(\hat M\) must be written in the same basis. Here, I will work in the basis formed by the Hamiltonian eigenstates (since everything is already written in that basis). The same calculation could also be done directly in bra-ket form. \begin{eqnarray} \langle \hat{M} \rangle &=&\left\langle {\psi(t)}\right|\hat{M}\left|{\psi(t)}\right\rangle \\ &=&\frac{1}{\sqrt2}\, \begin{pmatrix} e^{i\frac{E_1}{\hbar}t}& e^{i\frac{E_2}{\hbar}t} \end{pmatrix} \begin{pmatrix} 0& c\\ c& 0 \end{pmatrix} \frac{1}{\sqrt2}\, \begin{pmatrix} e^{-i\frac{E_1}{\hbar}t}\\ e^{-i\frac{E_2}{\hbar}t} \end{pmatrix} \\ &=&\frac{c}{2}\left(e^{i\frac{E_1}{\hbar}t}e^{-i\frac{E_2}{\hbar}t} +e^{i\frac{E_2}{\hbar}t}e^{-i\frac{E_1}{\hbar}t}\right) \\ &=&\frac{c}{2}\left(e^{-i\frac{E_2-E_1}{\hbar}t} +e^{i\frac{E_2-E_1}{\hbar}t}\right) \\ &=&c\cos\left(\frac{(E_2-E_1)t}{\hbar}\right) \end{eqnarray} The frequency of this oscillation is \(\omega=\frac{E_2-E_1}{\hbar}\), which is the Bohr frequency.
- In this problem, the Hamiltionian is already diagonal, so we know its eigenvalues are just the diagonal elements \(E_1\) and \(E_2\) and its basis is the standard basis: \begin{equation} \left|{E_1}\right\rangle \doteq \begin{pmatrix} 1\\ 0 \end{pmatrix}, \left|{E_2}\right\rangle \doteq \begin{pmatrix} 0\\ 1 \end{pmatrix} \end{equation}
- To calculate the state of the system after a time \(t\), we need to write the state \(\left|{m_1}\right\rangle \) in terms of the energy eigenstates \(|E_n\rangle\). Since we have a matrix representation for the observable \(M\) in this Hamiltonian basis, all we have to do is find the eigenvectors and eigenvalues of \(\hat{M}\) and we can read off the coefficients of \(\left|{m_1}\right\rangle \). To find the eigenvalues, we solve the characteristic equation: \begin{eqnarray} 0 &=&\left\vert\ \begin{matrix} -\lambda & c\\ c & -\lambda \end{matrix} \right\vert \\ &=&\lambda^2-c^2 \\ &\Rightarrow& \lambda=\pm c \end{eqnarray} Since \(c>0\), the larger eigenvalue is \(+c\), so \(\left|{m_1}\right\rangle \) satisfies \(\hat M|m_1\rangle = c|m_1\rangle\). The eigenvector corresponding to \(\lambda=+c\) is found by: \begin{equation} \begin{pmatrix} 0 & c\\ c & 0 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix} = c \begin{pmatrix} x\\ y \end{pmatrix} \end{equation} This implies \(cy=cx\). If we set \(x=1\), then \(y=1\). And the state we want is given by: \begin{equation} \left|{m_1}\right\rangle \doteq N \begin{pmatrix} 1\\ 1\\ \end{pmatrix} \end{equation} The state is normalized if we choose \(N=\frac{1}{\sqrt{2}}\). In bra-ket language, we have: \begin{equation} \left|{\psi(0)}\right\rangle =\left|{m_1}\right\rangle =\frac{1}{\sqrt{2}}\left|{E_1}\right\rangle +\frac{1}{\sqrt{2}}\left|{E_2}\right\rangle \end{equation}
- Now we multiply each eigenstate coefficient by the phase factor \(e^{-i\frac{E_n}{\hbar} t}\), where \(E_n\) is the eigenvalue of the Hamiltonian corresponding to that eigenstate. \begin{equation} \left|{\psi(t)}\right\rangle =\frac{1}{\sqrt2}\, e^{-i\frac{E_1}{\hbar}t}\, \left|{E_1}\right\rangle +\frac{1}{\sqrt2}\, e^{-i\frac{E_2}{\hbar}t}\, \left|{E_2}\right\rangle \end{equation}
Consider a spin-1/2 particle with a magnetic moment. At time \(t=0\), the state of the particle is \(\left|{\psi(t=0)}\right\rangle =\left|{+}\right\rangle \).
If the observable \(S_x\) is measured at time \(t=0\), what are the possible measurement values, and what is the probability of obtaining each value?
The question “what are the possible results of a measurement” is always a question asking us to find the eigenvalues of the relevant observable. In this case, the observable is \(S_x\). We have found the eigenvalues of the spin operators numerous times, so we know the possible results of the measurement are \(\pm\frac{\hbar}{2}\).
The calculation of a probability is always of the form \(\vert \left\langle {\hbox{out}}\middle|{\hbox{in}}\right\rangle \vert^2\). In this case, the in-state is \(\left|{\psi(0)}\right\rangle \) which is just \(\left|{+}\right\rangle \) and the out-states are \(\left|{+}\right\rangle _x\) and \(\left|{-}\right\rangle _x\). So these two probabilities are given by: \begin{eqnarray} {\cal P}_{\left|{+}\right\rangle _x} &=& \vert \left\langle {\hbox{out}}\middle|{\hbox{in}}\right\rangle \vert^2\\ &=& \vert {}_x\left\langle {+}\middle|{+}\right\rangle \vert^2\\ &=& \frac{1}{2} \end{eqnarray} where I have used the Spins Reference sheet (or my memory) for the value of this particular projection in the last line. In a more general case, if we had not already found this projection many times, I would need either to change \(\left|{+}\right\rangle \) to the \(x\)-basis or \(\left|{+}\right\rangle _x\) to the \(z\)-basis, explicitly.
Similarly, \({\cal P}_{\left|{-}\right\rangle _x}= \vert {}_x\left\langle {-}\middle|{+}\right\rangle \vert^2=\frac{1}{2}\).
Instead of performing the above measurement, the system is allowed to evolve in a uniform magnetic field \(\vec{B}=B_0\, \hat y\). The Hamiltonian for a system in a uniform magnetic field \(\vec B=B_0\, \hat y\) is \(\hat H=\omega_0\, \hat S_y\). (You can treat \(\omega_0\) as a given parameter in your answers to the following two questions.)
What is the state of the system at the later time t (i.e., find \(|\psi(t)\rangle\))? Express your answer as a superposition of the \(S_z\) eigenstates \(|+\rangle_z\) and \(|-\rangle_z\).
To find the state at the later time \(t\), we follow the prescription at the end of section 3.1.
Now we are asked explicitly to switch back to the \(z\)-basis. Again using the Spins Reference sheet, we find: \begin{eqnarray} \left|{\psi(t)}\right\rangle &=&\frac{1}{\sqrt2}\, e^{-i\frac{\omega_0}{2}t}\, \left|{+}\right\rangle _y +\frac{1}{\sqrt2}\, e^{i\frac{\omega_0}{2}t}\, \left|{-}\right\rangle _y\\ &=&\frac{1}{\sqrt2}\, e^{-i\frac{\omega_0}{2}t}\, \left\{\frac{1}{\sqrt2}\left|{+}\right\rangle +i\frac{1}{\sqrt2}\left|{-}\right\rangle \right\}\\ &&+\frac{1}{\sqrt2}\, e^{i\frac{\omega_0}{2}t}\, \left\{\frac{1}{\sqrt2}\left|{+}\right\rangle -i\frac{1}{\sqrt2}\left|{-}\right\rangle \right\}\\ &=& \cos\left(\frac{\omega_0 t}{2}\right)\left|{+}\right\rangle + \sin\left(\frac{\omega_0 t}{2}\right)\left|{-}\right\rangle \end{eqnarray}
- “Diagonalize” the Hamiltionian, i.e. find its eigenvalues and eigenvectors. The Hamiltonian for a system in a uniform magnetic field \(\vec B=B_0\, \hat y\) is \(H=\omega_0\, S_y\), i.e. it is proportional to \(S_y\). Here, we have introduced a standard notation, \(\omega_0\equiv\frac{eB_o}{m_e}\). Therefore the eigenvectors of \(H\) are \(\left|{\pm}\right\rangle _y\) with eigenvalues \(\pm\frac{\hbar\omega_0}{2}\).
- To calculate the state of the system after a time \(t\), we need to switch to the basis of the Hamiltonian. Again, since we are are using the eigenbases of the spin components, we can read this relationship off the Spins Reference sheet. By adding the two equations from the Spins Reference sheet for the kets \(\left|{\pm}\right\rangle _y\), and simplifying, we find that: \begin{equation} \left|{+}\right\rangle =\frac{1}{\sqrt2}\left|{+}\right\rangle _y+\frac{1}{\sqrt2}\left|{-}\right\rangle _y \end{equation}
- Now we multiply each eigenstate coefficient by the phase factor \(e^{-i\frac{E_n}{\hbar} t}\), where \(E_n\) is the eigenvalue of the Hamiltonian corresponding to that eigenstate. \begin{equation} \left|{\psi(t)}\right\rangle =\frac{1}{\sqrt2}\, e^{-i\frac{\hbar\omega_0}{2}t}\, \left|{+}\right\rangle _y +\frac{1}{\sqrt2}\, e^{i\frac{\hbar\omega_0}{2}t}\, \left|{-}\right\rangle _y \end{equation}
Again, the calculation of a probability is always of the form \(\vert \left\langle {\hbox{out}}\middle|{\hbox{in}}\right\rangle \vert^2\). In this case, the in-state is \(\left|{\psi(t)}\right\rangle \) which we found in the previous part and the out-state is the \(\left|{+}\right\rangle _x\) which corresponds to the eigenvalue \(\frac{\hbar\omega_0}{2}\) of \(S_x\). Since we found \(\left|{\psi(t)}\right\rangle \) in the \(z\)-basis, we use the Spins Reference sheet to write the out-state also in the \(z\)-basis. \begin{eqnarray} {\cal P}_{\left|{+}\right\rangle _x} &=& \vert \left\langle {\hbox{out}}\middle|{\hbox{in}}\right\rangle \vert^2\\ &=& \vert {}_x\left\langle {+}\middle|{\psi(t)}\right\rangle \vert^2\\ &=& \left\vert\left(\frac{1}{\sqrt{2}}\left\langle {+}\right| +\frac{1}{\sqrt{2}}\left\langle {-}\right|\right)\right.\\ &&\left.\left(\cos\left(\frac{\omega_0 t}{2}\right)\left|{+}\right\rangle + \sin\left(\frac{\omega_0 t}{2}\right)\left|{-}\right\rangle \right)\right\vert^2\\ &=&\left\vert\frac{1}{\sqrt2}\left(\cos\left(\frac{\omega_0 t}{2}\right) + \sin\left(\frac{\omega_0 t}{2}\right)\right)\right\vert^2\\ &=&\frac{1}{2}\left(1+2\cos\left(\frac{\omega_0 t}{2}\right) \sin\left(\frac{\omega_0 t}{2}\right)\right)\\ &=&\frac{1}{2}\left(1+\sin\left(\omega_0 t\right)\right) \end{eqnarray}
Show that the probability of a measurement of the energy is time independent for a general state:
\[\left|{\psi(t)}\right\rangle = \sum_n c_n(t) \left|{E_n}\right\rangle \]
that evolves due to a time-independent Hamiltonian.
For a time independent Hamiltonian, the time evolution of a quantum state is:
\begin{align*} \left|{\psi(t)}\right\rangle = \sum_n c_n(t) \left|{E_n}\right\rangle \\ \left|{\psi(t)}\right\rangle = \sum_n c_n(0) e^{iE_n t/\hbar} \left|{E_n}\right\rangle \\ \end{align*}
To find the probability of measuring the \(m\)th energy.
\begin{align*} \mathcal{P}(E_m) &= \Big| \left\langle {E_m}\middle|{\psi(t)}\right\rangle \Big|^2\\ &= \Big| \left\langle {E_m}\right| \Big( \sum_n c_n(0)e^{iE_n t/\hbar} \left|{E_n}\right\rangle \Big) \Big|^2\\ &= \Big| \Big( \sum_n c_n(0)e^{iE_n t/\hbar} \cancelto{\delta_{m,n}}{\left\langle {E_m}\middle|{E_n}\right\rangle } \Big)|^2\\ &= \Big| \Big( \sum_n c_n(0)e^{iE_n t/\hbar} \delta_{m,n} \Big)|^2\\ &= \Big| c_m(0) e^{iE_m t/\hbar}|^2\\ &= \Big( c_m^{*}(0) e^{-iE_m t/\hbar} \Big) \Big( c_m(0) e^{iE_m t/\hbar} \Big)\\ &= |c_m(0)|^2 \end{align*}
This probability is time-independent.
Observables that commute with the Hamiltonian share a common set of eigenstates with the Hamiltonian. For example, if observable \(A\), with eigenstates \(\left|{a_j}\right\rangle \) so that
\[\hat{A} \left|{a_j}\right\rangle = a_j \left|{a_j}\right\rangle \]
commutes with the Hamiltonian, then \(\left|{a_j}\right\rangle \) is also an eigenstate of the Hamiltonian:
\[\hat{H} \left|{a_j}\right\rangle = E_j \left|{a_j}\right\rangle = E_j \left|{E_j}\right\rangle \]
and
\[\left\langle {a_j}\middle|{E_n}\right\rangle = \delta_{j,n}\]
and the probability of measuring \(a_j\) is:
\begin{align*} \mathcal{P}(a_j) &= \Big| \left\langle {a_j}\middle|{\psi(t)}\right\rangle \Big|^2\\ &= \Big| \left\langle {a_j}\right| \Big( \sum_n c_n(0)e^{iE_n t/\hbar} \left|{E_n}\right\rangle \Big) \Big|^2\\ &= \Big| \Big( \sum_n c_n(0)e^{iE_n t/\hbar} \cancelto{\delta_{j,n}}{\left\langle {a_j}\middle|{E_n}\right\rangle } \Big)|^2\\ &= \Big| \Big( \sum_n c_n(0)e^{iE_n t/\hbar} \delta_{j,n} \Big)|^2\\ &= \Big| c_j(0) e^{iE_j t/\hbar}|^2\\ &= \Big( c_j^{*}(0) e^{-iE_j t/\hbar} \Big) \Big( c_j(0) e^{iE_j t/\hbar} \Big)\\ &= |c_j(0)|^2 \end{align*}
This probability is also time-independent.