Two-state quantum system

In quantum mechanics, a two-state system (also known as a TLS or two-level system) is a system which has two possible states. More formally, the Hilbert space of a two-state system has two degrees of freedom, so a complete basis spanning the space must consist of two independent states. An example of a two-state system is the spin of a spin-1/2 particle such as an electron, whose spin can have values +½ or -½ in units of ħ.

The physics of a quantum mechanical two-state system is trivial if both states are degenerate, that is, if the states have the same energy. However, if there is an energy difference between the two states, then nontrivial dynamics can occur.

1 Two-state dynamics
2 Dynamics of a two state system interacting with a field
3 Examples of two-state quantum systems
4 See also

Two-state dynamics

If the time-independent Hamiltonian is $H\,\!$ , and we label the two levels as $\left.|a\right\rangle\,\!$ and $\left.|b\right\rangle\,\!$ with corresponding orthonormal energy eigenvalues $E_a\,\!$ and $E_b\,\!$ , then the dynamics of the system can be specified as follows:

At some time $t_0\,\!$ , let the system be in an arbitrary (and completely general) state, $\left.|\psi(t_0)\right\rangle=c_a\left.|a\right\rangle%2Bc_b\left.|b\right\rangle \,\!$ then after evolving under $H\,\!$ , at time $t\,\!$ , the state will be

$\left.|\psi(t)\right\rangle = c_a e^{-i\frac{E_a(t-t_0)}{\hbar}}\left|a\right\rangle%2B c_b e^{-i\frac{E_b(t-t_0)}{\hbar}}\left|b\right\rangle \,\!$

The physics of two state systems can be usefully applied to multi-state systems where the system is known to have only enough energy available to excite the lowest two states, thus effectively creating a two state system. In fact, in nature, it is difficult to identify any true two-state systems; merely systems where the energetics of the circumstances isolate two particular states.

The set of all states in a two-level system can be mapped onto a representation known as the Bloch sphere.

Dynamics of a two state system interacting with a field

Free Precession

The interaction of a two state system with an external field results in the precession of the state vector. The ability to carefully control the position of the state vector on the Bloch sphere is central to the realization of the qubit. Consider the case of a spin-1/2 particle in a magnetic field. The interaction Hamiltonian for this system is

$H=-\vec{\mu}\cdot\vec{B}=-\mu\vec{\sigma}\cdot\vec{B}$

where $\mu$ is the magnitude of the particle's magnetic moment, $\vec{B}$ is the magnetic field, and $\vec{\sigma}$ is the vector of Pauli matrices. In analogy to related systems, $\mu$ is a coupling constant, and $\vec{B}=B\hat{n}$ is an external field. Solving the time dependent Schrödinger equation $i\hbar \partial_t \psi = H\psi=-\mu B \vec{\sigma}\cdot \hat{n}\psi$ yields the solution

$\psi = e^{i\omega t \vec{\sigma}\cdot \hat{n}} \psi_0$

where $\omega = \frac{\mu B}{\hbar}$ , $\psi_0$ is the state vector at time $t=0$ , and $e^{i\omega t \vec{\sigma}\cdot \hat{n}} = \cos{\left(\omega t\right)} I %2B i \hat{n}\cdot\vec{\sigma} \sin{\left(\omega t\right)}$ . Physically, this corresponds to the Bloch vector precessing around $\hat{n}$ with angular frequency $2\omega$ . Without loss of generality, assume the field points in $\hat{z}$ , such that

$e^{i\omega t \vec{\sigma}\cdot \hat{n}} = \begin{pmatrix} e^{i\omega t} & 0 \\ 0 & e^{-i\omega t} \end{pmatrix}$

The representation on the Bloch sphere for a state vector $\psi_0$ will simply be the vector of expectation values $\vec{R}=\left(\langle \sigma_x \rangle,\langle \sigma_y \rangle,\langle \sigma_z \rangle \right)$ . As an example, consider a state vector $\psi_0$ that is a normalized superposition of $|\uparrow\rangle$ and $|\downarrow\rangle$ , that is, a vector that can be represented in the $\sigma_z$ basis as

$\psi_0 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$

its components on the Bloch sphere will simply be $\vec{R}=\left(\cos{2\omega t},-\sin{2\omega t},0\right)$ . This is a unit vector that begins pointing in $\hat{x}$ and precesses around $\hat{z}$ in a left-handed manner. In general, by a rotation around $\hat{z}$ , any state vector $\psi_0$ can be represented as $a|\uparrow\rangle %2B b|\downarrow\rangle$ with real coefficients $a$ and $b$ . Such a state vector corresponds to a Bloch vector in the $xz$ plane making an angle $\tan{\frac{\theta}{2}} =\frac{b}{a}$ with the $z$ -axis. This vector will proceed to precess around $\hat{z}$ . In theory, by allowing the system to interact with the field of a particular direction and strength for precise durations, it is possible to obtain any orientation of the Bloch vector, which is equivalent to obtaining any complex superposition. This is the basis for numerous technologies including quantum computing and MRI.

NMR

NMR is an important example in the dynamics of two-state systems because it is involves the exact solution to a time dependent Hamiltonian. Consider a spin-1/2 particle in a magnetic field

$\vec{B}= \begin{pmatrix} B_1 \cos\omega_r t \\ B_1 \sin\omega_r t\\ B_0 \end{pmatrix}.$

This corresponds physically to a situation where there is a time independent field of strength $B_0$ pointing in $\hat{z}$ , and a field of strength $B_1$ rotating in a right-handed manner with an angular frequency $\omega_r$ . The dynamics of this situation are found as above by solving the time dependent Schrödinger equation

$i \hbar \frac{\partial\psi}{\partial t} = -\mu\vec{\sigma}\cdot\vec{B}\psi$

Which, after dividing through by $i \hbar$ and expanding out the dot product is

$\frac{\partial\psi}{\partial t} = i\left(\omega_1\sigma_x \cos{\omega_r t} %2B \omega_1\sigma_y \sin{\omega_r t} %2B \omega_0 \sigma_z\right)\psi$

where $\omega_i = \frac{\mu B_i}{\hbar}$ . To remove the time dependence from the problem, the wave function is transformed according to $\psi \rightarrow e^{-i \sigma_z \omega_r t/2}\psi$ . The time dependent Schrödinger equation becomes

$-i \sigma_z \frac{\omega_r}{2}e^{-i \sigma_z \omega_r t/2}\psi %2B e^{-i \sigma_z \omega_r t/2}\frac{\partial \psi}{\partial t}=i\left(\omega_1\sigma_x \cos{\omega_r t} %2B \omega_1\sigma_y \sin{\omega_r t} %2B \omega_0 \sigma_z\right)e^{-i \sigma_z \omega_r t/2}\psi$

Which after some rearrangement yields

$\frac{\partial \psi}{\partial t}=ie^{i \sigma_z \omega_r t/2}\left(\omega_1\sigma_x \cos{\omega_r t} %2B \omega_1\sigma_y \sin{\omega_r t} %2B \left(\omega_0%2B\frac{\omega_r}{2}\right) \sigma_z\right)e^{-i \sigma_z \omega_r t/2}\psi$

Evaluating each term on the right hand side of the equation

$e^{i \sigma_z \omega_r t/2}\sigma_x e^{-i \sigma_z \omega_r t/2} = \begin{pmatrix} e^{i\omega_r t/2} & 0 \\ 0 & e^{-i\omega_r t/2} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} e^{-i\omega_r t/2} & 0 \\ 0 & e^{i\omega_r t/2} \end{pmatrix}= \begin{pmatrix} 0 & e^{i\omega_r t} \\ e^{-i\omega_r t} & 0 \end{pmatrix}$

$e^{i \sigma_z \omega_r t/2}\sigma_y e^{-i \sigma_z \omega_r t/2} = \begin{pmatrix} e^{i\omega_r t/2} & 0 \\ 0 & e^{-i\omega_r t/2} \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} e^{-i\omega_r t/2} & 0 \\ 0 & e^{i\omega_r t/2} \end{pmatrix}= \begin{pmatrix} 0 & -i e^{i\omega_r t} \\ i e^{-i\omega_r t} & 0 \end{pmatrix}$

$e^{i \sigma_z \omega_r t/2}\sigma_z e^{-i \sigma_z \omega_r t/2} = \begin{pmatrix} e^{i\omega_r t/2} & 0 \\ 0 & e^{-i\omega_r t/2} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} e^{-i\omega_r t/2} & 0 \\ 0 & e^{i\omega_r t/2} \end{pmatrix}=\sigma_z$

The equation now reads

$\frac{\partial \psi}{\partial t} = i\left(\omega_1 \begin{pmatrix} 0 & e^{i\omega_r t}\left(\cos{\omega_r t} - i\sin{\omega_r t}\right) \\ e^{-i\omega_r t}\left(\cos{\omega_r t} %2B i\sin{\omega_r t}\right) & 0 \end{pmatrix}%2B\left(w_0%2B\frac{\omega_r}{2}\right)\sigma_z\right)\psi$

Which by Euler's identity becomes

$\frac{\partial \psi}{\partial t}=i\left(\omega_1 \sigma_x %2B \left(w_0%2B\frac{\omega_r}{2}\right)\sigma_z\right)\psi$

As per the previous section, the solution to this equation has the Bloch vector precessing around $(\omega_1,0,\omega_0 %2B \omega_r/2)$ with a frequency that is twice the magnitude of the vector. If $\omega_0$ is sufficiently strong, some proportion of the spins will be pointing directly down prior to the introduction of the rotating field. If the angular frequency of the rotating magnetic field is chosen such that $\omega_r = -2 \omega_0$ , in the rotating frame the state vector will precess around $\hat{x}$ with frequency $2 \omega_1$ , and will thus flip from down to up releasing energy in the form of detectable photons. This is the fundamental basis for NMR, and in practice is accomplished by scanning $\omega_r$ until the resonant frequency is found at which point the sample will emit light. Similar calculations are done in atomic physics, and in the case that the field is not rotating, but oscillating with a complex amplitude, use is made of the rotating wave approximation in deriving such results.

Relation to Bloch equations

The Bloch equations for a collection of spin-1/2 particles can be derived from the time dependent Schrödinger equation for a two level system. Starting with the previously stated Hamiltonian $i\hbar\partial_t\psi=-\mu\vec{\sigma}\cdot\vec{B}\psi$ , it can be written in summation notation after some rearrangement as

$\frac{\partial\psi}{\partial t}=i\frac{\mu}{\hbar}\sigma_i B_i \psi$

Multiplying by a Pauli matrix $\sigma_i$ and the conjugate transpose of the wavefunction, and subsequently expanding the product of two Pauli matrices yields

$\psi^\dagger \sigma_j \frac{\partial\psi}{\partial t}=i\frac{\mu}{\hbar}\psi^\dagger\sigma_j\sigma_i B_i \psi =i\frac{\mu}{\hbar}\psi^\dagger \left(I\delta_{ij} - i \sigma_k \varepsilon_{ijk}\right)B_i\psi=\frac{\mu}{\hbar}\psi^\dagger \left(iI\delta_{ij} %2B \sigma_k \varepsilon_{ijk}\right)B_i\psi$

Adding this equation to its own conjugate transpose yields a left hand side of the form

$\psi^\dagger \sigma_j \frac{\partial\psi}{\partial t} %2B \frac{\partial\psi^\dagger}{\partial t} \sigma_j \psi = \frac{\partial \left( \psi^\dagger \sigma_j \psi\right)}{\partial t}$

And a right hand side of the form

$\frac{\mu}{\hbar}\psi^\dagger \left(iI\delta_{ij} %2B \sigma_k \varepsilon_{ijk}\right)B_i\psi %2B \frac{\mu}{\hbar}\psi^\dagger \left(-iI\delta_{ij} %2B \sigma_k \varepsilon_{ijk}\right)B_i\psi = \frac{2\mu}{\hbar} \left( \psi^\dagger\sigma_k\psi\right) B_i \varepsilon_{ijk}$

As previously mentioned, the expectation value of each Pauli matrix is a component of the Bloch vector, $\langle \sigma_i \rangle = \psi^\dagger\sigma_i\psi = R_i$ . Equating the left and right hand sides, and noting that $\frac{2\mu}{\hbar}$ is the gyromagnetic ratio $\gamma$ , yields another form for the equations of motion of the Bloch vector

$\frac{\partial R_j}{\partial t} = \gamma R_k B_i \varepsilon_{kij}$

where the fact that $\varepsilon_{ijk} = \varepsilon_{kij}$ has been used. In vector form these three equations can be expressed in terms of a cross product

$\frac{\partial \vec{R}}{\partial t} = \gamma \vec{R} \times \vec{B}$

Classically, this equation describes the dynamics of a spin in a magnetic field. An ideal magnet consists of a collection of identical spins behaving independently, and thus the total magnetization $\vec{M}$ is proportional to the Bloch vector $\vec{R}$ . All that is left to obtain the final form of the Bloch equations is the inclusion of the phenomenological relaxation terms.

As a final aside, the above equation can be derived by considering the time evolution of the angular momentum operator in the Heisenberg picture.

$i\hbar\frac{d\sigma_j}{dt}=[\sigma_j,H]=[\sigma_j, -\mu \sigma_i B_i]=-\mu\left(\sigma_j\sigma_i B_i - \sigma_i\sigma_j B_i\right)=\mu[\sigma_i,\sigma_j]B_i = 2\mu i \varepsilon_{ijk}\sigma_k B_i$

Which, when coupled with the fact that $\vec{R_i} = \langle \sigma_i \rangle$ , is the same equation as before.

Examples of two-state quantum systems

Spin-1/2 particles are two-state quantum systems when only the spin degree of freedom is considered.
The "inverting" degree of freedom in an ammonia molecule; the nitrogen at the vertex of an ammonia molecule exhibits two molecular states - "up" and "down", corresponding to opposite positions with respect to the plane of the three hydrogen atoms. In an electric field, these two states are non-degenerate.
Two-level systems are important in the field of quantum computing as they are used to implement qubits.