Representation

Supposing the two available basis states of the system are ${\displaystyle |1\rangle }$ and ${\displaystyle |2\rangle }$, in general the state can be written as a superposition of these two states with probability amplitudes ${\displaystyle c_{1},c_{2}}$,

$${\displaystyle |\psi \rangle =c_{1}|1\rangle +c_{2}|2\rangle .}$$

Since the basis states are orthonormal, ${\displaystyle \langle i|j\rangle =\delta _{ij}}$ where ${\displaystyle i,j\in {1,2}}$ and ${\displaystyle \delta _{ij}}$ is the Kronecker delta, so ${\displaystyle c_{i}=\langle i|\psi \rangle }$. These two complex numbers may be considered coordinates in a two-dimensional complex Hilbert space.^[1] Thus the state vector corresponding to the state ${\displaystyle |\psi \rangle }$ is

$${\displaystyle |\psi \rangle \equiv {\begin{pmatrix}\langle 1|\psi \rangle \\\langle 2|\psi \rangle \end{pmatrix}}={\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}}=c_{1}{\begin{pmatrix}1\\0\end{pmatrix}}+c_{2}{\begin{pmatrix}0\\1\end{pmatrix}}=\mathbf {c} ,}$$

and the basis states correspond to the basis vectors,

$${\displaystyle |1\rangle \equiv {\begin{pmatrix}\langle 1|1\rangle \\\langle 2|1\rangle \end{pmatrix}}={\begin{pmatrix}1\\0\end{pmatrix}}}$$

and

$${\displaystyle |2\rangle \equiv {\begin{pmatrix}\langle 1|2\rangle \\\langle 2|2\rangle \end{pmatrix}}={\begin{pmatrix}0\\1\end{pmatrix}}.}$$

If the state ${\displaystyle |\psi \rangle }$ is normalized, the norm of the state vector is unity, i.e. ${\displaystyle {|c_{1}|}^{2}+{|c_{2}|}^{2}=1}$.

All observable physical quantities, such as energy, are associated with hermitian operators. In the case of energy and the corresponding Hamiltonian, H, this means

$${\displaystyle H_{ij}=\langle i|H|j\rangle =\langle j|H|i\rangle ^{*}=H_{ji}^{*},}$$

i.e. ${\displaystyle H_{11}}$ and ${\displaystyle H_{22}}$ are real, and ${\displaystyle H_{12}=H_{21}^{*}}$. Thus, these four matrix elements ${\displaystyle H_{ij}}$ produce a 2×2 hermitian matrix,

$${\displaystyle \mathbf {H} ={\begin{pmatrix}\langle 1|H|1\rangle &\langle 1|H|2\rangle \\\langle 2|H|1\rangle &\langle 2|H|2\rangle \end{pmatrix}}={\begin{pmatrix}H_{11}&H_{12}\\H_{12}^{*}&H_{22}\end{pmatrix}}.}$$

The time-independent Schrödinger equation states that ${\displaystyle H|\psi \rangle =E|\psi \rangle }$; substituting for ${\displaystyle |\psi \rangle }$ in terms of the basis states from above, and multiplying both sides by ${\displaystyle \langle 1|}$ or ${\displaystyle \langle 2|}$ produces a system of two linear equations that can be written in matrix form,

$${\displaystyle {\begin{pmatrix}H_{11}&H_{12}\\H_{12}^{*}&H_{22}\end{pmatrix}}{\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}}=E{\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}},}$$

or ${\displaystyle \mathbf {Hc} =E\mathbf {c} }$ which is a 2×2 matrix eigenvalues and eigenvectors problem. As mentioned above, this equation comes from plugging a general state into the time-independent Schrödinger equation. Remember that the time-independent Schrödinger equation is a restrictive condition used to specify the eigenstates. Therefore, when plugging a general state into it, we are seeing what form the general state must take to be an eigenstate. Doing so, and distributing, we get ${\displaystyle c_{1}H|1\rangle +c_{2}H|2\rangle =c_{1}E|1\rangle +c_{2}E|2\rangle }$, which requires ${\displaystyle c_{1}}$ or ${\displaystyle c_{2}}$ to be zero (${\displaystyle E}$ cannot be equal to both ${\displaystyle \varepsilon _{1}}$ and ${\displaystyle \varepsilon _{2}}$, the energies of the individual states, which are by definition different). Upon setting ${\displaystyle c_{1}}$ or ${\displaystyle c_{2}}$ to be 0, only one state remains, and ${\displaystyle E}$ is the energy of the surviving state. This result is a redundant reminder that the time-independent Schrödinger equation is only satisfied by eigenstates of H, which are (by definition of the state vector) the states where all except one coefficient are zero. Now, if we follow the same derivation, but before acting with the Hamiltonian on the individual states, we multiply both sides by ${\displaystyle \langle 1|}$ or ${\displaystyle \langle 2|}$, we get a system of two linear equations that can be combined into the above matrix equation. Like before, this can only be satisfied if ${\displaystyle c_{1}}$ or ${\displaystyle c_{2}}$ is zero, and when this happens, the constant ${\displaystyle E}$ will be the energy of the remaining state. The above matrix equation should thus be interpreted as a restrictive condition on a general state vector to yield an eigenvector of ${\displaystyle H}$, exactly analogous to the time-independent Schrödinger equation.

Of course, in general, commuting the matrix with a state vector will not result in the same vector multiplied by a constant E. For general validity, one has to write the equation in the form

$${\displaystyle {\begin{pmatrix}H_{11}&H_{12}\\H_{12}^{*}&H_{22}\end{pmatrix}}{\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}}={\begin{pmatrix}\varepsilon _{1}c_{1}\\\varepsilon _{2}c_{2}\end{pmatrix}},}$$

with the individual eigenstate energies still inside the product vector. In either case, the Hamiltonian matrix can be derived using the method specified above, or via the more traditional method of constructing a matrix using boundary conditions; specifically, by using the requirement that when it acts on either basis state, it must return that state multiplied by the energy of that state. (There are no boundary conditions on how it acts on a general state.) This results in a diagonal matrix with the diagonal elements being the energies of the eigenstates and the off-diagonal elements being zero. The form of the matrix above that uses bra-ket-enclosed Hamiltonians is a more generalized version of this matrix.

One might ask why it is necessary to write the Hamiltonian matrix in such a general form with bra-ket-enclosed Hamiltonians, since ${\displaystyle H_{ij},i\neq j}$ should always equal zero and ${\displaystyle H_{ii}}$ should always equal ${\displaystyle \varepsilon _{i}}$. The reason is that, in some more complex problems, the state vectors may not be eigenstates of the Hamiltonian used in the matrix. One place where this occurs is in degenerate perturbation theory, where the off-diagonal elements are nonzero until the problem is solved by diagonalization.

Because of the hermiticity of ${\displaystyle \mathbf {H} }$ the eigenvalues are real; or, rather, conversely, it is the requirement that the energies are real that implies the hermiticity of ${\displaystyle \mathbf {H} }$. The eigenvectors represent the stationary states, i.e., those for whom the absolute magnitude of the squares of the probability amplitudes do not change with time.

Eigenvalues of the Hamiltonian

The most general form of a 2×2 Hermitian matrix such as the Hamiltonian of a two-state system is given by

$${\displaystyle \mathbf {H} ={\begin{pmatrix}\varepsilon _{1}&\beta -i\gamma \\\beta +i\gamma &\varepsilon _{2}\end{pmatrix}},}$$

where ${\displaystyle \varepsilon _{1},\varepsilon _{2},\beta }$ and γ are real numbers with units of energy. The allowed energy levels of the system, namely the eigenvalues of the Hamiltonian matrix, can be found in the usual way.

Equivalently, this matrix can be decomposed as,

$${\displaystyle \mathbf {H} =\alpha \cdot \sigma _{0}+\beta \cdot \sigma _{1}+\gamma \cdot \sigma _{2}+\delta \cdot \sigma _{3}={\begin{pmatrix}\alpha +\delta &\beta -i\gamma \\\beta +i\gamma &\alpha -\delta \end{pmatrix}}.}$$

Here, ${\textstyle \alpha ={\frac {1}{2}}\left(\varepsilon _{1}+\varepsilon _{2}\right)}$ and ${\textstyle \delta ={\frac {1}{2}}\left(\varepsilon _{1}-\varepsilon _{2}\right)}$ are real numbers. The matrix ${\displaystyle \sigma _{0}}$ is the 2×2 identity matrix and the matrices ${\displaystyle \sigma _{k}}$ with ${\displaystyle k=1,2,3}$ are the Pauli matrices. This decomposition simplifies the analysis of the system, especially in the time-independent case, where the values of ${\displaystyle \alpha ,\beta ,\gamma }$ and ${\displaystyle \delta }$ are constants.

The Hamiltonian can be further condensed as

$${\displaystyle \mathbf {H} =\alpha \cdot \sigma _{0}+\mathbf {r} \cdot {\boldsymbol {\sigma }}.}$$

The vector ${\displaystyle \mathbf {r} }$ is given by ${\displaystyle (\beta ,\gamma ,\delta )}$ and ${\displaystyle \sigma }$ is given by ${\displaystyle (\sigma _{1},\sigma _{2},\sigma _{3})}$. This representation simplifies the analysis of the time evolution of the system and is easier to use with other specialized representations such as the Bloch sphere.

If the two-state system's time-independent Hamiltonian H is defined as above, then its eigenvalues are given by ${\displaystyle E_{\pm }=\alpha \pm |\mathbf {r} |}$. Evidently, α is the average energy of the two levels, and the norm of ${\displaystyle \mathbf {r} }$ is the splitting between them. The corresponding eigenvectors are denoted as ${\displaystyle |+\rangle }$ and ${\displaystyle |-\rangle }$.

Time dependence

We now assume that the probability amplitudes are time-dependent, though the basis states are not. The Time-dependent Schrödinger equation states ${\textstyle i\hbar \partial _{t}|\psi \rangle =H|\psi \rangle }$, and proceeding as before (substituting for ${\displaystyle |\psi \rangle }$ and premultiplying by ${\displaystyle \langle 1|,\langle 2|}$ again produces a pair of coupled linear equations, but this time they are first order partial differential equations: ${\textstyle i\hbar \partial _{t}\mathbf {c} =\mathbf {Hc} }$. If ${\displaystyle \mathbf {H} }$ is time independent there are several approaches to find the time dependence of ${\displaystyle c_{1},c_{2}}$, such as normal modes. The result is that

$${\displaystyle \mathbf {c} (t)=e^{-i\mathbf {H} t/\hbar }\mathbf {c} _{0}=\mathbf {U} (t)\mathbf {c} _{0}.}$$

where ${\displaystyle \mathbf {c} _{0}=\mathbf {c} (0)}$ is the statevector at ${\displaystyle t=0}$. Here the exponential of a matrix may be found from the series expansion. The matrix ${\displaystyle \mathbf {U} (t)}$ is called the time evolution matrix (which comprises the matrix elements of the corresponding time evolution operator ${\displaystyle U(t)}$). It is easily proved that ${\displaystyle \mathbf {U} (t)}$ is unitary, meaning that ${\displaystyle \mathbf {U} ^{\dagger }\mathbf {U} =1}$.

It can be shown that

$${\displaystyle \mathbf {U} (t)=e^{-i\mathbf {H} t/\hbar }=e^{-i\alpha t/\hbar }\left(\cos \left({\frac {|\mathbf {r} |}{\hbar }}t\right)\sigma _{0}-i\sin \left({\frac {|\mathbf {r} |}{\hbar }}t\right){\hat {r}}\cdot {\boldsymbol {\sigma }}\right),}$$

where ${\textstyle {\hat {r}}={\frac {\mathbf {r} }{|\mathbf {r} |}}.}$

When one changes the basis to the eigenvectors of the Hamiltonian, in other words, if the basis states ${\displaystyle |1\rangle ,|2\rangle }$ are chosen to be the eigenvectors, then ${\displaystyle \epsilon _{1}=H_{11}=\langle 1|H|1\rangle =E_{1}\langle 1|1\rangle =E_{1}}$ and ${\displaystyle \beta +i\gamma =H_{21}=\langle 2|H|1\rangle =E_{1}\langle 2|1\rangle =0}$ and so the Hamiltonian is diagonal, i.e. ${\displaystyle |\mathbf {r} |=\delta }$ and is of the form,

$${\displaystyle \mathbf {H} ={\begin{pmatrix}E_{1}&0\\0&E_{2}\end{pmatrix}}.}$$

Now, the unitary time evolution operator ${\displaystyle U}$ is easily seen to be given by:

$${\displaystyle \mathbf {U} (t)=e^{-i\mathbf {H} t/\hbar }={\begin{pmatrix}e^{-iE_{1}t/\hbar }&0\\0&e^{-iE_{2}vt/\hbar }\end{pmatrix}}=e^{-i\alpha t/\hbar }{\begin{pmatrix}e^{-i\delta t/\hbar }&0\\0&e^{i\delta t/\hbar }\end{pmatrix}}=e^{-i\alpha t/\hbar }\left(\cos \left({\frac {\delta }{\hbar }}t\right)\sigma _{0}-i\sin \left({\frac {\delta }{\hbar }}t\right){\boldsymbol {\sigma }}_{3}\right).}$$

The ${\displaystyle e^{-i\alpha t/\hbar }}$ factor merely contributes to the overall phase of the operator, and can usually be ignored to yield a new time evolution operator that is physically indistinguishable from the original operator. Moreover, any perturbation to the system (which will be of the same form as the Hamiltonian) can be added to the system in the eigenbasis of the unperturbed Hamiltonian and analysed in the same way as above. Therefore, for any perturbation the new eigenvectors of the perturbed system can be solved for exactly, as mentioned in the introduction.

Rabi formula for a static perturbation

Suppose that the system starts in one of the basis states at ${\displaystyle t=0}$, say ${\displaystyle |1\rangle }$ so that ${\textstyle \mathbf {c} _{0}={\begin{pmatrix}1\\0\end{pmatrix}}}$, and we are interested in the probability of occupation of each of the basis states as a function of time when ${\displaystyle \mathbf {H} }$ is the time-independent Hamiltonian.

$${\displaystyle \mathbf {c} (t)=\mathbf {U} (t)\mathbf {c} _{0}={\begin{pmatrix}U_{11}(t)&U_{12}(t)\\U_{21}(t)&U_{22}(t)\end{pmatrix}}{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}U_{11}(t)\\U_{21}(t)\end{pmatrix}}.}$$

The probability of occupation of state i is ${\displaystyle P_{i}(t)=|c_{i}(t)|^{2}=|U_{i1}(t)|^{2}}$. In the case of the starting state, ${\displaystyle P_{1}(t)=|c_{1}(t)|^{2}=|U_{11}(t)|^{2}}$, and from above,

$${\displaystyle U_{11}(t)=e^{\frac {-i\alpha t}{\hbar }}\left(\cos \left({\frac {|\mathbf {r} |}{\hbar }}t\right)-i\sin \left({\frac {|\mathbf {r} |}{\hbar }}t\right){\frac {\delta }{|\mathbf {r} |}}\right).}$$

Hence,

$${\displaystyle P_{1}(t)=\cos ^{2}(\Omega t)+\sin ^{2}(\Omega t){\frac {\Delta ^{2}}{\Omega ^{2}}}.}$$

Obviously, ${\displaystyle P_{1}(0)=1}$ due to the initial condition. The frequency

$${\displaystyle \Omega ={\frac {|\mathbf {r} |}{\hbar }}={\frac {1}{\hbar }}{\sqrt {\beta ^{2}+\gamma ^{2}+\delta ^{2}}}={\sqrt {|\Omega _{R}|^{2}+\Delta ^{2}}}}$$

is called the generalised Rabi frequency, ${\displaystyle \Omega _{R}=(\beta +i\gamma )/\hbar }$ is called the Rabi frequency, and ${\displaystyle \Delta =\delta /\hbar }$ is called the detuning.

At zero detuning, ${\displaystyle P_{1}(t)=\cos ^{2}(|\Omega _{R}|t)}$, i.e., there is Rabi flopping from guaranteed occupation of state 1, to guaranteed occupation of state 2, and back to state 1, etc., with frequency ${\displaystyle |\Omega _{R}|}$. As the detuning is increased away from zero, the frequency of the flopping increases (to Ω) and the amplitude of exciting the electron decreases to ${\displaystyle \Omega ^{2}/\Delta ^{2}}$.

For time dependent Hamiltonians induced by light waves, see the articles on Rabi cycle and rotating wave approximation.

Precession in a field

Consider the case of a spin-1/2 particle in a magnetic field ${\displaystyle \mathbf {B} =B\mathbf {\hat {n}} }$. The interaction Hamiltonian for this system is

$${\displaystyle H=-{\boldsymbol {\mu }}\cdot \mathbf {B} =-\mu {\boldsymbol {\sigma }}\cdot \mathbf {B} ,}$$

where ${\displaystyle \mu }$ is the magnitude of the particle's magnetic moment and ${\displaystyle {\boldsymbol {\sigma }}}$ is the vector of Pauli matrices. Solving the time dependent Schrödinger equation ${\displaystyle H\psi =i\hbar \partial _{t}\psi }$ yields

$${\displaystyle \psi (t)=e^{i\omega t{\boldsymbol {\sigma }}\cdot \mathbf {\hat {n}} }\psi (0),}$$

where ${\displaystyle \omega =\mu B/\hbar }$ and ${\displaystyle e^{i\omega t{\boldsymbol {\sigma }}\cdot \mathbf {\hat {n}} }=\cos {\left(\omega t\right)}I+i\;\mathbf {\hat {n}} \cdot {\boldsymbol {\sigma }}\sin {\left(\omega t\right)}}$. Physically, this corresponds to the Bloch vector precessing around ${\displaystyle \mathbf {\hat {n}} }$ with angular frequency ${\displaystyle 2\omega }$. Without loss of generality, assume the field is uniform points in ${\displaystyle \mathbf {\hat {z}} }$, so that the time evolution operator is given as

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

It can be seen that such a time evolution operator acting on a general spin state of a spin-1/2 particle will lead to the precession about the axis defined by the applied magnetic field (this is the quantum mechanical equivalent of Larmor precession)^[2]

The above method can be applied to the analysis of any generic two-state system that is interacting with some field (equivalent to the magnetic field in the previous case) if the interaction is given by an appropriate coupling term that is analogous to the magnetic moment. The precession of the state vector (which need not be a physical spinning as in the previous case) can be viewed as the precession of the state vector on the Bloch sphere.

The representation on the Bloch sphere for a state vector ${\displaystyle \psi (0)}$ will simply be the vector of expectation values ${\displaystyle \mathbf {R} =\left(\langle \sigma _{x}\rangle ,\langle \sigma _{y}\rangle ,\langle \sigma _{z}\rangle \right)}$. As an example, consider a state vector ${\displaystyle \psi (0)}$ that is a normalized superposition of ${\displaystyle \left|\uparrow \right\rangle }$ and ${\displaystyle \left|\downarrow \right\rangle }$, that is, a vector that can be represented in the ${\displaystyle \sigma _{z}}$ basis as

$${\displaystyle \psi (0)={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\1\end{pmatrix}}}$$

The components of ${\displaystyle \psi (t)}$ on the Bloch sphere will simply be ${\displaystyle \mathbf {R} =\left(\cos {2\omega t},-\sin {2\omega t},0\right)}$. This is a unit vector that begins pointing along ${\displaystyle \mathbf {\hat {x}} }$ and precesses around ${\displaystyle \mathbf {\hat {z}} }$ in a left-handed manner. In general, by a rotation around ${\displaystyle \mathbf {\hat {z}} }$, any state vector ${\displaystyle \psi (0)}$ can be represented as ${\displaystyle a\left|\uparrow \right\rangle +b\left|\downarrow \right\rangle }$ with real coefficients ${\displaystyle a}$ and ${\displaystyle b}$. Such a state vector corresponds to a Bloch vector in the xz-plane making an angle ${\displaystyle \tan(\theta /2)=b/a}$ with the z-axis. This vector will proceed to precess around ${\displaystyle \mathbf {\hat {z}} }$. 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.

Evolution in a time-dependent field: Nuclear magnetic resonance

Nuclear magnetic resonance (NMR) is an important example in the dynamics of two-state systems because it involves the exact solution to a time dependent Hamiltonian. The NMR phenomenon is achieved by placing a nucleus in a strong, static field B₀ (the "holding field") and then applying a weak, transverse field B₁ that oscillates at some radiofrequency ω_r.^[3] Explicitly, consider a spin-1/2 particle in a holding field ${\displaystyle B_{0}\mathbf {\hat {z}} }$ and a transverse rf field B₁ rotating in the xy-plane in a right-handed fashion around B₀:

$${\displaystyle \mathbf {B} ={\begin{pmatrix}B_{1}\cos \omega _{\mathrm {r} }t\\B_{1}\sin \omega _{\mathrm {r} }t\\B_{0}\end{pmatrix}}.}$$

As in the free precession case, the Hamiltonian is ${\displaystyle H=-\mu {\boldsymbol {\sigma }}\cdot \mathbf {B} }$, and the evolution of a state vector ${\displaystyle \psi (t)}$ is found by solving the time-dependent Schrödinger equation ${\displaystyle H\psi =i\hbar \,\partial \psi /\partial t}$. After some manipulation (given in the collapsed section below), it can be shown that the Schrödinger equation becomes

$${\displaystyle {\frac {\partial \psi }{\partial t}}=i\left(\omega _{1}\sigma _{x}+\left(w_{0}+{\frac {\omega _{r}}{2}}\right)\sigma _{z}\right)\psi ,}$$

where ${\displaystyle \omega _{0}=\mu B_{0}/\hbar }$ and ${\displaystyle \omega _{1}=\mu B_{1}/\hbar }$.

As per the previous section, the solution to this equation has the Bloch vector precessing around ${\displaystyle (\omega _{1},0,\omega _{0}+\omega _{r}/2)}$ with a frequency that is twice the magnitude of the vector. If ${\displaystyle \omega _{0}}$ 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 ${\displaystyle \omega _{r}=-2\omega _{0}}$, in the rotating frame the state vector will precess around ${\displaystyle {\hat {x}}}$ with frequency ${\displaystyle 2\omega _{1}}$, 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 ${\displaystyle \omega _{r}}$ 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.

Derivation of above expression for the NMR Schrödinger equation

Here the Schrödinger equation reads

$${\displaystyle -\mu {\boldsymbol {\sigma }}\cdot \mathbf {B} \psi =i\hbar {\frac {\partial \psi }{\partial t}}.}$$

Expanding the dot product and dividing by ${\displaystyle i\hbar }$ yields

$${\displaystyle {\frac {\partial \psi }{\partial t}}=i\left(\omega _{1}\sigma _{x}\cos {\omega _{r}t}+\omega _{1}\sigma _{y}\sin {\omega _{r}t}+\omega _{0}\sigma _{z}\right)\psi .}$$

To remove the time dependence from the problem, the wave function is transformed according to ${\displaystyle \psi \rightarrow e^{-i\sigma _{z}\omega _{r}t/2}\psi }$. The time dependent Schrödinger equation becomes

$${\displaystyle -i\sigma _{z}{\frac {\omega _{r}}{2}}e^{-i\sigma _{z}\omega _{r}t/2}\psi +e^{-i\sigma _{z}\omega _{r}t/2}{\frac {\partial \psi }{\partial t}}=i\left(\omega _{1}\sigma _{x}\cos {\omega _{r}t}+\omega _{1}\sigma _{y}\sin {\omega _{r}t}+\omega _{0}\sigma _{z}\right)e^{-i\sigma _{z}\omega _{r}t/2}\psi ,}$$

which after some rearrangement yields

$${\displaystyle {\frac {\partial \psi }{\partial t}}=ie^{i\sigma _{z}\omega _{r}t/2}\left(\omega _{1}\sigma _{x}\cos {\omega _{r}t}+\omega _{1}\sigma _{y}\sin {\omega _{r}t}+\left(\omega _{0}+{\frac {\omega _{r}}{2}}\right)\sigma _{z}\right)e^{-i\sigma _{z}\omega _{r}t/2}\psi }$$

Evaluating each term on the right hand side of the equation

$${\displaystyle e^{i\sigma _{z}\omega _{r}t/2}\sigma _{x}e^{-i\sigma _{z}\omega _{r}t/2}={\begin{pmatrix}e^{i\omega _{r}t/2}&0\\0&e^{-i\omega _{r}t/2}\end{pmatrix}}{\begin{pmatrix}0&1\\1&0\end{pmatrix}}{\begin{pmatrix}e^{-i\omega _{r}t/2}&0\\0&e^{i\omega _{r}t/2}\end{pmatrix}}={\begin{pmatrix}0&e^{i\omega _{r}t}\\e^{-i\omega _{r}t}&0\end{pmatrix}}}$$

$${\displaystyle e^{i\sigma _{z}\omega _{r}t/2}\sigma _{y}e^{-i\sigma _{z}\omega _{r}t/2}={\begin{pmatrix}e^{i\omega _{r}t/2}&0\\0&e^{-i\omega _{r}t/2}\end{pmatrix}}{\begin{pmatrix}0&-i\\i&0\end{pmatrix}}{\begin{pmatrix}e^{-i\omega _{r}t/2}&0\\0&e^{i\omega _{r}t/2}\end{pmatrix}}={\begin{pmatrix}0&-ie^{i\omega _{r}t}\\ie^{-i\omega _{r}t}&0\end{pmatrix}}}$$

$${\displaystyle e^{i\sigma _{z}\omega _{r}t/2}\sigma _{z}e^{-i\sigma _{z}\omega _{r}t/2}={\begin{pmatrix}e^{i\omega _{r}t/2}&0\\0&e^{-i\omega _{r}t/2}\end{pmatrix}}{\begin{pmatrix}1&0\\0&-1\end{pmatrix}}{\begin{pmatrix}e^{-i\omega _{r}t/2}&0\\0&e^{i\omega _{r}t/2}\end{pmatrix}}=\sigma _{z}}$$

The equation now reads

$${\displaystyle {\frac {\partial \psi }{\partial t}}=i\left(\omega _{1}{\begin{pmatrix}0&e^{i\omega _{r}t}\left(\cos {\omega _{r}t}-i\sin {\omega _{r}t}\right)\\e^{-i\omega _{r}t}\left(\cos {\omega _{r}t}+i\sin {\omega _{r}t}\right)&0\end{pmatrix}}+\left(w_{0}+{\frac {\omega _{r}}{2}}\right)\sigma _{z}\right)\psi ,}$$

which by Euler's identity becomes

$${\displaystyle {\frac {\partial \psi }{\partial t}}=i\left(\omega _{1}\sigma _{x}+\left(w_{0}+{\frac {\omega _{r}}{2}}\right)\sigma _{z}\right)\psi }$$

Relation to Bloch equations

The optical 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 ${\displaystyle i\hbar \partial _{t}\psi =-\mu {\boldsymbol {\sigma }}\cdot \mathbf {B} \psi }$, it can be written in summation notation after some rearrangement as

$${\displaystyle {\frac {\partial \psi }{\partial t}}=i{\frac {\mu }{\hbar }}\sigma _{i}B_{i}\psi }$$

Multiplying by a Pauli matrix ${\displaystyle \sigma _{i}}$ and the conjugate transpose of the wavefunction, and subsequently expanding the product of two Pauli matrices yields

$${\displaystyle \psi ^{\dagger }\sigma _{j}{\frac {\partial \psi }{\partial t}}=i{\frac {\mu }{\hbar }}\psi ^{\dagger }\sigma _{j}\sigma _{i}B_{i}\psi =i{\frac {\mu }{\hbar }}\psi ^{\dagger }\left(I\delta _{ij}-i\sigma _{k}\varepsilon _{ijk}\right)B_{i}\psi ={\frac {\mu }{\hbar }}\psi ^{\dagger }\left(iI\delta _{ij}+\sigma _{k}\varepsilon _{ijk}\right)B_{i}\psi }$$

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

$${\displaystyle \psi ^{\dagger }\sigma _{j}{\frac {\partial \psi }{\partial t}}+{\frac {\partial \psi ^{\dagger }}{\partial t}}\sigma _{j}\psi ={\frac {\partial \left(\psi ^{\dagger }\sigma _{j}\psi \right)}{\partial t}}}$$

And a right hand side of the form

$${\displaystyle {\frac {\mu }{\hbar }}\psi ^{\dagger }\left(iI\delta _{ij}+\sigma _{k}\varepsilon _{ijk}\right)B_{i}\psi +{\frac {\mu }{\hbar }}\psi ^{\dagger }\left(-iI\delta _{ij}+\sigma _{k}\varepsilon _{ijk}\right)B_{i}\psi ={\frac {2\mu }{\hbar }}\left(\psi ^{\dagger }\sigma _{k}\psi \right)B_{i}\varepsilon _{ijk}}$$

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

$${\displaystyle {\frac {\partial R_{j}}{\partial t}}=\gamma R_{k}B_{i}\varepsilon _{kij}}$$

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

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

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 ${\displaystyle \mathbf {M} }$ is proportional to the Bloch vector ${\displaystyle \mathbf {R} }$. All that is left to obtain the final form of the optical 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.

$${\displaystyle i\hbar {\frac {d\sigma _{j}}{dt}}=\left[\sigma _{j},H\right]=\left[\sigma _{j},-\mu \sigma _{i}B_{i}\right]=-\mu \left(\sigma _{j}\sigma _{i}B_{i}-\sigma _{i}\sigma _{j}B_{i}\right)=\mu [\sigma _{i},\sigma _{j}]B_{i}=2\mu i\varepsilon _{ijk}\sigma _{k}B_{i}}$$

When coupled with the fact that ${\displaystyle \mathbf {R} _{i}=\langle \sigma _{i}\rangle }$, this equation is the same equation as before.