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Two-state quantum system

(Redirected fromTwo-level system)

Inquantum mechanics, atwo-state system (also known as atwo-level system) is aquantum system that can exist in anyquantum superposition of two independent (physically distinguishable)quantum states. TheHilbert space describing such a system is two-dimensional. Therefore, a completebasis spanning the space will consist of two independent states.[1] Any two-state system can also be seen as aqubit.

An electrically neutral silver atom beams throughStern–Gerlach experiment's inhomogeneous magnetic field splits into two, each of which corresponds to one possible spin value of the outermost electron of the silver atom.

Two-state systems are the simplest quantum systems that are of interest, since the dynamics of a one-state system is trivial (as there are no other states in which the system can exist). The mathematical framework required for the analysis of two-state systems is that oflinear differential equations andlinear algebra of two-dimensional spaces. As a result, the dynamics of a two-state system can be solved analytically without any approximation. The generic behavior of the system is that the wavefunction's amplitude oscillates between the two states.

A well known example of a two-state system is thespin of aspin-1/2 particle such as an electron, whose spin can have values +ħ/2 or −ħ/2, whereħ is thereduced Planck constant.

The two-state system cannot be used as a description of absorption or decay, because such processes require coupling to a continuum. Such processes would involve exponential decay of the amplitudes, but the solutions of the two-state system are oscillatory.

Analytical solutions for stationary state energies and time-dependence

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Representation

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Supposing the two available basis states of the system are|1{\displaystyle |1\rangle }  and|2{\displaystyle |2\rangle } , in general the state can be written as asuperposition of these two states withprobability amplitudesc1,c2{\displaystyle c_{1},c_{2}} ,|ψ=c1|1+c2|2.{\displaystyle |\psi \rangle =c_{1}|1\rangle +c_{2}|2\rangle .} 

Since the basis states areorthonormal,i|j=δij{\displaystyle \langle i|j\rangle =\delta _{ij}}  wherei,j1,2{\displaystyle i,j\in {1,2}}  andδij{\displaystyle \delta _{ij}}  is theKronecker delta, soci=i|ψ{\displaystyle c_{i}=\langle i|\psi \rangle } . These twocomplex numbers may be considered coordinates in a two-dimensionalcomplex Hilbert space.[2] Thus thestate vector corresponding to the state|ψ{\displaystyle |\psi \rangle }  is|ψ(1|ψ2|ψ)=(c1c2)=c1(10)+c2(01)=c,{\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,|1(1|12|1)=(10){\displaystyle |1\rangle \equiv {\begin{pmatrix}\langle 1|1\rangle \\\langle 2|1\rangle \end{pmatrix}}={\begin{pmatrix}1\\0\end{pmatrix}}}  and|2(1|22|2)=(01).{\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|ψ{\displaystyle |\psi \rangle }  isnormalized, thenorm of the state vector is unity, i.e.|c1|2+|c2|2=1{\displaystyle {|c_{1}|}^{2}+{|c_{2}|}^{2}=1} .

Allobservable physical quantities, such as energy, are associated withhermitian operators. In the case of energy and the correspondingHamiltonian,H, this meansHij=i|H|j=j|H|i=Hji,{\displaystyle H_{ij}=\langle i|H|j\rangle =\langle j|H|i\rangle ^{*}=H_{ji}^{*},}  i.e.H11{\displaystyle H_{11}}  andH22{\displaystyle H_{22}}  are real, andH12=H21{\displaystyle H_{12}=H_{21}^{*}} . Thus, these four matrix elementsHij{\displaystyle H_{ij}}  produce a 2×2hermitian matrix,H=(1|H|11|H|22|H|12|H|2)=(H11H12H12H22).{\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}}.} 

Thetime-independent Schrödinger equation states thatH|ψ=E|ψ{\displaystyle H|\psi \rangle =E|\psi \rangle } ; substituting for|ψ{\displaystyle |\psi \rangle }  in terms of the basis states from above, and multiplying both sides by1|{\displaystyle \langle 1|}  or2|{\displaystyle \langle 2|}  produces asystem of two linear equations that can be written in matrix form,(H11H12H12H22)(c1c2)=E(c1c2),{\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}},} orHc=Ec{\displaystyle \mathbf {Hc} =E\mathbf {c} }  which is a 2×2 matrixeigenvalues 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 getc1H|1+c2H|2=c1E|1+c2E|2{\displaystyle c_{1}H|1\rangle +c_{2}H|2\rangle =c_{1}E|1\rangle +c_{2}E|2\rangle } , which requiresc1{\displaystyle c_{1}}  orc2{\displaystyle c_{2}}  to be zero (E{\displaystyle E}  cannot be equal to bothε1{\displaystyle \varepsilon _{1}}  andε2{\displaystyle \varepsilon _{2}} , the energies of the individual states, which are by definition different). Upon settingc1{\displaystyle c_{1}}  orc2{\displaystyle c_{2}}  to be 0, only one state remains, andE{\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 by1|{\displaystyle \langle 1|}  or2|{\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 ifc1{\displaystyle c_{1}}  orc2{\displaystyle c_{2}}  is zero, and when this happens, the constantE{\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 ofH{\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 constantE. For general validity, one has to write the equation in the form(H11H12H12H22)(c1c2)=(ε1c1ε2c2),{\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, theHamiltonian 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, sinceHij,ij{\displaystyle H_{ij},i\neq j}  should always equal zero andHii{\displaystyle H_{ii}}  should always equalεi{\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 indegenerate perturbation theory, where the off-diagonal elements are nonzero until the problem is solved bydiagonalization.

Because of the hermiticity ofH{\displaystyle \mathbf {H} }  the eigenvalues are real; or, rather, conversely, it is the requirement that the energies are real that implies the hermiticity ofH{\displaystyle \mathbf {H} } . The eigenvectors represent thestationary 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

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The most general form of a 2×2 Hermitian matrix such as the Hamiltonian of a two-state system is given byH=(ε1βiγβ+iγε2),{\displaystyle \mathbf {H} ={\begin{pmatrix}\varepsilon _{1}&\beta -i\gamma \\\beta +i\gamma &\varepsilon _{2}\end{pmatrix}},} whereε1,ε2,β{\displaystyle \varepsilon _{1},\varepsilon _{2},\beta }  andγ are real numbers with units of energy. The allowed energy levels of the system, namely theeigenvalues of the Hamiltonian matrix, can be found in the usual way.

Equivalently, this matrix can be decomposed as,H=ασ0+βσ1+γσ2+δσ3=(α+δβiγβ+iγαδ).{\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,α=12(ε1+ε2){\textstyle \alpha ={\frac {1}{2}}\left(\varepsilon _{1}+\varepsilon _{2}\right)}  andδ=12(ε1ε2){\textstyle \delta ={\frac {1}{2}}\left(\varepsilon _{1}-\varepsilon _{2}\right)}  are real numbers. The matrixσ0{\displaystyle \sigma _{0}}  is the 2×2 identity matrix and the matricesσk{\displaystyle \sigma _{k}}  withk=1,2,3{\displaystyle k=1,2,3}  are thePauli matrices. This decomposition simplifies the analysis of the system, especially in the time-independent case, where the values ofα,β,γ{\displaystyle \alpha ,\beta ,\gamma }  andδ{\displaystyle \delta }  are constants.

The Hamiltonian can be further condensed asH=ασ0+rσ.{\displaystyle \mathbf {H} =\alpha \cdot \sigma _{0}+\mathbf {r} \cdot {\boldsymbol {\sigma }}.} 

The vectorr{\displaystyle \mathbf {r} }  is given by(β,γ,δ){\displaystyle (\beta ,\gamma ,\delta )}  andσ{\displaystyle \sigma }  is given by(σ1,σ2,σ3){\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 theBloch sphere.

If the two-state system's time-independent HamiltonianH is defined as above, then itseigenvalues are given byE±=α±|r|{\displaystyle E_{\pm }=\alpha \pm |\mathbf {r} |} . Evidently,α is the average energy of the two levels, and thenorm ofr{\displaystyle \mathbf {r} }  is the splitting between them. The corresponding eigenvectors are denoted as|+{\displaystyle |+\rangle }  and|{\displaystyle |-\rangle } .

Time dependence

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We now assume that theprobability amplitudes are time-dependent, though the basis states are not. TheTime-dependent Schrödinger equation statesit|ψ=H|ψ{\textstyle i\hbar \partial _{t}|\psi \rangle =H|\psi \rangle } , and proceeding as before (substituting for|ψ{\displaystyle |\psi \rangle }  and premultiplying by1|,2|{\displaystyle \langle 1|,\langle 2|}  again produces a pair of coupled linear equations, but this time they are first order partial differential equations:itc=Hc{\textstyle i\hbar \partial _{t}\mathbf {c} =\mathbf {Hc} } . IfH{\displaystyle \mathbf {H} }  is time independent there are several approaches to find the time dependence ofc1,c2{\displaystyle c_{1},c_{2}} , such asnormal modes. The result is thatc(t)=eiHt/c0=U(t)c0.{\displaystyle \mathbf {c} (t)=e^{-i\mathbf {H} t/\hbar }\mathbf {c} _{0}=\mathbf {U} (t)\mathbf {c} _{0}.}  wherec0=c(0){\displaystyle \mathbf {c} _{0}=\mathbf {c} (0)}  is the statevector att=0{\displaystyle t=0} . Here theexponential of a matrix may be found from the series expansion. The matrixU(t){\displaystyle \mathbf {U} (t)}  is called the time evolution matrix (which comprises the matrix elements of the corresponding time evolution operatorU(t){\displaystyle U(t)} ). It is easily proved thatU(t){\displaystyle \mathbf {U} (t)}  isunitary, meaning thatUU=1{\displaystyle \mathbf {U} ^{\dagger }\mathbf {U} =1} .

It can be shown thatU(t)=eiHt/=eiαt/(cos(|r|t)σ0isin(|r|t)r^σ),{\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),}  wherer^=r|r|.{\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|1,|2{\displaystyle |1\rangle ,|2\rangle }  are chosen to be the eigenvectors, thenϵ1=H11=1|H|1=E11|1=E1{\displaystyle \epsilon _{1}=H_{11}=\langle 1|H|1\rangle =E_{1}\langle 1|1\rangle =E_{1}}  andβ+iγ=H21=2|H|1=E12|1=0{\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.|r|=δ{\displaystyle |\mathbf {r} |=\delta }  and is of the form,H=(E100E2).{\displaystyle \mathbf {H} ={\begin{pmatrix}E_{1}&0\\0&E_{2}\end{pmatrix}}.} 

Now, the unitary time evolution operatorU{\displaystyle U}  is easily seen to be given by:U(t)=eiHt/=(eiE1t/00eiE2vt/)=eiαt/(eiδt/00eiδt/)=eiαt/(cos(δt)σ0isin(δt)σ3).{\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).} Theeiαt/{\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, anyperturbation 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

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Suppose that the system starts in one of the basis states att=0{\displaystyle t=0} , say|1{\displaystyle |1\rangle }  so thatc0=(10){\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 whenH{\displaystyle \mathbf {H} }  is the time-independent Hamiltonian.c(t)=U(t)c0=(U11(t)U12(t)U21(t)U22(t))(10)=(U11(t)U21(t)).{\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 statei isPi(t)=|ci(t)|2=|Ui1(t)|2{\displaystyle P_{i}(t)=|c_{i}(t)|^{2}=|U_{i1}(t)|^{2}} . In the case of the starting state,P1(t)=|c1(t)|2=|U11(t)|2{\displaystyle P_{1}(t)=|c_{1}(t)|^{2}=|U_{11}(t)|^{2}} , and from above,U11(t)=eiαt(cos(|r|t)isin(|r|t)δ|r|).{\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,P1(t)=cos2(Ωt)+sin2(Ωt)Δ2Ω2.{\displaystyle P_{1}(t)=\cos ^{2}(\Omega t)+\sin ^{2}(\Omega t){\frac {\Delta ^{2}}{\Omega ^{2}}}.} 

Obviously,P1(0)=1{\displaystyle P_{1}(0)=1}  due to the initial condition. The frequencyΩ=|r|=1β2+γ2+δ2=|ΩR|2+Δ2{\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,ΩR=(β+iγ)/{\displaystyle \Omega _{R}=(\beta +i\gamma )/\hbar }  is called the Rabi frequency, andΔ=δ/{\displaystyle \Delta =\delta /\hbar }  is called the detuning.

At zero detuning,P1(t)=cos2(|ΩR|t){\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|ΩR|{\displaystyle |\Omega _{R}|} . As the detuning is increased away from zero, the frequency of the flopping increases (toΩ) and the amplitude of exciting the electron decreases toΩ2/Δ2{\displaystyle \Omega ^{2}/\Delta ^{2}} .

For time dependent Hamiltonians induced by light waves, see the articles onRabi cycle androtating wave approximation.

Some important two-state systems

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Precession in a field

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Consider the case of aspin-1/2 particle in a magnetic fieldB=Bn^{\displaystyle \mathbf {B} =B\mathbf {\hat {n}} } . The interaction Hamiltonian for this system isH=μB=μσB,{\displaystyle H=-{\boldsymbol {\mu }}\cdot \mathbf {B} =-\mu {\boldsymbol {\sigma }}\cdot \mathbf {B} ,}  whereμ{\displaystyle \mu }  is the magnitude of the particle'smagnetic moment andσ{\displaystyle {\boldsymbol {\sigma }}}  is the vector ofPauli matrices. Solving the time dependent Schrödinger equationHψ=itψ{\displaystyle H\psi =i\hbar \partial _{t}\psi }  yieldsψ(t)=eiωtσn^ψ(0),{\displaystyle \psi (t)=e^{i\omega t{\boldsymbol {\sigma }}\cdot \mathbf {\hat {n}} }\psi (0),}  whereω=μB/{\displaystyle \omega =\mu B/\hbar }  andeiωtσn^=cos(ωt)I+in^σsin(ωt){\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 theBloch vector precessing aroundn^{\displaystyle \mathbf {\hat {n}} }  with angular frequency2ω{\displaystyle 2\omega } . Without loss of generality, assume the field is uniform and points inz^{\displaystyle \mathbf {\hat {z}} } , so that the time evolution operator is given aseiωtσn^=(eiωt00eiωt).{\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 ofLarmor precession)[3]

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 theBloch sphere.

The representation on the Bloch sphere for a state vectorψ(0){\displaystyle \psi (0)}  will simply be the vector of expectation valuesR=(σx,σy,σz){\displaystyle \mathbf {R} =\left(\langle \sigma _{x}\rangle ,\langle \sigma _{y}\rangle ,\langle \sigma _{z}\rangle \right)} . As an example, consider a state vectorψ(0){\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 theσz{\displaystyle \sigma _{z}}  basis asψ(0)=12(11){\displaystyle \psi (0)={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\1\end{pmatrix}}} 

The components ofψ(t){\displaystyle \psi (t)}  on the Bloch sphere will simply beR=(cos2ωt,sin2ωt,0){\displaystyle \mathbf {R} =\left(\cos {2\omega t},-\sin {2\omega t},0\right)} . This is a unit vector that begins pointing alongx^{\displaystyle \mathbf {\hat {x}} }  and precesses aroundz^{\displaystyle \mathbf {\hat {z}} }  in a left-handed manner. In general, by a rotation aroundz^{\displaystyle \mathbf {\hat {z}} } , any state vectorψ(0){\displaystyle \psi (0)}  can be represented asa|+b|{\displaystyle a\left|\uparrow \right\rangle +b\left|\downarrow \right\rangle }  with real coefficientsa{\displaystyle a}  andb{\displaystyle b} . Such a state vector corresponds to aBloch vector in thexz-plane making an angletan(θ/2)=b/a{\displaystyle \tan(\theta /2)=b/a}  with thez-axis. This vector will proceed to precess aroundz^{\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 theBloch vector, which is equivalent to obtaining any complex superposition. This is the basis for numerous technologies includingquantum computing andMRI.

Evolution in a time-dependent field: Nuclear magnetic resonance

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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 fieldB0 (the "holding field") and then applying a weak, transverse fieldB1 that oscillates at some radiofrequencyωr.[4] Explicitly, consider aspin-1/2 particle in a holding fieldB0z^{\displaystyle B_{0}\mathbf {\hat {z}} }  and a transverse rf fieldB1 rotating in thexy-plane in a right-handed fashion aroundB0:B=(B1cosωrtB1sinωrtB0).{\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 isH=μσB{\displaystyle H=-\mu {\boldsymbol {\sigma }}\cdot \mathbf {B} } , and the evolution of a state vectorψ(t){\displaystyle \psi (t)}  is found by solving the time-dependent Schrödinger equationHψ=iψ/t{\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 becomesψt=i(ω1σx+(w0+ωr2)σz)ψ,{\displaystyle {\frac {\partial \psi }{\partial t}}=i\left(\omega _{1}\sigma _{x}+\left(w_{0}+{\frac {\omega _{r}}{2}}\right)\sigma _{z}\right)\psi ,}  whereω0=μB0/{\displaystyle \omega _{0}=\mu B_{0}/\hbar }  andω1=μB1/{\displaystyle \omega _{1}=\mu B_{1}/\hbar } .

As per the previous section, the solution to this equation has theBloch vector precessing around(ω1,0,ω0+ωr/2){\displaystyle (\omega _{1},0,\omega _{0}+\omega _{r}/2)}  with a frequency that is twice the magnitude of the vector. Ifω0{\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 thatωr=2ω0{\displaystyle \omega _{r}=-2\omega _{0}} , in the rotating frame the state vector will precess aroundx^{\displaystyle {\hat {x}}}  with frequency2ω1{\displaystyle 2\omega _{1}} , and will thus flip from down to up releasing energy in the form of detectable photons.[citation needed] This is the fundamental basis forNMR, and in practice is accomplished by scanningωr{\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 therotating wave approximation in deriving such results.

Derivation of above expression for the NMR Schrödinger equation

Here the Schrödinger equation readsμσBψ=iψt.{\displaystyle -\mu {\boldsymbol {\sigma }}\cdot \mathbf {B} \psi =i\hbar {\frac {\partial \psi }{\partial t}}.} 

Expanding the dot product and dividing byi{\displaystyle i\hbar }  yieldsψt=i(ω1σxcosωrt+ω1σysinωrt+ω0σz)ψ.{\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 toψeiσzωrt/2ψ{\displaystyle \psi \rightarrow e^{-i\sigma _{z}\omega _{r}t/2}\psi } . The time dependent Schrödinger equation becomesiσzωr2eiσzωrt/2ψ+eiσzωrt/2ψt=i(ω1σxcosωrt+ω1σysinωrt+ω0σz)eiσzωrt/2ψ,{\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 yieldsψt=ieiσzωrt/2(ω1σxcosωrt+ω1σysinωrt+(ω0+ωr2)σz)eiσzωrt/2ψ{\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 equationeiσzωrt/2σxeiσzωrt/2=(eiωrt/200eiωrt/2)(0110)(eiωrt/200eiωrt/2)=(0eiωrteiωrt0){\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}}} eiσzωrt/2σyeiσzωrt/2=(eiωrt/200eiωrt/2)(0ii0)(eiωrt/200eiωrt/2)=(0ieiωrtieiωrt0){\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}}} eiσzωrt/2σzeiσzωrt/2=(eiωrt/200eiωrt/2)(1001)(eiωrt/200eiωrt/2)=σz{\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 readsψt=i(ω1(0eiωrt(cosωrtisinωrt)eiωrt(cosωrt+isinωrt)0)+(w0+ωr2)σz)ψ,{\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 byEuler's identity becomesψt=i(ω1σx+(w0+ωr2)σz)ψ{\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

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Theoptical Bloch equations for a collection ofspin-1/2 particles can be derived from the time dependent Schrödinger equation for a two level system. Starting with the previously stated Hamiltonianitψ=μσBψ{\displaystyle i\hbar \partial _{t}\psi =-\mu {\boldsymbol {\sigma }}\cdot \mathbf {B} \psi } , it can be written in summation notation after some rearrangement asψt=iμσiBiψ{\displaystyle {\frac {\partial \psi }{\partial t}}=i{\frac {\mu }{\hbar }}\sigma _{i}B_{i}\psi } 

Multiplying by aPauli matrixσi{\displaystyle \sigma _{i}}  and the conjugate transpose of the wavefunction, and subsequently expanding the product of two Pauli matrices yieldsψσjψt=iμψσjσiBiψ=iμψ(Iδijiσkεijk)Biψ=μψ(iIδij+σkεijk)Biψ{\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 formψσjψt+ψtσjψ=(ψσjψ)t{\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 formμψ(iIδij+σkεijk)Biψ+μψ(iIδij+σkεijk)Biψ=2μ(ψσkψ)Biεijk{\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 eachPauli matrix is a component of theBloch vector,σi=ψσiψ=Ri{\displaystyle \langle \sigma _{i}\rangle =\psi ^{\dagger }\sigma _{i}\psi =R_{i}} . Equating the left and right hand sides, and noting that2μ{\displaystyle {\frac {2\mu }{\hbar }}}  is thegyromagnetic ratioγ{\displaystyle \gamma } , yields another form for the equations of motion of theBloch vectorRjt=γRkBiεkij{\displaystyle {\frac {\partial R_{j}}{\partial t}}=\gamma R_{k}B_{i}\varepsilon _{kij}}  where the fact thatεijk=εkij{\displaystyle \varepsilon _{ijk}=\varepsilon _{kij}}  has been used. In vector form these three equations can be expressed in terms of across productRt=γR×B{\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 totalmagnetizationM{\displaystyle \mathbf {M} }  is proportional to theBloch vectorR{\displaystyle \mathbf {R} } . All that is left to obtain the final form of theoptical Bloch equations is the inclusion of the phenomenologicalrelaxation terms.

As a final aside, the above equation can be derived by considering the time evolution of theangular momentum operator in theHeisenberg picture.idσjdt=[σj,H]=[σj,μσiBi]=μ(σjσiBiσiσjBi)=μ[σi,σj]Bi=2μiεijkσkBi{\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 thatRi=σi{\displaystyle \mathbf {R} _{i}=\langle \sigma _{i}\rangle } , this equation is the same equation as before.

Validity

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Two-state systems are the simplest non-trivial quantum systems that occur in nature, but the above-mentioned methods of analysis are not just valid for simple two-state systems. Any general multi-state quantum system can be treated as a two-state system as long as the observable of interest has two eigenvalues. For example, a spin-1/2 particle may in reality have additional translational or even rotational degrees of freedom, but those degrees of freedom are irrelevant to the preceding analysis. Mathematically, the neglected degrees of freedom correspond to the degeneracy of the spin eigenvalues.

Another case where the effective two-state formalism is valid is when the system under consideration has two levels that are effectively decoupled from the system. This is the case in the analysis of the spontaneous or stimulated emission of light by atoms and that ofcharge qubits. In this case it should be kept in mind that the perturbations (interactions with an external field) are in the right range and do not cause transitions to states other than the ones of interest.

Significance and other examples

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Pedagogically, the two-state formalism is among the simplest of mathematical techniques used for the analysis of quantum systems. It can be used to illustrate fundamental quantum mechanical phenomena such as theinterference exhibited by particles of the polarization states of the photon,[5] but also more complex phenomena such asneutrino oscillation or theneutral K-meson oscillation.

Two-state formalism can be used to describe simple mixing of states, which leads to phenomena such asresonance stabilization and otherlevel crossing related symmetries. Such phenomena have a wide variety of application in chemistry. Phenomena with tremendous industrial applications such as themaser andlaser can be explained using the two-state formalism.

The two-state formalism also forms the basis ofquantum computing.Qubits, which are the building blocks of a quantum computer, are nothing but two-state systems. Any quantum computational operation is a unitary operation that rotates the state vector on the Bloch sphere.

Further reading

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See also

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References

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  1. ^Viola, Lorenza; Lloyd, Seth (October 1998)."Dynamical suppression of decoherence in two-state quantum systems".Physical Review A.58 (4). American Physical Society:2733–2744.arXiv:quant-ph/9803057.doi:10.1103/PhysRevA.58.2733.
  2. ^Griffiths, David (2005).Introduction to Quantum Mechanics (2nd ed.). p. 353.
  3. ^Feynman, R.P. (1965). "7-5 and 10-7".The Feynman Lectures on Physics: Volume 3. Addison Wesley.
  4. ^Griffiths, p. 377.
  5. ^Feynman, R.P. (1965). "11-4".The Feynman Lectures on Physics: Volume 3. Addison Wesley.

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