S-matrix

Consider a localized one dimensional potential barrier V(x), subjected to a beam of quantum particles with energy E. These particles are incident on the potential barrier from left to right.

The solutions of Schrödinger's equation outside the potential barrier are plane waves given by

$${\displaystyle \psi _{\rm {L}}(x)=Ae^{ikx}+Be^{-ikx}}$$

for the region to the left of the potential barrier, and

$${\displaystyle \psi _{\rm {R}}(x)=Ce^{ikx}+De^{-ikx}}$$

for the region to the right to the potential barrier, where

$${\displaystyle k={\sqrt {2mE/\hbar ^{2}}}}$$

is the wave vector. The time dependence is not needed in our overview and is hence omitted. The term with coefficient A represents the incoming wave, whereas term with coefficient C represents the outgoing wave. B stands for the reflecting wave. Since we set the incoming wave moving in the positive direction (coming from the left), D is zero and can be omitted.

The "scattering amplitude", i.e., the transition overlap of the outgoing waves with the incoming waves is a linear relation defining the S-matrix,

$${\displaystyle {\begin{pmatrix}B\\C\end{pmatrix}}={\begin{pmatrix}S_{11}&S_{12}\\S_{21}&S_{22}\end{pmatrix}}{\begin{pmatrix}A\\D\end{pmatrix}}.}$$

The above relation can be written as

$${\displaystyle \Psi _{\rm {out}}=S\Psi _{\rm {in}}}$$

where

$${\displaystyle \Psi _{\rm {out}}={\begin{pmatrix}B\\C\end{pmatrix}},\quad \Psi _{\rm {in}}={\begin{pmatrix}A\\D\end{pmatrix}},\qquad S={\begin{pmatrix}S_{11}&S_{12}\\S_{21}&S_{22}\end{pmatrix}}.}$$

The elements of S completely characterize the scattering properties of the potential barrier V(x).

The unitary property of the S-matrix is directly related to the conservation of the probability current in quantum mechanics.

The probability current density J of the wave function ψ(x) is defined as

$${\displaystyle J={\frac {\hbar }{2mi}}\left(\psi ^{*}{\frac {\partial \psi }{\partial x}}-\psi {\frac {\partial \psi ^{*}}{\partial x}}\right).}$$

The probability current density ${\displaystyle J_{\rm {L}}(x)}$ of ${\displaystyle \psi _{\rm {L}}(x)}$ to the left of the barrier is

$${\displaystyle J_{\rm {L}}(x)={\frac {\hbar k}{m}}\left(|A|^{2}-|B|^{2}\right),}$$

while the probability current density ${\displaystyle J_{\rm {R}}(x)}$ of ${\displaystyle \psi _{\rm {R}}(x)}$ to the right of the barrier is

$${\displaystyle J_{\rm {R}}(x)={\frac {\hbar k}{m}}\left(|C|^{2}-|D|^{2}\right).}$$

For conservation of the probability current, J_L = J_R. This implies the S-matrix is a unitary matrix.

Proof

$${\displaystyle {\begin{aligned}&J_{\rm {L}}=J_{\rm {R}}\\&|A|^{2}-|B|^{2}=|C|^{2}-|D|^{2}\\&|B|^{2}+|C|^{2}=|A|^{2}+|D|^{2}\\&\Psi _{\text{out}}^{\dagger }\Psi _{\text{out}}=\Psi _{\text{in}}^{\dagger }\Psi _{\text{in}}\\&\Psi _{\text{in}}^{\dagger }S^{\dagger }S\Psi _{\text{in}}=\Psi _{\text{in}}^{\dagger }\Psi _{\text{in}}\\&S^{\dagger }S=I\\\end{aligned}}}$$

If the potential V(x) is real, then the system possesses time-reversal symmetry. Under this condition, if ψ(x) is a solution of Schrödinger's equation, then ψ*(x) is also a solution.

The time-reversed solution is given by

$${\displaystyle \psi _{\rm {L}}^{*}(x)=A^{*}e^{-ikx}+B^{*}e^{ikx}}$$

for the region to the left to the potential barrier, and

$${\displaystyle \psi _{\rm {R}}^{*}(x)=C^{*}e^{-ikx}+D^{*}e^{ikx}}$$

for the region to the right to the potential barrier, where the terms with coefficient B*, C* represent incoming wave, and terms with coefficient A*, D* represent outgoing wave.

They are again related by the S-matrix,

$${\displaystyle {\begin{pmatrix}A^{*}\\D^{*}\end{pmatrix}}={\begin{pmatrix}S_{11}&S_{12}\\S_{21}&S_{22}\end{pmatrix}}{\begin{pmatrix}B^{*}\\C^{*}\end{pmatrix}}\,}$$

that is,

$${\displaystyle \Psi _{\rm {in}}^{*}=S\Psi _{\rm {out}}^{*}.}$$

Now, the relations

$${\displaystyle \Psi _{\rm {in}}^{*}=S\Psi _{\rm {out}}^{*},\quad \Psi _{\rm {out}}=S\Psi _{\rm {in}}}$$

together yield a condition

$${\displaystyle S^{*}S=I}$$

This condition, in conjunction with the unitarity relation, implies that the S-matrix is symmetric, as a result of time reversal symmetry,

$${\displaystyle S^{T}=S.}$$

By combining the symmetry and the unitarity, the S-matrix can be expressed in the form:

$${\displaystyle {\begin{pmatrix}S_{11}&S_{12}\\S_{21}&S_{22}\end{pmatrix}}={\begin{pmatrix}e^{i\varphi }e^{i\delta }\cdot r&e^{i\varphi }{\sqrt {1-r^{2}}}\\e^{i\varphi }{\sqrt {1-r^{2}}}&-e^{i\varphi }e^{-i\delta }\cdot r\end{pmatrix}}=e^{i\varphi }{\begin{pmatrix}e^{i\delta }\cdot r&{\sqrt {1-r^{2}}}\\{\sqrt {1-r^{2}}}&-e^{-i\delta }\cdot r\end{pmatrix}}}$$

with ${\displaystyle \delta ,\varphi \in [0;2\pi ]}$ and ${\displaystyle r\in [0;1]}$. So the S-matrix is determined by three real parameters.

The transfer matrix ${\displaystyle M}$ relates the plane waves ${\displaystyle Ce^{ikx}}$ and ${\displaystyle De^{-ikx}}$ on the right side of scattering potential to the plane waves ${\displaystyle Ae^{ikx}}$ and ${\displaystyle Be^{-ikx}}$ on the left side:[3]

$${\displaystyle {\begin{pmatrix}C\\D\end{pmatrix}}={\begin{pmatrix}M_{11}&M_{12}\\M_{21}&M_{22}\end{pmatrix}}{\begin{pmatrix}A\\B\end{pmatrix}}}$$

and its components can be derived from the components of the S-matrix via:[4] ${\displaystyle M_{11}=1/S_{12}^{*}=1/S_{21}^{*}{,}\ M_{22}=M_{11}^{*}}$ and ${\displaystyle M_{12}=-S_{11}^{*}/S_{12}^{*}=S_{22}/S_{12}{,}\ M_{21}=M_{12}^{*}}$, whereby time-reversal symmetry is assumed.

In the case of time-reversal symmetry, the transfer matrix ${\displaystyle \mathbf {M} }$ can be expressed by three real parameters:

$${\displaystyle M={\frac {1}{\sqrt {1-r^{2}}}}{\begin{pmatrix}e^{i\varphi }&-r\cdot e^{-i\delta }\\-r\cdot e^{i\delta }&e^{-i\varphi }\end{pmatrix}}}$$

with ${\displaystyle \delta ,\varphi \in [0;2\pi [}$ and ${\displaystyle r\in [0;1[}$ (in case r = 1 there would be no connection between the left and the right side)

The one-dimensional, non-relativistic problem with time-reversal symmetry of a particle with mass m that approaches a (static) finite square well, has the potential function V with

$${\displaystyle V(x)={\begin{cases}-V_{0}&{\text{for}}~~x\leq |a|~~(V_{0}>0)\quad {\text{and}}\\[1ex]0&{\text{for}}~~|x|>a\end{cases}}}$$

The scattering can be solved by decomposing the wave packet of the free particle into plane waves ${\displaystyle A_{k}\exp(ikx)}$ with wave numbers ${\displaystyle k>0}$ for a plane wave coming (faraway) from the left side or likewise ${\displaystyle D_{k}\exp(-ikx)}$ (faraway) from the right side.

The S-matrix for the plane wave with wave number k has the solution:[4]

$${\displaystyle S_{12}=S_{21}={\frac {\exp(-2ika)}{\cos(2la)-i\sin(2la){\frac {l^{2}+k^{2}}{2kl}}}}}$$

and ${\displaystyle S_{11}=S_{12}\cdot i\sin(2la){\frac {l^{2}-k^{2}}{2kl}}}$  ; hence ${\displaystyle e^{i\delta }=\pm i}$ and therefore ${\displaystyle -e^{-i\delta }=e^{i\delta }}$ and ${\displaystyle S_{22}=S_{11}}$ in this case.

Whereby ${\displaystyle l={\sqrt {k^{2}+{\frac {2mV_{0}}{\hbar ^{2}}}}}}$ is the (increased) wave number of the plane wave inside the square well, as the energy eigenvalue ${\displaystyle E_{k}}$ associated with the plane wave has to stay constant: ${\displaystyle E_{k}={\frac {\hbar ^{2}k^{2}}{2m}}={\frac {\hbar ^{2}l^{2}}{2m}}-V_{0}}$

The transmission is ${\displaystyle T_{k}=|S_{21}|^{2}=|S_{12}|^{2}={\frac {1}{(\cos(2la))^{2}+(\sin(2la))^{2}{\frac {(l^{2}+k^{2})^{2}}{4k^{2}l^{2}}}}}={\frac {1}{1+(\sin(2la))^{2}{\frac {(l^{2}-k^{2})^{2}}{4k^{2}l^{2}}}}}}$

In the case of ${\displaystyle \sin(2la)=0}$ then ${\displaystyle \cos(2la)=\pm 1}$ and therefore ${\displaystyle S_{11}=S_{22}=0}$ and ${\displaystyle |S_{21}|=|S_{12}|=1}$ i.e. a plane wave with wave number k passes the well without reflection if ${\displaystyle k^{2}+{\frac {2mV_{0}}{\hbar ^{2}}}={\frac {n^{2}\pi ^{2}}{4a^{2}}}}$ for a ${\displaystyle n\in \mathbb {N} }$

The square barrier is similar to the square well with the difference that ${\displaystyle V(x)=+V_{0}>0}$ for ${\displaystyle |x|\leq a}$.

There are three different cases depending on the energy eigenvalue ${\displaystyle E_{k}={\frac {\hbar ^{2}k^{2}}{2m}}}$ of the plane waves (with wave numbers k resp. −k) far away from the barrier:

The transmission coefficient from the left of the potential barrier is, when D = 0,

$${\displaystyle T_{\rm {L}}={\frac {|C|^{2}}{|A|^{2}}}=|S_{21}|^{2}.}$$

The reflection coefficient from the left of the potential barrier is, when D = 0,

$${\displaystyle R_{\rm {L}}={\frac {|B|^{2}}{|A|^{2}}}=|S_{11}|^{2}.}$$

Similarly, the transmission coefficient from the right of the potential barrier is, when A = 0,

$${\displaystyle T_{\rm {R}}={\frac {|B|^{2}}{|D|^{2}}}=|S_{12}|^{2}.}$$

The reflection coefficient from the right of the potential barrier is, when A = 0,

$${\displaystyle R_{\rm {R}}={\frac {|C|^{2}}{|D|^{2}}}=|S_{22}|^{2}.}$$

The relations between the transmission and reflection coefficients are

$${\displaystyle T_{\rm {L}}+R_{\rm {L}}=1}$$

and

$${\displaystyle T_{\rm {R}}+R_{\rm {R}}=1.}$$

This identity is a consequence of the unitarity property of the S-matrix.

With time-reversal symmetry, the S-matrix is symmetric and hence ${\displaystyle T_{\rm {L}}=|S_{21}|^{2}=|S_{12}|^{2}=T_{\rm {R}}}$ and ${\displaystyle R_{\rm {L}}=R_{\rm {R}}}$.

In the case of free particles V(x) = 0, the S-matrix is[6]

$${\displaystyle S={\begin{pmatrix}0&1\\1&0\end{pmatrix}}.}$$

Whenever V(x) is different from zero, however, there is a departure of the S-matrix from the above form, to

$${\displaystyle S={\begin{pmatrix}2ir&1+2it\\1+2it&2ir^{*}{\frac {1+2it}{1-2it^{*}}}\end{pmatrix}}.}$$

This departure is parameterized by two complex functions of energy, r and t. From unitarity there also follows a relationship between these two functions,

$${\displaystyle |r|^{2}+|t|^{2}=\operatorname {Im} (t).}$$

The analogue of this identity in three dimensions is known as the optical theorem.

A straightforward way to define the S-matrix begins with considering the interaction picture.[7] Let the Hamiltonian H be split into the free part H₀ and the interaction V, H = H₀ + V. In this picture, the operators behave as free field operators and the state vectors have dynamics according to the interaction V. Let

$${\displaystyle \left|\Psi (t)\right\rangle }$$

denote a state that has evolved from a free initial state

$${\displaystyle \left|\Phi _{\rm {i}}\right\rangle .}$$

The S-matrix element is then defined as the projection of this state on the final state

$${\displaystyle \left\langle \Phi _{\rm {f}}\right|.}$$

Thus

$${\displaystyle S_{\rm {fi}}\equiv \lim _{t\rightarrow +\infty }\left\langle \Phi _{\rm {f}}|\Psi (t)\right\rangle \equiv \left\langle \Phi _{\rm {f}}\right|S\left|\Phi _{\rm {i}}\right\rangle ,}$$

where S is the S-operator. The great advantage of this definition is that the time-evolution operator U evolving a state in the interaction picture is formally known,[8]

$${\displaystyle U(t,t_{0})=Te^{-i\int _{t_{0}}^{t}d\tau V(\tau )},}$$

where T denotes the time-ordered product. Expressed in this operator,

$${\displaystyle S_{\rm {fi}}=\lim _{t_{2}\rightarrow +\infty }\lim _{t_{1}\rightarrow -\infty }\left\langle \Phi _{\rm {f}}\right|U(t_{2},t_{1})\left|\Phi _{\rm {i}}\right\rangle ,}$$

from which

$${\displaystyle S=U(\infty ,-\infty ).}$$

Expanding using the knowledge about U gives a Dyson series,

$${\displaystyle S=\sum _{n=0}^{\infty }{\frac {(-i)^{n}}{n!}}\int _{-\infty }^{\infty }dt_{1}\cdots \int _{-\infty }^{\infty }dt_{n}T\left[V(t_{1})\cdots V(t_{n})\right],}$$

or, if V comes as a Hamiltonian density,

$${\displaystyle S=\sum _{n=0}^{\infty }{\frac {(-i)^{n}}{n!}}\int _{-\infty }^{\infty }dx_{1}^{4}\cdots \int _{-\infty }^{\infty }dx_{n}^{4}T\left[{\mathcal {H}}(t_{1})\cdots {\mathcal {H}}(t_{n})\right].}$$

Being a special type of time-evolution operator, S is unitary. For any initial state and any final state one finds

$${\displaystyle S_{\rm {fi}}=\left\langle \Phi _{\rm {f}}|S|\Phi _{\rm {i}}\right\rangle =\left\langle \Phi _{\rm {f}}\left|\sum _{n=0}^{\infty }{\frac {(-i)^{n}}{n!}}\int _{-\infty }^{\infty }dx_{1}^{4}\cdots \int _{-\infty }^{\infty }dx_{n}^{4}T\left[{\mathcal {H}}(t_{1})\cdots {\mathcal {H}}(t_{n})\right]\right|\Phi _{\rm {i}}\right\rangle .}$$

This approach is somewhat naïve in that potential problems are swept under the carpet.[9] This is intentional. The approach works in practice and some of the technical issues are addressed in the other sections.

Here a slightly more rigorous approach is taken in order to address potential problems that were disregarded in the interaction picture approach of above. The final outcome is, of course, the same as when taking the quicker route. For this, the notions of in and out states are needed. These will be developed in two ways, from vacua, and from free particle states. Needless to say, the two approaches are equivalent, but they illuminate matters from different angles.

If a^†(k) is a creation operator, its hermitian adjoint is an annihilation operator and destroys the vacuum,

$${\displaystyle a(k)\left|*,0\right\rangle =0.}$$

In Dirac notation, define

$${\displaystyle |*,0\rangle }$$

as a vacuum quantum state, i.e. a state without real particles. The asterisk signifies that not all vacua are necessarily equal, and certainly not equal to the Hilbert space zero state 0. All vacuum states are assumed Poincaré invariant, invariance under translations, rotations and boosts,[9] formally,

$${\displaystyle P^{\mu }|*,0\rangle =0,\quad M^{\mu \nu }|*,0\rangle =0}$$

where P^μ is the generator of translation in space and time, and M^μν is the generator of Lorentz transformations. Thus the description of the vacuum is independent of the frame of reference. Associated to the in and out states to be defined are the in and out field operators (aka fields) Φ_i and Φ_o. Attention is here focused to the simplest case, that of a scalar theory in order to exemplify with the least possible cluttering of the notation. The in and out fields satisfy

$${\displaystyle (\Box ^{2}+m^{2})\phi _{\rm {i,o}}(x)=0,}$$

the free Klein–Gordon equation. These fields are postulated to have the same equal time commutation relations (ETCR) as the free fields,

$${\displaystyle {\begin{aligned}{[\phi _{\rm {i,o}}(x),\pi _{\rm {i,o}}(y)]}_{x_{0}=y_{0}}&=i\delta (\mathbf {x} -\mathbf {y} ),\\{[\phi _{\rm {i,o}}(x),\phi _{\rm {i,o}}(y)]}_{x_{0}=y_{0}}&={[\pi _{\rm {i,o}}(x),\pi _{\rm {i,o}}(y)]}_{x_{0}=y_{0}}=0,\end{aligned}}}$$

where π_i,j is the field canonically conjugate to Φ_i,j. Associated to the in and out fields are two sets of creation and annihilation operators, a^†_i(k) and a^†_f (k), acting in the same Hilbert space,[10] on two distinct complete sets (Fock spaces; initial space i, final space f). These operators satisfy the usual commutation rules,

$${\displaystyle {\begin{aligned}{[a_{\rm {i,o}}(\mathbf {p} ),a_{\rm {i,o}}^{\dagger }(\mathbf {p} ')]}&=i\delta (\mathbf {p} -\mathbf {p'} ),\\{[a_{\rm {i,o}}(\mathbf {p} ),a_{\rm {i,o}}(\mathbf {p'} )]}&={[a_{\rm {i,o}}^{\dagger }(\mathbf {p} ),a_{\rm {i,o}}^{\dagger }(\mathbf {p'} )]}=0.\end{aligned}}}$$

The action of the creation operators on their respective vacua and states with a finite number of particles in the in and out states is given by

$${\displaystyle {\begin{aligned}\left|\mathrm {i} ,k_{1}\ldots k_{n}\right\rangle &=a_{i}^{\dagger }(k_{1})\cdots a_{\rm {i}}^{\dagger }(k_{n})\left|i,0\right\rangle ,\\\left|\mathrm {f} ,p_{1}\ldots p_{n}\right\rangle &=a_{\rm {f}}^{\dagger }(p_{1})\cdots a_{f}^{\dagger }(p_{n})\left|f,0\right\rangle ,\end{aligned}}}$$

where issues of normalization have been ignored. See the next section for a detailed account on how a general n-particle state is normalized. The initial and final spaces are defined by

$${\displaystyle {\mathcal {H}}_{\rm {i}}=\operatorname {span} \{\left|\mathrm {i} ,k_{1}\ldots k_{n}\right\rangle =a_{\rm {i}}^{\dagger }(k_{1})\cdots a_{\rm {i}}^{\dagger }(k_{n})\left|\mathrm {i} ,0\right\rangle \},}$$

$${\displaystyle {\mathcal {H}}_{\rm {f}}=\operatorname {span} \{\left|\mathrm {f} ,p_{1}\ldots p_{n}\right\rangle =a_{\rm {f}}^{\dagger }(p_{1})\cdots a_{\rm {f}}^{\dagger }(p_{n})\left|\mathrm {f} ,0\right\rangle \}.}$$

The asymptotic states are assumed to have well defined Poincaré transformation properties, i.e. they are assumed to transform as a direct product of one-particle states.[11] This is a characteristic of a non-interacting field. From this follows that the asymptotic states are all eigenstates of the momentum operator P^μ,[9]

$${\displaystyle P^{\mu }\left|\mathrm {i} ,k_{1}\ldots k_{m}\right\rangle =k_{1}^{\mu }+\cdots +k_{m}^{\mu }\left|\mathrm {i} ,k_{1}\ldots k_{m}\right\rangle ,\quad P^{\mu }\left|\mathrm {f} ,p_{1}\ldots p_{n}\right\rangle =p_{1}^{\mu }+\cdots +p_{n}^{\mu }\left|\mathrm {f} ,p_{1}\ldots p_{n}\right\rangle .}$$

In particular, they are eigenstates of the full Hamiltonian,

$${\displaystyle H=P^{0}.}$$

The vacuum is usually postulated to be stable and unique,[9][nb 1]

$${\displaystyle |\mathrm {i} ,0\rangle =|\mathrm {f} ,0\rangle =|*,0\rangle \equiv |0\rangle .}$$

The interaction is assumed adiabatically turned on and off.

The Heisenberg picture is employed henceforth. In this picture, the states are time-independent. A Heisenberg state vector thus represents the complete spacetime history of a system of particles.[11] The labeling of the in and out states refers to the asymptotic appearance. A state Ψ_{α, in} is characterized by that as t → −∞ the particle content is that represented collectively by α. Likewise, a state Ψ_{β, out} will have the particle content represented by β for t → +∞. Using the assumption that the in and out states, as well as the interacting states, inhabit the same Hilbert space and assuming completeness of the normalized in and out states (postulate of asymptotic completeness[9]), the initial states can be expanded in a basis of final states (or vice versa). The explicit expression is given later after more notation and terminology has been introduced. The expansion coefficients are precisely the S-matrix elements to be defined below.

While the state vectors are constant in time in the Heisenberg picture, the physical states they represent are not. If a system is found to be in a state Ψ at time t = 0, then it will be found in the state U(τ)Ψ = e^−iHτΨ at time t = τ. This is not (necessarily) the same Heisenberg state vector, but it is an equivalent state vector, meaning that it will, upon measurement, be found to be one of the final states from the expansion with nonzero coefficient. Letting τ vary one sees that the observed Ψ (not measured) is indeed the Schrödinger picture state vector. By repeating the measurement sufficiently many times and averaging, one may say that the same state vector is indeed found at time t = τ as at time t = 0. This reflects the expansion above of an in state into out states.

For this viewpoint, one should consider how the archetypical scattering experiment is performed. The initial particles are prepared in well defined states where they are so far apart that they don't interact. They are somehow made to interact, and the final particles are registered when they are so far apart that they have ceased to interact. The idea is to look for states in the Heisenberg picture that in the distant past had the appearance of free particle states. This will be the in states. Likewise, an out state will be a state that in the distant future has the appearance of a free particle state.[11]

The notation from the general reference for this section, Weinberg (2002) will be used. A general non-interacting multi-particle state is given by

$${\displaystyle \Psi _{p_{1}\sigma _{1}n_{1};p_{2}\sigma _{2}n_{2};\cdots },}$$

where

• p is momentum,
• σ is spin z-component or, in the massless case, helicity,
• n is particle species.

These states are normalized as

$${\displaystyle \left(\Psi _{p_{1}'\sigma _{1}'n_{1}';p_{2}'\sigma _{2}'n_{2}';\cdots },\Psi _{p_{1}\sigma _{1}n_{1};p_{2}\sigma _{2}n_{2};\cdots }\right)=\delta ^{3}(\mathbf {p} _{1}'-\mathbf {p} _{1})\delta _{\sigma _{1}'\sigma _{1}}\delta _{n_{1}'n_{1}}\delta ^{3}(\mathbf {p} _{2}'-\mathbf {p} _{2})\delta _{\sigma _{2}'\sigma _{2}}\delta _{n_{2}'n_{2}}\cdots \quad \pm {\text{ permutations}}.}$$

Permutations work as such; if s ∈ S_k is a permutation of k objects (for a k-particle state) such that

$${\displaystyle n_{s(i)}'=n_{i},\quad 1\leq i\leq k,}$$

then a nonzero term results. The sign is plus unless s involves an odd number of fermion transpositions, in which case it is minus. The notation is usually abbreviated letting one Greek letter stand for the whole collection describing the state. In abbreviated form the normalization becomes

$${\displaystyle \left(\Psi _{\alpha '},\Psi _{\alpha }\right)=\delta (\alpha '-\alpha ).}$$

When integrating over free-particle states one writes in this notation

$${\displaystyle d\alpha \cdots \equiv \sum _{n_{1}\sigma _{1}n_{2}\sigma _{2}\cdots }\int d^{3}p_{1}d^{3}p_{2}\cdots ,}$$

where the sum includes only terms such that no two terms are equal modulo a permutation of the particle type indices. The sets of states sought for are supposed to be complete. This is expressed as

$${\displaystyle \Psi =\int d\alpha \ \Psi _{\alpha }\left(\Psi _{\alpha },\Psi \right),}$$

which could be paraphrased as

$${\displaystyle \int d\alpha \ \left|\Psi _{\alpha }\right\rangle \left\langle \Psi _{\alpha }\right|=1,}$$

where for each fixed α, the right hand side is a projection operator onto the state α. Under an inhomogeneous Lorentz transformation (Λ, a), the field transforms according to the rule

$${\displaystyle U(\Lambda ,a)\Psi _{p_{1}\sigma _{1}n_{1};p_{2}\sigma _{2}n_{2}\cdots }=e^{-ia_{\mu }((\Lambda p_{1})^{\mu }+(\Lambda p_{2})^{\mu }+\cdots )}{\sqrt {\frac {(\Lambda p_{1})^{0}(\Lambda p_{2})^{0}\cdots }{p_{1}^{0}p_{2}^{0}\cdots }}}\sum _{\sigma _{1}'\sigma _{2}'\cdots }D_{\sigma _{1}'\sigma _{1}}^{(j_{1})}(W(\Lambda ,p_{1}))D_{\sigma _{2}'\sigma _{2}}^{(j_{2})}(W(\Lambda ,p_{2}))\cdots \Psi _{\Lambda p_{1}\sigma _{1}'n_{1};\Lambda p_{2}\sigma _{2}'n_{2}\cdots },}$$

(1)

where W(Λ, p) is the Wigner rotation and D^(j) is the (2j + 1)-dimensional representation of SO(3). By putting Λ = 1, a = (τ, 0, 0, 0), for which U is exp(iHτ), in (1), it immediately follows that

$${\displaystyle H\Psi =E_{\alpha }\Psi ,\quad E_{\alpha }=p_{1}^{0}+p_{2}^{0}+\cdots ,}$$

so the in and out states sought after are eigenstates of the full Hamiltonian that are necessarily non-interacting due to the absence of mixed particle energy terms. The discussion in the section above suggests that the in states Ψ⁺ and the out states Ψ⁻ should be such that

$${\displaystyle e^{-iH\tau }\int d\alpha \ g(\alpha )\Psi _{\alpha }^{\pm }=\int d\alpha \ e^{-iE_{\alpha }\tau }g(\alpha )\Psi _{\alpha }^{\pm }}$$

for large positive and negative τ has the appearance of the corresponding package, represented by g, of free-particle states, g assumed smooth and suitably localized in momentum. Wave packages are necessary, else the time evolution will yield only a phase factor indicating free particles, which cannot be the case. The right hand side follows from that the in and out states are eigenstates of the Hamiltonian per above. To formalize this requirement, assume that the full Hamiltonian H can be divided into two terms, a free-particle Hamiltonian H₀ and an interaction V, H = H₀ + V such that the eigenstates Φ_γ of H₀ have the same appearance as the in- and out-states with respect to normalization and Lorentz transformation properties,

$${\displaystyle H_{0}\Phi _{\alpha }=E_{\alpha }\Phi _{\alpha },}$$

$${\displaystyle (\Phi _{\alpha }',\Phi _{\alpha })=\delta (\alpha '-\alpha ).}$$

The in and out states are defined as eigenstates of the full Hamiltonian,

$${\displaystyle H\Psi _{\alpha }^{\pm }=E_{\alpha }\Psi _{\alpha }^{\pm },}$$

satisfying

$${\displaystyle e^{-iH\tau }\int d\alpha \ g(\alpha )\Psi _{\alpha }^{\pm }\rightarrow e^{-iH_{0}\tau }\int d\alpha \ g(\alpha )\Phi _{\alpha }.}$$

for τ → −∞ or τ → +∞ respectively. Define

$${\displaystyle \Omega (\tau )\equiv e^{+iH\tau }e^{-iH_{0}\tau },}$$

then

$${\displaystyle \Psi _{\alpha }^{\pm }=\Omega (\mp \infty )\Phi _{\alpha }.}$$

This last expression will work only using wave packages.From these definitions follow that the in and out states are normalized in the same way as the free-particle states,

$${\displaystyle (\Psi _{\beta }^{+},\Psi _{\alpha }^{+})=(\Phi _{\beta },\Phi _{\alpha })=(\Psi _{\beta }^{-},\Psi _{\alpha }^{-})=\delta (\beta -\alpha ),}$$

and the three sets are unitarily equivalent. Now rewrite the eigenvalue equation,

$${\displaystyle (E_{\alpha }-H_{0}\pm i\epsilon )\Psi _{\alpha }^{\pm }=\pm i\epsilon \Psi _{\alpha }^{\pm }+V\Psi _{\alpha }^{\pm },}$$

where the ±iε terms has been added to make the operator on the LHS invertible. Since the in and out states reduce to the free-particle states for V → 0, put

$${\displaystyle i\epsilon \Psi _{\alpha }^{\pm }=i\epsilon \Phi _{\alpha }}$$

on the RHS to obtain

$${\displaystyle \Psi _{\alpha }^{\pm }=\Phi _{\alpha }+(E_{\alpha }-H_{0}\pm i\epsilon )^{-1}V\Psi _{\alpha }^{\pm }.}$$

Then use the completeness of the free-particle states,

$${\displaystyle V\Psi _{\alpha }^{\pm }=\int d\beta \ (\Phi _{\beta },V\Psi _{\alpha }^{\pm })\Phi _{\beta }\equiv \int d\beta \ T_{\beta \alpha }^{\pm }\Phi _{\beta },}$$

to finally obtain

$${\displaystyle \Psi _{\alpha }^{\pm }=\Phi _{\alpha }+\int d\beta \ {\frac {T_{\beta \alpha }^{\pm }\Phi _{\beta }}{E_{\alpha }-E_{\beta }\pm i\epsilon }}.}$$

Here H₀ has been replaced by its eigenvalue on the free-particle states. This is the Lippmann–Schwinger equation.

The initial states can be expanded in a basis of final states (or vice versa). Using the completeness relation,

$${\displaystyle \Psi _{\alpha }^{-}=\int d\beta (\Psi _{\beta }^{+},\Psi _{\alpha }^{-})\Psi _{\beta }^{+}=\int d\beta |\Psi _{\beta }^{+}\rangle \langle \Psi _{\beta }^{+}|\Psi _{\alpha }^{-}\rangle =\sum _{n_{1}\sigma _{1}n_{2}\sigma _{2}\cdots }\int d^{3}p_{1}d^{3}p_{2}\cdots (\Psi _{\beta }^{+},\Psi _{\alpha }^{-})\Psi _{\beta }^{+},}$$

$${\displaystyle \Psi _{\alpha }^{-}=\left|\mathrm {i} ,k_{1}\ldots k_{n}\right\rangle =C_{0}\left|\mathrm {f} ,0\right\rangle \ +\sum _{m=1}^{\infty }\int {d^{4}p_{1}\ldots d^{4}p_{m}C_{m}(p_{1}\ldots p_{m})\left|\mathrm {f} ,p_{1}\ldots p_{m}\right\rangle }~,}$$

where |C_m|² is the probability that the interaction transforms

$${\displaystyle \left|\mathrm {i} ,k_{1}\ldots k_{n}\right\rangle =\Psi _{\alpha }^{-}}$$

into

$${\displaystyle \left|\mathrm {f} ,p_{1}\ldots p_{m}\right\rangle =\Psi _{\beta }^{+}.}$$

By the ordinary rules of quantum mechanics,

$${\displaystyle C_{m}(p_{1}\ldots p_{m})=\left\langle \mathrm {f} ,p_{1}\ldots p_{m}\right|\mathrm {i} ,k_{1}\ldots k_{n}\rangle =(\Psi _{\beta }^{+},\Psi _{\alpha }^{-})}$$

and one may write

$${\displaystyle \left|\mathrm {i} ,k_{1}\ldots k_{n}\right\rangle =C_{0}\left|\mathrm {f} ,0\right\rangle \ +\sum _{m=1}^{\infty }\int {d^{4}p_{1}\ldots d^{4}p_{m}\left|\mathrm {f} ,p_{1}\ldots p_{m}\right\rangle }\left\langle \mathrm {f} ,p_{1}\ldots p_{m}\right|\mathrm {i} ,k_{1}\ldots k_{n}\rangle ~.}$$

The expansion coefficients are precisely the S-matrix elements to be defined below.

The S-matrix is now defined by[11]

$${\displaystyle S_{\beta \alpha }=\langle \Psi _{\beta }^{-}|\Psi _{\alpha }^{+}\rangle =\langle \mathrm {f} ,\beta |\mathrm {i} ,\alpha \rangle ,\qquad |\mathrm {f} ,\beta \rangle \in {\mathcal {H}}_{\rm {f}},\quad |\mathrm {i} ,\alpha \rangle \in {\mathcal {H}}_{\rm {i}}.}$$

Here α and β are shorthands that represent the particle content but suppresses the individual labels. Associated to the S-matrix there is the S-operator S defined by[11]

$${\displaystyle \langle \Phi _{\beta }|S|\Phi _{\alpha }\rangle \equiv S_{\beta \alpha },}$$

where the Φ_γ are free particle states.[11][nb 2] This definition conforms with the direct approach used in the interaction picture. Also, due to unitary equivalence,

$${\displaystyle \langle \Psi _{\beta }^{+}|S|\Psi _{\alpha }^{+}\rangle =S_{\beta \alpha }=\langle \Psi _{\beta }^{-}|S|\Psi _{\alpha }^{-}\rangle .}$$

As a physical requirement, S must be a unitary operator. This is a statement of conservation of probability in quantum field theory. But

$${\displaystyle \langle \Psi _{\beta }^{-}|S|\Psi _{\alpha }^{-}\rangle =S_{\beta \alpha }=\langle \Psi _{\beta }^{-}|\Psi _{\alpha }^{+}\rangle .}$$

By completeness then,

$${\displaystyle S|\Psi _{\alpha }^{-}\rangle =|\Psi _{\alpha }^{+}\rangle ,}$$

so S is the unitary transformation from in-states to out states. Lorentz invariance is another crucial requirement on the S-matrix.[11][nb 3] The S-operator represents the quantum canonical transformation of the initial in states to the final out states. Moreover, S leaves the vacuum state invariant and transforms in-space fields to out-space fields,[nb 4]

$${\displaystyle S\left|0\right\rangle =\left|0\right\rangle }$$

$${\displaystyle \phi _{\mathrm {f} }=S\phi _{\mathrm {i} }S^{-1}~.}$$

In terms of creation and annihilation operators, this becomes

$${\displaystyle a_{\rm {f}}(p)=Sa_{\rm {i}}(p)S^{-1},a_{\rm {f}}^{\dagger }(p)=Sa_{\rm {i}}^{\dagger }(p)S^{-1},}$$

hence

$${\displaystyle {\begin{aligned}S|\mathrm {i} ,k_{1},k_{2},\ldots ,k_{n}\rangle &=Sa_{\rm {i}}^{\dagger }(k_{1})a_{\rm {i}}^{\dagger }(k_{2})\cdots a_{\rm {i}}^{\dagger }(k_{n})|0\rangle =Sa_{\rm {i}}^{\dagger }(k_{1})S^{-1}Sa_{\rm {i}}^{\dagger }(k_{2})S^{-1}\cdots Sa_{\rm {i}}^{\dagger }(k_{n})S^{-1}S|0\rangle \\[1ex]&=a_{\rm {o}}^{\dagger }(k_{1})a_{\rm {o}}^{\dagger }(k_{2})\cdots a_{\rm {o}}^{\dagger }(k_{n})S|0\rangle =a_{\rm {o}}^{\dagger }(k_{1})a_{\rm {o}}^{\dagger }(k_{2})\cdots a_{\rm {o}}^{\dagger }(k_{n})|0\rangle =|\mathrm {o} ,k_{1},k_{2},\ldots ,k_{n}\rangle .\end{aligned}}}$$

A similar expression holds when S operates to the left on an out state. This means that the S-matrix can be expressed as

$${\displaystyle S_{\beta \alpha }=\langle \mathrm {o} ,\beta |\mathrm {i} ,\alpha \rangle =\langle \mathrm {i} ,\beta |S|\mathrm {i} ,\alpha \rangle =\langle \mathrm {o} ,\beta |S|\mathrm {o} ,\alpha \rangle .}$$

If S describes an interaction correctly, these properties must be also true:

• If the system is made up with a single particle in momentum eigenstate |k⟩, then S|k⟩ = |k⟩. This follows from the calculation above as a special case.
• The S-matrix element may be nonzero only where the output state has the same total momentum as the input state. This follows from the required Lorentz invariance of the S-matrix.