S Matrix

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In Quantum field theory, the S Matrix is an operator mapping free particle in-states to free particle out-states in the Heisenberg picture. This is very userful because we cannot describe exactly the interaction (at least, the most interesting ones). In Dirac notation, we define \left |0\right\rangle as void Quantum state. If a^{\dagger}(k) is a creation operator, its hermitian conjugate (destruction operator) acts on the void as follows:

a(K)\left |0\right\rangle = 0

Now, we define two kind of creation/destruction operators, acting on different Hilbert spaces (IN space i, OUT space f), a_i^\dagger (k) and a_f^\dagger (k).

So now:

\mathcal H_{IN} = Span\{ \left| I, k_1...k_n \right\rangle = a_i^\dagger (k_1)...a_i^\dagger (k_n)\left| I, 0\right\rangle\} \mathcal H_{OUT} = Span\{ \left| F, p_1...p_n \right\rangle = a_f^\dagger (p_1)...a_f^\dagger (p_n)\left| F, 0\right\rangle\}

It is possible to prove that \left| I, 0\right\rangle and \left| F, 0\right\rangle are both invariant under translation and that the states \left| I, k_1...k_n \right\rangle and \left| F, p_1...p_n \right\rangle are eingenstates of \mathcal P^\mu operator (momentum). In Heisenberg picture the states are time-indipendent, so we can expand initial states on a basis of final states (or vice versa) as following:

\left| I, k_1...k_n \right\rangle = C_0 + \sum_{m=1}^\infty \int{d^4p_1...d^4p_mC_m(p_1...p_m)\left| F, p_1...p_n \right\rangle}

Where | Cm | 2 is the probability that the interaction transforms \left| I, k_1...k_n \right\rangle in \left| F, p_1...p_n \right\rangle

According to Wigner's theorem, S must be an unitary operator such that \left \langle I,\beta \right |S\left | I,\alpha\right\rangle = S_{\alpha\beta} = \left \langle F,\beta | I,\alpha\right\rangle. Moreover, S leaves the void invariant and transforms IN-space fields in OUT-space fields:

S\left|0\right\rangle = \left|0\right\rangle

φf = S − 1φfS

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 \left| k\right\rangle, then S\left| k\right\rangle=\left| k\right\rangle

The S-Matrix element must be non zero if and only if momentum is conserved.

S-Matrix and Evolution Operator U

a(k,t) = U − 1(t)ai(k)U(t)

\phi_f=U^{-1}(\infty)\phi_i U(\infty)=S^{-1}\phi_i S

So we have S=e^{i\alpha}U(\infty) where e^{i\alpha}=\left\langle 0|U(\infty)|0\right\rangle because S\left|0\right\rangle = \left|0\right\rangle. Substituting the explicit expression for U we obtain:

S=\frac{1}{\left\langle 0|U(\infty)|0\right\rangle}\mathcal T e^{-i\int{d\tau V_i(\tau)}}

You can see that this formula is not explicitly covariant.

L.S.Z. (Lehman, Symanzik, Zimmermann) Reduction Formula

F_n(x_1...x_n)=\left\langle 0|\mathcal T\phi(x_1)...\phi(x_n)|0\right\rangle

The task is to find an expression for the S-Matrix element using the reduction formula. Before starting to accomplish this, it is useful to show the following trick:

\left(\lim_{x_0\to \infty} - \lim_{x_0 \to \infty}\right)f^*\partial_0^{\leftrightarrow}\phi=\int_{-\infty}^{\infty}{dx_0\left( f^*\ddot\phi-\ddot f\phi*\right)}

\lim_{t_1,t_2\to\infty}\int_{t_1}^{t_2}{d\tau\frac{\partial}{\partial t}\int{d^3x\psi(x,t)}}=\left(\lim_{x_0\to \infty} - \lim_{x_0 \to \infty}\right)\int{d^3x\psi(x,t)}

We will use this in the following calculation:

S_{fi}=\left \langle F,k_1, k_2 | I,p_1,p_2\right\rangle=\left \langle F,k_1, k_2 | a_i^\dagger(p_2)|I,p_1\right\rangle

This operation is called particle extraction.

=\left \langle F,k_1, k_2 | a_i^\dagger(p_2)-a_f^\dagger(p_2)|I,p_1\right\rangle

This is true because p is not equal to k.

=-i\int{d^3xf^*(p_2,x)\partial_0^\leftrightarrow \left \langle F,k_1, k_2 | \phi_i(x)-\phi_f(x)|I,p_1\right\rangle}
=i\left(\lim_{t\to \infty} - \lim_{t \to \infty}\right)\int{d^3xf^*(p_2,t)\partial_0^\leftrightarrow \left \langle F,k_1, k_2 | \phi(x)|I,p_1\right\rangle}
=i\int{d^4x\left \langle F,k_1, k_2 | f*\ddot \phi - \ddot f^*\phi|I,p_1\right\rangle}

Remebering that f functions are solutions of Klein-Gordon equation:

\left( \Box + m^2 \right ) f^*=0=\ddot f^* - \nabla^2 f^* + m^2 f^* \Rightarrow \ddot f^*=\left( \nabla^2-m^2\right)f^*

Substituting this in previous equation we get (integrating by parts two times):

S_{fi}=i\int{d^4xf^*(p_2,x)\left(\Box_x+m^2\right )\left \langle F,k_1, k_2 | \phi(x)|I,p_1\right\rangle}

Now we repeat these operations for all the particle in the system, and finally we get:

S_{fi}=(i)^4\int{d^4x_1d^4x_2d^4y_1d^4y_2f^*(p_1,x_1)f^*(p_2,x_2)f(k_1,y_1)f(k_2,y_2)\left(\Box_{x_1}+m^2\right )\left(\Box_{x_2}+m^2\right )\left(\Box_{y_1}+m^2\right )\left(\Box_{y_2}+m^2\right )\left \langle 0|\mathcal T\phi(x_1)\phi(x_2)\phi(y_1)\phi(y_2)|0\right\rangle}

This is, of course, the simplest case with only four interacting particles.

Now we Fourier transform (it is not exactly a Fourier transformation) the reduction formula F and we get:

f_{mn}(q_1...1_{m+n})=\int{d^4x_1...d^4x_nd^4y_1...d^4y_m\frac{e^{-iq_1x_1}}{\sqrt{(2\pi)^32\omega_k}}...\frac{e^{-iq_{n+m}x_{n+m}}}{\sqrt{(2\pi)^32\omega_k}}F_{nm}(x_1...x_n,y_1...y_m}

There is a theorem that states (proof omitted) that the S-Matrix elements are the residuals of f calculated on mass-shell:

S_{fi}=(i)^{n+m}\lim_{q_i\to m^2}(m^2-q_1)...(m^2-q_{n+m})f_{nm}(q_1...1_{n+m})

The matter is that we do not have an explicit expression for φ(x), so we have to make a perturbative expansion with φi(x)

In the end, we obtain:

F_p(x)=\left \langle 0 |\mathcal T\phi(x_1)...\phi(x_p)|\right \rangle=\frac{\left \langle 0 |\mathcal T e^{-i\int{d\tau V_i(\tau)\phi_i(x_1)...\phi_i(x_p)}}|\right \rangle}{\left \langle 0 |e^{-i\int{d\tau V_i(\tau)}}|\right \rangle}

Wick's Theorem

Def Contraction

\mathcal C(x_1, x_2)=\left \langle 0 |\mathcal T\phi_i(x_1)\phi_i(x_2)|0\right \rangle=\overline{\phi_i(x_1)\phi_i(x_2)}=i\Delta_F(x_1-x_2)=i\int{\frac{d^4k}{(2\pi)^4}\frac{e^{-ik(x_1-x_2}}{(k^2-m^2)+i\epsilon}}

Which means that \overline{AB}=\mathcal TAB-:AB:

In the end, we approach at Wick's theorem:

T Wick's theorem

The T-product of a time-ordered free fields string can be expressed in the following manner:

\mathcal T\Pi_{k=1}^m\phi(x_k)=:\Pi\phi_i(x_k):+\sum_{\alpha,\beta}\overline{\phi(x_\alpha)\phi(x_\beta)}:\Pi_{k\not=\alpha,\beta}\phi_i(x_k):+\sum_{(\alpha,\beta),(\gamma,\delta)}\overline{\phi(x_\alpha)\phi(x_\beta)}\;\overline{\phi(x_\gamma)\phi(x_\delta)}:\Pi_{k\not=\alpha,\beta,\gamma,\delta}\phi_i(x_k):+...

Applying this theorem to S-Matrix elements, we discover that normal-ordered terms acting on void state give a null contribute to the sum. We conculde that m is even and only completely contracted terms remain.

F_m^i(x)=\left \langle 0 |\mathcal T\phi_i(x_1)\phi_i(x_2)|0\right \rangle=\sum_{pairs}\overline{\phi(x_1)\phi(x_2)}...\overline{\phi(x_{m-1})\phi(x_m})

G_p^{(n)}=\left \langle 0 |\mathcal T:v_i(y_1):...:v_i(y_n):\phi_i(x_1)...\phi_i(x_p)|0\right \rangle

Where p is the number of interaction fields (or, equivalently, the number of interacting particles) and n is the develpment order (or the number of vertices of interaction). For example, if v=gy^4 \Rightarrow :v_i(y_1):=:\phi_i(y_1)\phi_i(y_1)\phi_i(y_1)\phi_i(y_1):

See also: Feynman diagrams

See also: S Matrix, Dirac notation, Feynman diagrams, Heisenberg picture, Hilbert space, Klein-Gordon equation, Operator (physics), Quantum field theory