In quantum field theory, the Lehmann–Symanzik–Zimmermann (LSZ) reduction formula is a method to calculate S-matrix elements (the scattering amplitudes) from the time-ordered correlation functions of a quantum field theory. It is a step of the path that starts from the Lagrangian of some quantum field theory and leads to prediction of measurable quantities. It is named after the three German physicists Harry Lehmann, Kurt Symanzik and Wolfhart Zimmermann.[1]
Although the LSZ reduction formula cannot handle bound states, massless particles and topological solitons, it can be generalized to cover bound states, by use of composite fields which are often nonlocal. Furthermore, the method, or variants thereof, have turned out to be also fruitful in other fields of theoretical physics. For example, in statistical physics they can be used to get a particularly general formulation of the fluctuation-dissipation theorem.
S-matrix elements are amplitudes of transitions between in states and out states.[2] [3] [4] [5] [6] An in state
|\{p\} in\rangle
|\{p\} out\rangle
In and out states are states in Heisenberg picture so they should not be thought to describe particles at a definite time, but rather to describe the system of particles in its entire evolution, so that the S-matrix element:
S\rm=\langle\{q\} out|\{p\} in\rangle
is the probability amplitude for a set of particles which were prepared with definite momenta to interact and be measured later as a new set of particles with momenta
The easy way to build in and out states is to seek appropriate field operators that provide the right creation and annihilation operators. These fields are called respectively in and out fields:
Just to fix ideas, suppose we deal with a Klein–Gordon field that interacts in some way which doesn't concern us:
lL=
1 | |
2 |
\partial\mu\varphi\partial\mu\varphi-
1 | |
2 |
2 | |
m | |
0 |
\varphi2+lLint
lLint
g \varphi\bar\psi\psi
2\right)\varphi(x)=j | |
\left(\partial | |
0(x) |
where, if
lLint
j | ||||
|
We may expect the in field to resemble the asymptotic behaviour of the free field as, making the assumption that in the far away past interaction described by the current is negligible, as particles are far from each other. This hypothesis is named the adiabatic hypothesis. However self interaction never fades away and, besides many other effects, it causes a difference between the Lagrangian mass and the physical mass of the boson. This fact must be taken into account by rewriting the equation of motion as follows:
\left(\partial2+m
2\right)\varphi(x)=j(x) | |
0 |
This equation can be solved formally using the retarded Green's function of the Klein–Gordon operator
\partial2+m2
\Deltaret(x)=i\theta\left(x0\right)\int
d3k | |
(2\pi)32\omegak |
\left(e-ik ⋅ -eik ⋅
\right) | |||||||
|
2+m | |
\omega | |
k=\sqrt{k |
2}
allowing us to split interaction from asymptotic behaviour. The solution is:
\varphi(x)=\sqrtZ\varphiin(x)+\intd4y\Deltaret(x-y)j(y)
The factor is a normalization factor that will come handy later, the field is a solution of the homogeneous equation associated with the equation of motion:
\left(\partial2+m2\right)\varphiin(x)=0,
and hence is a free field which describes an incoming unperturbed wave, while the last term of the solution gives the perturbation of the wave due to interaction.
The field is indeed the in field we were seeking, as it describes the asymptotic behaviour of the interacting field as, though this statement will be made more precise later. It is a free scalar field so it can be expanded in plane waves:
\varphiin(x)=\intd3k\left\{fk(x)ain
* | |
(k)+f | |
k(x) |
\dagger | |
a | |
in |
(k)\right\}
where:
f | ||||||||||||||||||||||
|
\right| | |||||||
|
The inverse function for the coefficients in terms of the field can be easily obtained and put in the elegant form:
ain(k)=i\intd3x
* | |
f | |
k(x)\overleftrightarrow{\partial |
0}\varphiin(x)
where:
{g
The Fourier coefficients satisfy the algebra of creation and annihilation operators:
[ain(p),ain(q)]=0; [ain
\dagger | |
(p),a | |
in |
(q)]=\delta3(p-q);
and they can be used to build in states in the usual way:
\left|k1,\ldots,kn in\right\rangle=\sqrt{2\omega
k1 |
The relation between the interacting field and the in field is not very simple to use, and the presence of the retarded Green's function tempts us to write something like:
\varphi(x)\sim\sqrtZ\varphiin(x) as x0\to-infty
implicitly making the assumption that all interactions become negligible when particles are far away from each other. Yet the current contains also self interactions like those producing the mass shift from to . These interactions do not fade away as particles drift apart, so much care must be used in establishing asymptotic relations between the interacting field and the in field.
The correct prescription, as developed by Lehmann, Symanzik and Zimmermann, requires two normalizable states
|\alpha\rangle
|\beta\rangle
(\partial2+m2)f(x)=0
\lim | |
x0\to-infty |
\intd3x\langle\alpha|f(x)\overleftrightarrow{\partial0}\varphi(x)|\beta\rangle=\sqrtZ\intd3x\langle\alpha|f(x)\overleftrightarrow{\partial0}\varphiin(x)|\beta\rangle
The second member is indeed independent of time as can be shown by differentiating and remembering that both and satisfy the Klein–Gordon equation.
With appropriate changes the same steps can be followed to construct an out field that builds out states. In particular the definition of the out field is:
\varphi(x)=\sqrtZ\varphiout(x)+\intd4y\Deltaadv(x-y)j(y)
where is the advanced Green's function of the Klein–Gordon operator. The weak asymptotic relation between out field and interacting field is:
\lim | |
x0\toinfty |
\intd3x\langle\alpha|f(x)\overleftrightarrow{\partial0}\varphi(x)|\beta\rangle=\sqrtZ\int
3x \langle\alpha|f(x)\overleftrightarrow{\partial | |
d | |
0}\varphi |
out(x)|\beta\rangle
The asymptotic relations are all that is needed to obtain the LSZ reduction formula. For future convenience we start with the matrix element:
lM=\langle\beta out|T\varphi(y1)\ldots\varphi(yn)|\alpha in\rangle
which is slightly more general than an S-matrix element. Indeed,
lM
\varphi(y1) … \varphi(yn)
lM
lM=\sqrt{2\omegap} \left\langle\beta out|T\left[\varphi(y1)\ldots\varphi(yn)\right]
\dagger(p)|\alpha' in | |
a | |
in |
\right\rangle
where the prime on
\alpha
lM=\sqrt{2\omegap} \left\langle\beta out|\left\{T\left[\varphi(y1)\ldots\varphi(yn)\right]
\dagger | |
a | |
in |
\dagger(p)T\left[\varphi(y | |
(p)-a | |
1)\ldots\varphi(y |
n)\right]\right\} |\alpha' in\right\rangle
because
\dagger | |
a | |
out |
lM=-i\sqrt{2\omegap} \intd3xfp(x)\overleftrightarrow{\partial0}\left\langle\beta out|\left\{ T\left[\varphi(y1)\ldots\varphi(yn)\right]\varphiin(x)-\varphiout(x)T\left[\varphi(y1)\ldots\varphi(yn)\right]\right\} |\alpha' in\right\rangle
Now we can use the asymptotic condition to write:
lM=-i\sqrt{
2\omegap | |
Z |
Then we notice that the field can be brought inside the time-ordered product, since it appears on the right when and on the left when :
lM=-i\sqrt{ | 2\omegap |
Z |
In the following, dependence in the time-ordered product is what matters, so we set:
\langle\beta out|T\left[\varphi(x)\varphi(y1)\ldots\varphi(yn)\right]|\alpha' in\rangle=η(x)
It can be shown by explicitly carrying out the time integration that:
lM=i\sqrt{ | 2\omegap |
Z |
so that, by explicit time derivation, we have:
lM=i\sqrt{ | 2\omegap |
Z |
By its definition we see that is a solution of the Klein–Gordon equation, which can be written as:
2\right) | |
\partial | |
p(x)=\left(\Delta-m |
fp(x)
Substituting into the expression for
lM
lM=i\sqrt{ | 2\omegap |
Z |
That is:
lM= | i | |||||||||||||||
|
\intd4xe-ip ⋅ \left(\Box+m2\right)\langle\beta out|T\left[\varphi(x)\varphi(y1)\ldots\varphi(yn)\right]|\alpha' in\rangle
Starting from this result, and following the same path another particle can be extracted from the in state, leading to the insertion of another field in the time-ordered product. A very similar routine can extract particles from the out state, and the two can be iterated to get vacuum both on right and on left of the time-ordered product, leading to the general formula:
\langlep1,\ldots,pn out|q1,\ldots,qm in\rangle=\int
m | |
\prod | |
i=1 |
4x | |
\left\{d | |
i |
| ||||||||||||||||
|
\right\}
n | |
\prod | |
j=1 |
\left\{
4y | |
d | |
j |
| ||||||||||||||||
|
\right\}\langle\Omega|T\varphi(x1)\ldots\varphi(xm)\varphi(y1)\ldots\varphi(yn)|\Omega\rangle
Which is the LSZ reduction formula for Klein–Gordon scalars. It gains a much better looking aspect if it is written using the Fourier transform of the correlation function:
\Gamma\left(p1,\ldots,pn\right)=\int
n | |
\prod | |
i=1 |
4x | |
\left\{d | |
i |
ipi ⋅ xi | |
e |
\right\}\langle\Omega|T \varphi(x1)\ldots\varphi(xn)|\Omega\rangle
Using the inverse transform to substitute in the LSZ reduction formula, with some effort, the following result can be obtained:
\langlep1,\ldots,pn out|q1,\ldots,qm in\rangle=
m | ||
\prod | \left\{- | |
i=1 |
| ||||||||||||||||
|
\right\}
n | |
\prod | |
j=1 |
\left\{-
| ||||||||||||||||
|
\right\}\Gamma\left(p1,\ldots,pn;-q1,\ldots,-qm\right)
Leaving aside normalization factors, this formula asserts that S-matrix elements are the residues of the poles that arise in the Fourier transform of the correlation functions as four-momenta are put on-shell.
Recall that solutions to the quantized free-field Dirac equation may be written as
\Psi(x)=\sums=\pm
ip ⋅ x | |
\intd\tilde{p}(b | |
bf{p}e |
-ip ⋅ x | |
+d | |
bf{p}e |
),
where the metric signature is mostly plus,
s | |
b | |
bf{p} |
bf{p}
s=\pm
\daggers | |
d | |
bf{p} |
s
s | |
u | |
bf{p} |
s | |
v | |
bf{p} |
s | |
(p/+m)u | |
bf{p}=0 |
s | |
(p/-m)v | |
bf{p}= |
0
d\tilde{p}:=d3p/(2\pi)32\omegabf{p}
\omegabf{p}=\sqrt{bf{p}2+m2}
|\alpha in\rangle
|\beta out\rangle
l{M}=\langle\beta out|\alpha in\rangle,
where no extra time-ordered product of field operators has been inserted, for simplicity. The situation considered will be the scattering of
n
n'
n
\{bf{p}1,...,bf{p}n\}
\{s1,...,sn\}
\{bf{k}1,...,bf{k}n'\}
\{\sigma1,...,\sigman'\}
|\alpha in\rangle=
s1 | |
|bf{p} | |
1 |
sn | |
,...,bf{p} | |
n |
\rangle and |\beta out\rangle=
\sigma1 | |
|bf{k} | |
1 |
\sigman' | |
,...,bf{k} | |
n' |
\rangle.
Extracting an in particle from
|\alpha in\rangle
\daggers1 | |
b | |
bf{p |
1,in
l{M}=
\daggers1 | |
\langle\beta out|b | |
bf{p |
1,in
where the prime on
\alpha
\daggers | |
b | |
bf{p}= |
\intd3x eip ⋅ \bar{\Psi}(x)\gamma0
s | |
u | |
bf{p}, |
where
\bar{\Psi}(x)=\Psi\dagger(x)\gamma0
\Psiin
\Psiout
l{M}=
ip1 ⋅ x1 | |
\intd | |
1 e |
\langle\beta out|\bar{\Psi}in(x
0 | |
1)\gamma |
s1 | |
u | |
bf{p |
1}-\bar{\Psi}out(x
0 | |
1)\gamma |
s1 | |
u | |
bf{p |
1}|\alpha' in\rangle.
The weak asymptotic condition needed for a Dirac field, analogous to that for scalar fields, reads
\lim | |
x0 → -infty |
\intd3x\langle\beta|eip ⋅ \bar{\Psi}(x)\gamma0
s | |
u | |
bf{p |
and likewise for the out field. The scattering amplitude is then
l{M}=
1 | |
\sqrt{Z |
where now the interacting field appears in the inner product. Rewriting the limits in terms of the integral of a time derivative, we have
l{M}=-
1 | |
\sqrt{Z |
=- | 1 |
\sqrt{Z |
where the row vector of matrix elements of the barred Dirac field is written as
η(x1):=\langle\beta out|\bar{\Psi}(x1)|\alpha' in\rangle
eip ⋅
s | |
u | |
bf{p} |
(-i\partial/+m)eip ⋅
s | |
u | |
bf{p}=0. |
Solving for
0\partial | |
\gamma | |
0 |
eip ⋅
s | |
u | |
bf{p} |
l{M}=
i | |
\sqrt{Z |
l{M}=
i | |
\sqrt{Z |
Consider next the matrix element appearing in the integral. Extracting an out state creation operator and subtracting the corresponding in state operator, with the assumption that no incoming particle has the same momentum, we have
\langle\beta out|\bar{\Psi} | |
\alpha1 |
(x1)|\alpha' in\rangle=
\sigma1 | |
\langle\beta' out|b | |
bf{k |
1,out
Remembering that
(\bar{\Psi}\gamma0u
\dagger | |
bf{p}) |
=
0\Psi | |
\bar{u} | |
bf{p}\gamma |
\daggers | |
\bar{u} | |
bf{p}\beta |
\langle\beta out|\bar{\Psi} | |
\alpha1 |
(x1)|\alpha' in\rangle=
1 | |
\sqrt{Z |
Note that a time-ordering symbol has appeared, since the first term requires
\Psi | |
\beta1 |
(y1)
\langle\beta out|\bar{\Psi} | |
\alpha1 |
(x1)|\alpha' in\rangle=
i | |
\sqrt{Z |
The rest of the in and out states can then be extracted and reduced in the same way, ultimately resulting in
\langle
n | |
\beta out|\alpha in\rangle=\int\prod | |
j=1 |
d4xj
| |||||
\sqrt{Z |
The same procedure can be done for the scattering of d-type particles, for which
s | |
u | |
bf{p} |
s | |
v | |
bf{p} |
\Psi
\bar{\Psi}
The reason of the normalization factor in the definition of in and out fields can be understood by taking that relation between the vacuum and a single particle state
|p\rangle
\langle0|\varphi(x)|p\rangle=\sqrtZ\langle0|\varphiin(x)|p\rangle+\intd4y\Deltaret(x-y)\langle0|j(y)|p\rangle
Remembering that both and are scalar fields with their Lorentz transform according to:
\varphi(x)=eiP ⋅ \varphi(0)e-iP ⋅
where is the four-momentum operator, we can write:
e-ip ⋅ \langle0|\varphi(0)|p\rangle=\sqrtZe-ip ⋅ \langle0|\varphiin(0)|p\rangle+\intd4y\Deltaret(x-y)\langle0|j(y)|p\rangle
Applying the Klein–Gordon operator on both sides, remembering that the four-moment is on-shell and that is the Green's function of the operator, we obtain:
0=0+\intd4y\delta4(x-y)\langle0|j(y)|p\rangle; \Leftrightarrow \langle0|j(x)|p\rangle=0
So we arrive to the relation:
\langle0|\varphi(x)|p\rangle=\sqrtZ\langle0|\varphiin(x)|p\rangle
which accounts for the need of the factor . The in field is a free field, so it can only connect one-particle states with the vacuum. That is, its expectation value between the vacuum and a many-particle state is null. On the other hand, the interacting field can also connect many-particle states to the vacuum, thanks to interaction, so the expectation values on the two sides of the last equation are different, and need a normalization factor in between. The right hand side can be computed explicitly, by expanding the in field in creation and annihilation operators:
\langle0|\varphiin(x)|p\rangle=\int
d3q | ||||||||||||||||||
|
e-iq ⋅ \langle0|ain(q)|p\rangle=\int
d3q | |||||||
|
e-iq ⋅ \langle0|ain
\dagger | |
(q)a | |
in |
(p)|0\rangle
Using the commutation relation between and
\dagger | |
a | |
in |
\langle0|\varphiin(x)|p\rangle=
e-ip ⋅ | |||||||
|
leading to the relation:
\langle0|\varphi(0)|p\rangle=\sqrt
Z | |
(2\pi)3 |
by which the value of may be computed, provided that one knows how to compute
\langle0|\varphi(0)|p\rangle