multi-particle states

From Physics wiki

1 Introduction
2 One particle states
3 Multi-particle states
4 References

Introduction

Notions like states with definite particle number or the Fock-space representation that goes along with non-interacting field theory have to be re-evaluated for interacting field theories. At the very least, if we were to write the field $\phi\,$ in terms of creation and annihilation operators we would find that the number operator no longer commutes with the Hamiltonian. However, from what we've learned from the Källén-Lehmann spectral representation, we can still talk about one-particle states. Let us assume for the moment that there are no bound states.

One particle states

To start, let us define the one-particle state $\left|k\right\rangle\,$ to be a state, normalized according to $\left\langle k | k' \right\rangle = (2\pi)^3 2 \omega_k \delta^{(3)}(\mathbf{k} - \mathbf{k}')\,$ , with definite momentum $k^\mu\,$ , i.e., $P^\mu \left|k\right\rangle = k^\mu \left|k\right\rangle$ . By translational invariance of the theory,

$\left\langle k |\phi(x) | 0 \right\rangle = \left\langle k | e^{iP\cdot x}\phi(0)e^{-iP\cdot x} | 0 \right\rangle = e^{i k \cdot x} \left\langle k | \phi(0)| 0 \right\rangle\,$ ,

where we have made use of the fact that the vacuum transforms as a Lorentz scalar.

Let us re-define the field $\phi(x)\,$ to be

$Z^{\frac{1}{2}} \left[\phi(x) - \left\langle 0 | \phi(x) | 0 \right\rangle \right] \to \phi_r(x) \,$ ,

where $Z^{\frac{1}{2}} \equiv \left\langle k | \phi(0)| 0 \right\rangle\,$ refers to the field-strength renormalization. From the Källén-Lehmann spectral representation we know that in general $Z^{\frac{1}{2}} \neq 1\,$ . The new field has two nice properties

$\left\langle 0 | \phi_r(x) | 0\right\rangle = 0\,$ ,

$\left\langle k | \phi_r(x) | 0 \right\rangle = e^{i k \cdot x}\,$ .

In analogy with the non-interacting field, we interpret $\phi(x)\left|0\right\rangle\,$ as being the action of creating a single particle at the point $x\,$ . We can also define a state with some spread in momentum space according to

$\left|\varphi\right\rangle \equiv \int\!\frac{d^3k}{(2\pi)^3} \frac{1}{2\omega_k} \varphi(k) \left|k\right\rangle\,$ ,

where

$\varphi(x) \equiv \int\!\frac{d^3k}{(2\pi)^3} \frac{1}{2\omega_k} \varphi(k) e^{i k\cdot x}\,$ ,

is some (localized, square integrable) solution to the Klein-Gordon equation. We also define the operator

$a^{\dagger}_{\varphi}(t) \equiv \left\langle \varphi(x), \phi(x)_r\right\rangle\,$ ,

where the norm is defined as in that of the classical solutions. We recognize this as the creation operator corresponding to the wave-packet $\varphi(x)\,$ . Indeed,

$\left\langle 0 | a^{\dagger}_{\varphi}(t) | 0 \right\rangle = 0\,$ ,

and

$\left\langle k \| a^{\dagger}_{\varphi}(t) \| 0 \right\rangle\,$	$= \int\!\frac{d^3k'}{(2\pi)^3} \frac{1}{2\omega_k} \varphi^*(k') \left\langle k \| -i \int\!d^3x\ e^{-i k'\cdot x} \stackrel{\leftrightarrow}{\partial}_0 \phi_r(x) \| 0 \right\rangle\,$ ,
	$= -i \int\!d^3x \int\!\frac{d^3k'}{(2\pi)^3} \frac{1}{2\omega_k} \varphi^*(k') e^{-i k'\cdot x} \stackrel{\leftrightarrow}{\partial}_0 \left\langle k \| \phi_r(x) \| 0 \right\rangle\,$ ,
	$= -i \int\!d^3x \int\!\frac{d^3k'}{(2\pi)^3} \frac{1}{2\omega_k} \varphi^*(k') e^{-i k'\cdot x} \stackrel{\leftrightarrow}{\partial}_0 e^{i k \cdot x}\,$ ,
	$= \varphi^*(k)\,$ .

Similarly, we would find that $\left\langle k | a_{\varphi}(t) | 0 \right\rangle = 0\,$ , so that $\phi_f(x)\,$ behaves much like a creation operator when acting on the vacuum or one-particle states. However, note that $\left\langle 0 | a_{\varphi}(t) | 0 \right\rangle = 0\,$ and $\left\langle k | a_{\varphi}(t) | 0 \right\rangle = 0\,$ does not mean that $a_{\varphi}(t)\left| 0 \right\rangle= 0\,$ (as in the free theory). This would imply that the inner product vanishes against all states, which in general is not true. Note also that $a^{\dagger}_{\varphi}\,$ also obeys the correct algebra with $a_{\varphi}\,$ at equal times, even though $\phi\,$ is an interacting field.

Multi-particle states

The spectral density function receives contribution from the single-particle state of mass $m\,$ , which is usually isolated from the multi-particle spectrum. Therefore, let us write

$\rho(M^2)= Z (2\pi) \delta(M^2 - m^2) + \sum_{ \mathrm{multi\,\,particle}} (2\pi)\delta(M^2 - m_n^2)\left| \left\langle 0 | \phi(0) | n_\mathbf{0} \right\rangle \right|^2\,$ ,

where the sum now extends over multi-particle states $\left|n_\mathbf{0}\right\rangle_{mp}\,$ only. In general, there will be various $N\,$ -particle thresholds - energies at which states with $N\,$ particles appear. We see that a state $\left| n_\mathbf{k}\right\rangle_{mp}\,$ generally satisfies $k^0 > \omega_\mathbf{k}\,$ , since $k^0 = \sqrt{|\mathbf{k}|^2 + M^2}\,$ and $\omega_\mathbf{k} = \sqrt{|\mathbf{k}|^2 + m^2}\,$ and $M \ge 2m\,$ . Then

${}_{mp}\left\langle n_\mathbf{k} | a^{\dagger}_\varphi(t) | 0 \right\rangle = \frac{k^0 + \omega_\mathbf{k}}{2\omega_\mathbf{k}} A_{mp}(k)\varphi^*(k) e^{i (\omega_k - k^0) t}\,$ .

Here $A_{mp}(k)\,$ is defined by ${}_{mp}\left\langle n_\mathbf{k} |\phi_r(x) | 0 \right\rangle = {}_{mp}\left\langle n_\mathbf{0} | \phi_r(0)| 0 \right\rangle e^{i k \cdot x} \equiv A_{mp}(k) e^{i k \cdot x}\,$ . Now a physical multi-particle state $\left|\Psi_{mp}\right\rangle\,$ will be superposition of the states $\left|n_\mathbf{k}\right\rangle_{mp}\,$ , so that

$\left|\Psi_{mp}\right\rangle = \int\!d^3k \sum_{n} f(k) \left|n_\mathbf{k}\right\rangle\,$ .

We find then that

${}_{mp}\left\langle \Psi | a^{\dagger}_\varphi(t) | 0 \right\rangle = \sum_n \int\!d^3k\, \frac{k^0 + \omega_\mathbf{k}}{2\omega_\mathbf{k}} A_{mp,n}(k) f^*(k)\varphi^*(k) e^{i (\omega_\mathbf{k} - k^0) t}\,$ .

While this expression is certainly complicated, we can look at the limits $t \to \pm \infty\,$ . The integral will oscillate wildly since $k^0_{mp,n} > \omega_\mathbf{k}\,$ (recall $M \geq 2m > m\,$ ), which means that

$\lim_{t \to \pm \infty} \left\langle \Psi _{mp}| a^{\dagger}_\varphi(t) | 0 \right\rangle = 0\,$ ,

provided that the time-independent functions inside the integral are sufficiently well behaved, i.e., square integrable. This is a consequence of the Riemann-Lebesgue lemma. The same argument also shows that

$\lim_{t \to \pm \infty} \left\langle \Psi _{mp}| a_\varphi(t) | 0 \right\rangle = 0\,$ ,

however now the phase in the exponential is positive definite, so that the integral is always zero.

back to Källén-Lehmann spectral representation

on to scattering

References

Mark Srednicki, Quantum Field Theory, Cambridge University Press, Cambridge, 2007.

Retrieved from "http://www.physics.thetangentbundle.net/wiki/Quantum_field_theory/multi-particle_states"

$\left\langle k \| a^{\dagger}_{\varphi}(t) \| 0 \right\rangle\,$	$= \int\!\frac{d^3k'}{(2\pi)^3} \frac{1}{2\omega_k} \varphi^*(k') \left\langle k \| -i \int\!d^3x\ e^{-i k'\cdot x} \stackrel{\leftrightarrow}{\partial}_0 \phi_r(x) \| 0 \right\rangle\,$ ,
	$= -i \int\!d^3x \int\!\frac{d^3k'}{(2\pi)^3} \frac{1}{2\omega_k} \varphi^*(k') e^{-i k'\cdot x} \stackrel{\leftrightarrow}{\partial}_0 \left\langle k \| \phi_r(x) \| 0 \right\rangle\,$ ,
	$= -i \int\!d^3x \int\!\frac{d^3k'}{(2\pi)^3} \frac{1}{2\omega_k} \varphi^*(k') e^{-i k'\cdot x} \stackrel{\leftrightarrow}{\partial}_0 e^{i k \cdot x}\,$ ,
	$= \varphi^*(k)\,$ .

multi-particle states

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Contents

Introduction

One particle states

Multi-particle states

References

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