3
$\begingroup$

This question is a followup to my previous one "Why are momentum eigenstates in QFT plane waves? " as it made sense for me to ask this separately, in a self-contained manner.

In QFT we have one-particle states which are classified by the unitary irreducible representations of the Poincaré group. This classification is carried out in detail in Chapter 2 of Weinberg's "The Quantum Theory of Fields". The one-particle states are spanned by the (improper) basis $|p,\sigma\rangle$ where $p^2=-m^2$, $p^0> 0$ and where $\sigma$ is either the spin projection or helicity depending on whether $m\neq 0$ or $m =0$.

On the other hand we have free relativistic quantum fields $\psi_\ell(x)$, whose creation and annihilation parts are of the form $$\psi_\ell^+(x)=\sum_\sigma \int d^3p\ u_\ell(x;p,\sigma) a(p,\sigma),\quad \psi_\ell^-(x)=\sum_\sigma \int d^3p\ v_\ell(x;p,\sigma)a^\dagger(p,\sigma)\tag{1}$$ with coefficients chosen so that the fields transform into Lorentz representations $$U_0(\Lambda,a)\psi_\ell(x)U_0(\Lambda,a)^\dagger=\sum_{\bar \ell} D_{\bar \ell\ell}(\Lambda^{-1})\psi_{\bar\ell}(\Lambda x+a)\tag{2}.$$

Now let $|\alpha\rangle$ be some one-particle with momentum wavefunction $\alpha_{\sigma}(p)=\langle p,\sigma|\alpha\rangle$. We can define the corresponding wavefunction to be the classical field configuration $$\alpha_\ell(x):=\langle 0|\psi_\ell(x)|\alpha\rangle\tag{3}.$$

Using this we immediately see that if $|\alpha\rangle=|p,\sigma\rangle$ is a momentum eigenstate then we get essentially a plane wave. That is the point made in the great answer by @ChiralAnomaly to my previous question.

The followup question now consists of two parts:

  1. The definition of a wavefunction following (3) gives rise to a mapping between momentum-space wavefunctions $\alpha_\sigma(p)=\langle p,\sigma|\alpha\rangle$ and position-space wavefunctions $\alpha_\ell(x)=\langle 0|\psi_\ell(x)|\alpha\rangle$. Is there some sense in which the mapping $\alpha_\sigma(p)\to \alpha_\ell(x)$ gives rise to a Hilbert space isomorphism?

    In other words, can we equally well characterize a unitary irrep of the Poincare group by either the functions $\alpha_\sigma(p)$ as done in Weinberg and by solutions to some wave equation $\alpha_\ell(x)$? I'm aware that the positive frequency solutions to the KG equation and to the Maxwell equation have a natural inner product and form a Hilbert space. The question would be if this mapping $\alpha_\sigma(p)\to \alpha_\ell(x)$ bridges these two spaces and establishes one representation isomorphism for the Poincaré group.

  2. Even if we can characterize the states by the wavefunctions (3), why would we do so? What is it that motivates this definition and makes it useful? Of course we are free to define always whatever we want, but usually we define things because the definition is useful. What is the reason to characterize states by $\alpha_\ell(x)$ as in (3) instead of the traditional $\alpha_p(\sigma)$?

    Something tells me that this has to do with LSZ reduction, but I still do not have a complete understanding of how all of these things come together.

$\endgroup$
2
$\begingroup$

Let us first observe that the relationship $\langle p \vert \phi(x)\vert 0\rangle = \mathrm{e}^{\mathrm{i}xp}$ is precisely analogous to the non-relativistic expression $\langle \vec p\vert \vec x\rangle = \mathrm{e}^{\mathrm{i}\vec x \vec p}$, which results in the non-relativistic momentum-space wavefunctions $\psi(\vec p) = \langle p\vert \psi\rangle$ and $\psi(\vec x) = \langle p \vert \psi\rangle$ being Fourier transforms of each other. Might something similar work here?

The first hurdle is that the states $\phi(x)\vert 0\rangle$ are a bit strange - they depend on time $t$ through the argument $x = (\vec x, t)$ of $\phi$. So what we really have here is a collection of relationships $$ \langle p \vert \phi(\vec x, t_0)\vert 0\rangle = \mathrm{e}^{\mathrm{i}t_0p^0}\mathrm{e}^{\mathrm{i}\vec x \vec p}\tag{*}$$ for each time $t_0$ that expresses the fact that the position space function $\alpha(\vec x,t_0)$ at each instant of time is some sort of modified (by the $\mathrm{e}^{\mathrm{i}t_0p^0}$ - don't forget that $p^0$ is a function of $\vec p$ since we're on the mass shell) Fourier transform of the momentum-space function $\alpha(\vec p)$ (writing $\vec p$ instead of $p$ for the argument makes no difference precisely because of the aforementioned mass shell). Explicitly,

\begin{align} \alpha(\vec p) & = \langle p\vert \alpha\rangle = \int \langle p\vert \phi(x)\vert 0\rangle\langle 0\vert \phi(x)\vert \alpha\rangle\mathrm{d}^3\mathrm{x} \\ & = \int\mathrm{e}^{\mathrm{i}t_0p^0}\mathrm{e}^{\mathrm{i}\vec x \vec p}\alpha_{t_0}(\vec x)\mathrm{d}^3x, \end{align} which is just the Fourier transform of $\alpha(\vec x)$ with a pre-factor $\mathrm{e}^{\mathrm{i}t_0p^0}$. The inverse is \begin{align} \alpha_{t_0}(\vec x) & = \langle 0\vert \phi(\vec x,t_0)\vert \alpha\rangle = \int \langle 0\vert \phi(x)\vert p\rangle\langle p\vert \alpha \rangle \mathrm{d}^3 p\sqrt{2p^0} \\ & = \int \mathrm{e}^{-\mathrm{i}t_0p^0}\mathrm{e}^{-\mathrm{i}\vec x \vec p}\alpha(\vec p)\mathrm{d}^3 p, \end{align} which says that $\alpha_{t_0}(\vec x)$ is the Fourier transform of $\mathrm{e}^{-\mathrm{i}t_0p^0}\alpha(\vec p)$. So this means that you can choose one instant $t_0 = 0$ at which the $\alpha_{t_0}(\vec x)$ and $\alpha(\vec p)$ really are just Fourier transforms of each other. So this answers the question of an isomorphism, since the Fourier transform is in particular an isomorphism.

The usefulness of this representation is, however, extremely limited, since the arbitrary but distinguished time $t_0$ means the loss of manifest Lorentz covariance. It can thus only feature in intermediate steps in calculations - and indeed it does: In the course of one of the standard derivations of the LSZ formula, the relationship between $a,a^\dagger(\vec p)$ and $\phi(\vec x,t)$ is also inverted at an instant $t_0$ in order to express momentum amplitudes $\langle q\vert p\rangle$ in terms of correlators $\langle 0 \vert \prod_i\phi(x_i)\vert 0\rangle$ - the instant then drops out of all results. Note that this is morally very similar to switching from a basis of $\lvert p\rangle$ to a basis of $\phi(x)\lvert 0\rangle$.

$\endgroup$
1
$\begingroup$

In my view this characterization naturally appears when one studies LSZ Reduction. Let us label the unitary irreps of the universal cover of the Poincaré group by $(m,s)$ where $m\in [0,+\infty)$ is the mass and $s\in \frac{1}{2}\mathbb{Z}$ is the spin/helicity. Suppose you have some particle in the irrep $(m,s)$ whose state space is spanned by the improper momentum eigenstates $|p,\sigma\rangle$. An arbitrary state $|f\rangle$ of this particle will be $$|f\rangle=\sum_{\sigma}\int d\Omega(p) f_{\sigma}(p)|p,\sigma\rangle,\tag{1}$$

where $d\Omega(p)$ is the Lorentz invariant measure in the mass shell $H_m^+$. Such a state can be given one creation operator $a^\dagger(f)$ by simply defining $$a^\dagger(f)=\sum_\sigma \int d\Omega(p) f_\sigma(p) a^\dagger(p,\sigma)\tag{2}.$$

The associated annihilation operator $a(f)$ can be defined to be its adjoint. It is clear then that inside the Fock space $a^\dagger(f)|0\rangle=|f\rangle$ while $a(f)|0\rangle=0$ as it should be. Moreover we have the Fock algebra $$[a(f),a^\dagger(g)]_\pm=\langle f|g\rangle,\quad [a(f),a(g)]_\pm=[a^\dagger(f),a^\dagger(g)]_\pm=0,\tag{3}$$ where we have the commutator for bosons and anticommutator for fermions.

Now suppose you want to compute one scattering amplitude of $m$ incoming particles into $n$ outgoing ones: $${\cal A}_{n+m}(f_1,\dots, f_m; g_1,\dots, g_n)= \langle g_1,\dots, g_n|f_1,\dots, f_m\rangle\tag{4}$$

where I have left generic single particle states and where we work in the Heisenberg picture. The idea of LSZ Reduction is to remove particles from both incoming and outgoing states, replace by their creation and annihilation operators and rewrite such operators in terms of fields.

Let me focus on the incoming ones. The key aspect here is to ask: how can we write $a_{\rm in}^\dagger(f)$ in terms of Heisenberg picture fields? There are a few points towards that. I'm going to focus on scalar fields/spinless particles, but the only difference in the more general case lies in the choice of inner product on the space of solutions and in the introduction of polarization tensors/spinors:

  1. From (2) $a_{\rm in}^\dagger(f)$ is fully specified by $f$ and by $a_{\rm in}^\dagger(p)$ so we should focus on $a_{\rm in}^\dagger(p)$ really.

  2. If we study $a_{\rm in}^\dagger(p)$ we should recall that the incoming field $\phi_{\rm in}(x)$ contains such $a_{\rm in}^\dagger(p)$ in its plane wave expansion. Precisely we have something of the form $$\phi_{\rm in, \ell}(x)=\sum_\sigma \int d\Omega(p)[e^{ipx}a_{\rm in}(p)+e^{-ipx}a_{\rm in}^\dagger(p)]\tag{5}$$ In fact such an expansion defines the free, incoming and outgoing fields as explained by Weinberg in Chapter 5 of "The Quantum Theory of Fields".

  3. It turns out that such field obeys the Klein-Gordon equation. The space of solutions to the wave equation is equipped with a natural inner product: $$(\Phi_1,\Phi_2)=-i\int_{\Sigma}d\Sigma_\mu (\Phi_1\nabla_\mu \Phi_2^\ast - \Phi_{2}^\ast \nabla_\mu \Phi_1)\tag{6},$$ where $\Sigma$ is one arbitrary Cauchy surface. It turns out that when one uses such an inner product, defining $u_p(x) = e^{ipx}$ and $v_p(x)=e^{-ipx}$ with $p^2=-m^2$ and $p^0 > 0$, we have the orthogonality relations $$(u_p,u_{p'})=(2\pi)^3(2\omega_p)\delta^{3}(\vec p-\vec p'),\quad (u_p,v_p')=0,\quad (v_p,v_{p'})=-(2\pi)^3(2\omega_p)\delta^{(3)}(\vec p-\vec p').\tag{7}$$ The beautify of this is that it allows you to recover $a^\dagger_{\rm in}(p) = -(v_p,\phi_{\rm in})$.

  4. By taking a superposition it follows that $$a^\dagger_{\rm in}(f)=-\int d\Omega(p) f(p) (v_p,\phi_{\rm in}) =- (\psi_f^-, \phi_{\rm in}),\tag{8}$$ where we have defined the solution $$\psi_f^-(x) = \int d\Omega(p) f(p)e^{-ipx}\tag{9}$$

  5. We have worked with one incoming field but since the inner product uses one arbitrary Cauchy surface we can push to $t\to -\infty$ and use the fact that $\phi(x)\to \sqrt{Z}\phi_{\rm in}(x)$ inside mean values in that limit in order to relate to the Heisenberg picture interacting field.

Now the beauty in all of this is that you can extract $a^\dagger_{\rm in}(f)$ by taking the Klein-Gordon inner product of the quantum field with one position space wavefunction $\psi_f^-(x)$. A similar discussion shows that you can extract $a_{\rm out}(f)$ which in the scattering amplitude (4) gives you one outgoing particle, by one Klein-Gordon inner product of the quantum field with one position space wavefunction $\psi_f^+(x)$ in which we expand in terms of $e^{ipx}$ instead of $e^{-ipx}$. In my view of the issue, this is the main usefulness of such characterization of relativistic particle states in terms of position space wavefunctions.

A final remark: it is simple to show that the wavefunction I have defined here is the same as in equation (3) of the question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.