LOG#142. Basic QFT in curved ST(II).

Posted on by September 9, 2019

I. Back to QFT in flat Spacetime.

Any scalar quantum field has an action and lagrangian density:

S_\phi=\displaystyle{\int d^Dx\mathcal{L}}

in D spacetime dimensions. The lagrangian density is usually written as follows

\mathcal{L}_\phi=-\dfrac{1}{2}\eta^{\mu\nu}\partial_\mu\phi\partial_\nu\phi-\dfrac{1}{2}m^2\phi^2

The EOM (equation of motion) reads:

\square^2\phi-m^2\phi=0

A nice feynmanity! Indeed, it is the free Klein-Gordon equation! The metric is split into

ds^2=\eta_{\mu\nu}dx^\mu dx^\nu=-dt^2+dx^2

The canonical momentum is

\Pi=\dfrac{\partial \mathcal{L}}{\partial (\partial_0\phi)}

\Pi=\dot{\phi}

H=\displaystyle{\int d^{D-1}x\mathcal{H}}

\mathcal{H}(\phi, \Pi)=\Pi \dot{\phi}-\mathcal{L}=\dfrac{\Pi^2}{2}+\dfrac{1}{2}(\nabla\phi)^2+\dfrac{1}{2}m^2\phi^2

\phi (x^\mu)=\phi_0 e^{ik_\mu x^\mu}=\phi_0 e^{-i\omega t+ikx}

where

k^\mu=(\omega, k) with \omega^2=k^2+m^2

Important remark: \phi (x) is NOT a wavefunction, indeed it is a QUANTUM operator with some dynamics (usually written in terms of normal coordinates, i.e., in the creation/annihilation formalism!).

The solutions to the EOM for the free field are quite straightforward:

A) f_k(x)=\dfrac{e^{ik^\mu x_\mu}}{\left((2\pi)^{D-1}2\omega\right)^{1/2}}

B) (f_{k_1},f_{k_2})=\delta^{D-1}(k_1-k_2)

C) Positive frequency solutions:

\partial_t f_k=-i\omega f_k with \omega>0 implies (f_{k_1},f_{k_2}^*)=0

D) Negative frequency solutions:

\partial_t f_k=i\omega f_k with \omega>0 implies (f_{k_1}^*,f_{k_2}^*)=-\delta^{D-1}(k_1-k_2)

E) Equal time commutators and other useful relationships:

\left[ \phi (t,x),\phi (t, x')\right]=0

\left[ \phi ( t, x), \Pi (t, x')\right]=0

\left[ \phi (t,x), \Pi (t, x')\right]=i\delta^{D-1}(x-x')

\phi (t,x)=\displaystyle{\int d^{D-1} a_kf_k (t,x)+a_k^+f_k^*(t, x)}

\left[a_k, a_{k'}\right]=\left[a_k^+,a_{k'}^+\right]=0

\left[a_{k},a_{k'}^+\right]=\delta^{D-1}(k-k')

F) Creation/Annihilation filed operators.

a_k\vert 0\rangle=0 (vacuum)

\vert n_k\rangle=\dfrac{1}{\sqrt{n!}}(a_k^+)^{n_k}\vert 0\rangle (n-particle state)

\vert n_1,n_2,\ldots, n_j\rangle=\dfrac{1}{\sqrt{n_1!n_2!\cdots n_j!}}(a_{k_1}^+)^{n_1}\cdots (a_{k_j}^+)^{n_j}\vert 0\rangle

a_{k_i}\vert n_1,n_2,\ldots,n_j\rangle=\sqrt{n_i}\vert n_1,n_2,\ldots, n_i-1,\ldots,n_j\rangle

a^+_{k_i}\vert n_1,n_2,\dots,n_j\rangle=\sqrt{n_i+1}\vert n_1,n_2,\ldots, n_i+1,\ldots, n_j\rangle

N_k=a^+_ka_k

n_{k_i}\vert n_1,n_2,\ldots, n_i,\ldots, n_j\rangle= n_i\vert n_1,n_2,\ldots, n_i,\ldots, n_j\rangle

H=\displaystyle{\int d^{D-1}x\left[\dfrac{1}{2}\dot{\phi}^2+\dfrac{1}{2}(\nabla\phi)^2+\dfrac{1}{2}m^2\phi^2\right]=\dfrac{1}{2}\int d^{D-1}k\left[a_k^+a_k+a_ka_k^+\right]\omega}

IF \left[a_k,a_{k'}\right]=\delta^{D-1}(k-k'), then

H=\displaystyle{\int d^{D-1}k\left[n_k+\dfrac{1}{2}\delta^{D-1}(0)\right]\omega}

P^i=\displaystyle{\int d^{D-1}k n_k k^i}

Comments:

i) Thus, particles are excitations in certain Fock space (basis with multiparticle states) in QFT!

ii) The energy eigenstates are a set of particles with definite momenta.

iii) Normal  modes are plane waves extended on (phase) space!

Issue: curved spacetime WILL NOT have plane wave solutions in general. Moreover, frequency will not split into positive and negative solutions so easy, and then, the solutions are hardly understood as “particles”. Only in the asymptotic past and future we could “read off” a particle interpretation, thanks to certain S-matrix.

Issue(II): Vacuum energy. In flat space QFT, the groun state energy can be non-zero, and we can absorb or neglect an “infinite” constant (delta function singularity). Substracting infinities or infinite constants is generally made with the use of the “normal ordering” prescription. Thus, we can “renormalize” QFT in flat space “by the face”. However, this issue is not negligible in curved spacetime, since energy “gravitates”, so it is less clear why the vacuum energy that flat spacetime QFT predicts is “wrong” in comparation with the almost de Sitter (curved spacetime) we observe in Cosmology!

Issue(III): Cosmological constant and vacuum energy. Why the vacuum energy that flat spacetime QFT predicts is so close to zero? Flat spacetime QFT predicts that it should be very large (or “infinite”!). Nobody has solved this problem yet!

II. QFT in curved spacetime: a fast tour

We reboot the scalar field example in a curved spacetime now! We write

\mathcal{L}=\sqrt{-g}\left(-\dfrac{1}{2}g^{\mu\nu}\nabla_\mu \phi\nabla_\nu \phi-\dfrac{1}{2}m^2\phi^2-\xi R\phi^2\right)

Here, R is the curvature scalar and

\xi=\dfrac{D-2}{4(D-1)}

is the conformal parameter. It equals \xi=0 in the minimal coupling, and it yields \xi=\dfrac{1}{6} if the coupling with the metric is “conformal”. A conformal transformation of the metric is a map:

g_{\mu\nu}(x)\rightarrow \omega^2(x)g_{\mu\nu}=\Omega^2(x)g_{\mu\nu}

The quantization in the canonical formalism provides

\Pi=\dfrac{\partial\mathcal{L}}{\partial(\nabla_0\phi)}=\sqrt{-g}\nabla_0\phi

Remark: in curved spacetime, the metric is an arbitrary FUNCTION on the spacetime coordinates.

Moreover,

\left[\phi (t, x),\phi (t,x')\right]=\left[\Pi (t, x),\Pi (t, x')\right]=0

\left[ \phi (t,x),\Pi (t,x')\right]=\dfrac{i}{\sqrt{-g}}\delta^{D-1}(x-x')

The EOM are

\square^2-m^2\phi-\xi R\phi=0

Remark: note the violation of the equivalence principle in the case \xi\neq 0!

There, \square^2=\nabla_\mu\nabla^\mu

Now, we proceed with the semiclassical setting:

1st. We introduce positive frequency and negative frequency modes to form a complete bases for the solutions of the EOM.

2nd. We expand \phi in terms of these modes.

3rd. We interpretate the operator coefficients as creation/annihilation operators.

Some problems arise with this procedure:

1) NO time-like Killing vector exists in general.

2) It is not easy to solve the wave equations in curved spacetime. Modes are not in general positive/negative frequencies.

3) Key point: the vacuum, as an operator number, IS observer dependent!

4) The time coordinate is NOT unique!

5) Empty/full particle vacuum is distinguished:

\vert 0\rangle_{in}\neq \vert 0\rangle_{out}

A partial conservative solution that “works”:

1) Find a set of solutions so you can write f_+(x^\mu)

2) Use the so called Bogoliubov transformations for the equation

\ddot{\phi}_k+\omega_k^2(\eta)\phi=0

3) Write the Klein-Gordon current

j^\mu=-i(\phi_1\partial^\mu\phi^*_2-\phi_2\partial^\mu\phi_1)

\nabla_\mu j^\mu=0

( \phi_1,\phi_2)=-i\int \phi_1\overleftrightarrow{\partial^\mu}\phi_2\sqrt{-g}d\xi_\mu

4) Idea: make a map f_i\rightarrow F_i such as, e.g.,

\left[f_i^{in/out}\right]=\sqrt{\dfrac{1}{2m\omega_{in/out}}}e^{-i(\omega t)\vert_{t\rightarrow\pm \infty}}

Bogoliubov’s transformations are the transformations or relations between these two sets of creation/annihilation operators.

See you in other QFT in curved spacetime post!

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