What are fields? (diff between theory of field and theory of particles)

Why quantum field theory?

Desire (photon and phonon)

Marriage of QM and R. (# of part. can change)

Fields maybe more fundemantal.

Connection between QFT and classical thermal field theory at \(T>0\).

How to quantize? discretize, solve N-body, take limit.

\(\mathbf{x}_i(t) \Longleftrightarrow \phi(\mathbf{x}, t)\)

A relativistic one particle Shrödinger Equation: (Also describes acoustic phonons.) \[\hbar^2 (-\partial_{ct}^2 +\nabla^2) \psi = M^2 c^2 \psi\] \[-\hbar^2\partial^\mu\partial_\mu \psi = M^2c^2 \psi\]

Problem: \(|\psi|^2\) is no longer prob. density. Not conserved.

What is conserved? \(\rho = i(\psi^* \dot{\psi} - \dot{\psi}^*\psi)\). However, this can be negative.

Also, energy have negative solutions.

Classical limit: \(\ddot{\phi} = (c^2\nabla^2 -\frac{(Mc^2)^2}{\hbar^2})\phi \Longrightarrow \ddot{\phi} = (c^2\nabla^2 - \omega_{min}^2)\phi\)

\(M=0\) case: \(\ddot{\phi} = c^2 \nabla^2 \phi\). Dispersion relation: (Note, \(c\) is the wave speed, not speed of light.) \[\omega^2 = c^2 k^2\] Arbitratily low frequency waves \(\Longrightarrow\) “acoustic limit”

\(M\ne 0\) case: \(\ddot{\phi} = (c^2\nabla^2 - \omega_{min}^2)\phi\). Dispersion relation: (“optical” branches) \[\omega^2 = c^2 k^2 + \omega_{min}^2\]

Nomenclature: gap.

\(\ddot{\phi} = (\nabla^2 - m^2)\phi\)

Action: (Lagrangian density and Lagrangian) \[S[\phi(\mathbf{x},t)] = \int_{t_1}^{t_2} dt \int d^3 x \mathcal{L}(\phi(\mathbf{x},t), \dot{\phi}(\mathbf{x},t), \nabla\phi(\mathbf{x},t))\]

Example:

Small transverse fluctuations of a violin strings. \(\mathcal{L} = \frac{1}{2}\rho \dot{\phi}^2 - \frac{1}{2} T (\phi')^2\)

EM in \(A_0 = 0\) gauge. \(\mathcal{L} = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu}\)

Eular-Lagrangian Equation. Excise: Derive this from Hamilton’s principle, or variation principle. Cf. Page 7.

\[\partial_t \frac{\partial \mathcal{L}}{\partial (\partial_t \phi)} + \nabla\cdot \frac{\partial \mathcal{L}}{\partial (\nabla \phi)} = \frac{\partial \mathcal{L}}{\partial \phi}\]

Relativistic version:

\[\partial_\mu \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} = \frac{\partial \mathcal{L}}{\partial \phi}\]

Example on \(\partial_\mu\) and on a given Lagrangian.

Note: different people treat different terms as K.E. or P.E..

Lagrangian – Hamiltonian

Review: \(H = \sum_i p_i \dot{q}_i - L\). Quantization: \([\hat{p}_i, \hat{q}_j] = -i\hbar \delta_{ij}\)

Canonical “momentum” field conjugate to \(\phi(\mathbf{x})\): \[\pi(\mathbf{x})\equiv \frac{\partial \mathcal{L}}{\partial \dot{\phi}}\]

Moral: canonical momentum fields are often not related to what you think of as actual momentum.

\[H = \left(\int d^3 x \pi(\mathbf{x}, t)\dot{\phi}(\mathbf{x}, t)\right) - L = \int d^3 x \mathcal{H}\] with \[\mathcal{H} = \pi(\mathbf{x}, t)\dot{\phi}(\mathbf{x}, t) - \mathcal{L}\]

Quantization: (the rest are \(0\)) \[[\hat{\pi}(\mathbf{x}), \hat{\phi}(\mathbf{y})] = - i \hbar \delta^{(3)} (\mathbf{x} - \mathbf{y})\]

Example: For \(\mathcal{L} = \frac{1}{2}\dot{\phi}^2 - \frac{1}{2}(\nabla \phi)^2 - \frac{1}{2}m^2 \phi^2\): \[\mathcal{H} = \frac{1}{2} \pi^2 + \frac{1}{2} (\nabla \phi)^2 + \frac{1}{2} m^2 \phi^2\] Note: this is not Lorentz inv.

\[\hat{\phi}(\mathbf{x}) = \int \frac{d^3 k}{(2\pi)^3 } \hat{\tilde{\phi}}(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x}}\] \[\hat{\pi}(\mathbf{y}) = \int \frac{d^3 k}{(2\pi)^3 } \hat{\tilde{\pi}}(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{y}}\]

So (excise: Prove this): \[\hat{H} = \int \frac{d^3 k}{(2\pi)^3 } \left[ \frac{1}{2}\hat{\tilde{\pi}}(\mathbf{k})^\dagger \hat{\tilde{\pi}}(\mathbf{k}) +\frac{1}{2}\omega_\mathbf{k}^2 \hat{\tilde{\phi}}(\mathbf{k})^\dagger \hat{\tilde{\phi}}(\mathbf{k}) \right]\]

Note: These are complex harmonic oscillators. \(\tilde{\phi}(-\mathbf{k})=\tilde{\phi}(\mathbf{k})^*\). All \(\mathbf{k}\)’s are not independent.

\[[\hat{\tilde{\pi}}(\mathbf{k'}), \hat{\tilde{\phi}}(\mathbf{k})] = - i (2\pi)^3 \delta^{(3)} (\mathbf{k} + \mathbf{k'})\] Or: \[[\hat{\tilde{\pi}}(\mathbf{k'})^\dagger, \hat{\tilde{\phi}}(\mathbf{k})] = - i (2\pi)^3 \delta^{(3)} (\mathbf{k} - \mathbf{k'})\]

Defining uppering and lowering operators, we get: \[\hat{\phi}(\mathbf{x}) = \int \frac{d^3 k}{(2\pi)^3 } \frac{1}{\sqrt{2\omega_\mathbf{k}}}\left[ \hat{a}_\mathbf{k} e^{i\mathbf{k}\cdot\mathbf{x}} + \hat{a}_\mathbf{k}^\dagger e^{-i\mathbf{k}\cdot \mathbf{x}} \right]\] Note the sign difference, as \(\hat{\phi}(\mathbf{x})\) must be Hermitian.

\[[a_\mathbf{k}, a_\mathbf{k'}^\dagger] = (2\pi)^3 \delta(\mathbf{k} - \mathbf{k'})\] \[\hat{H} = \int \frac{d^3 k}{(2\pi)^3 } \omega_\mathbf{k} \left(a_\mathbf{k}^\dagger a_\mathbf{k'} + \frac{1}{2}\right)\]

Ex: check this. \(a_\mathbf{k}\) and \(a_\mathbf{k}^\dagger\) are also called annihilation and creation operators.

Note: the 1/2 in the above expression is actually \((1/2)\delta^{(3)}(0)\). There is some argument on discretized vs continuous normalization concerning this topic.

Aside: regularization and renormalization. Find tuning problem. Dirac sea. Supersymmetry.

Starting point: \(a_\mathbf{k}^\dagger\) should also increase the momentum of the system by \(\hbar \mathrm{k}\).

Lorentz transform: \[x^\mu \rightarrow \Lambda^\mu_{\,\,\nu}x^\nu\]

It can be a boost or a rotation or a combination of both. Inverse: \(\Lam