2. free field theory

Quantization of Harmonic Oscillator in the Heisenberg Picture

$\ddot{x}+{\omega }^{2}x=0$ classical solution: $x(t)=\frac{1}{\sqrt{2}}(a{e}^{-i\omega t}+a^{\ast }{e}^{-i\omega t})$

${\partial }_t^2\hat{x}+{\omega }^2\hat{x}=0$ quantum solution: $\hat{x}(t)=\frac{1}{\sqrt{2}}(\hat{a}{e}^{-i\omega t}+{\hat{a}}^{†}{e}^{i\omega t})$ canonical momentum: $\hat{p}={\partial }_t\hat{x}$ canonical quantization condition: $[\hat{x}(t),\hat{p}(t)]=i$ $\Rightarrow$ $[\hat{a},{\hat{a}}^{†}]=1$ Hilbert space: $\hat{a}|0⟩=0$ , $|\hat{n}⟩=\frac{1}{\sqrt{n!}}({\hat{a}}^{†}{)}^n|0⟩$

summary: Steps of Quantization

1. classical eom → quantum operator eqn
2. find most general solution to classical eom
3. promote integration consts ($a,a^{\ast }$) to constant quantum operators $\Rightarrow$ full time evolution at quantum level
4. impose canonical quantization conditions
5. constant operators, which can be used to generalize Hilbert space

For free scalar field theory:

Fourier transformation: $\varphi (x)=e^{-iEt+i\vec{k}\cdot \vec{x}}$ (plane wave) $E^2={\vec{k}}^2+m^2\equiv {\omega }_k^2$ $\Rightarrow$ $u_{\vec{k}}=e^{-i{\omega }_{\vec{k}}t+i\vec{k}\cdot \vec{x}}$ (positive energy soln) and $u_{\vec{k}}^{\ast }=e^{i{\omega }_{\vec{k}}t-i\vec{k}\cdot \vec{x}}$ (negative energy soln) $\varphi (x)=\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}\frac{1}{\sqrt{2{\omega }_{\vec{k}}}}(a_{\vec{k}}{u}_{\vec{k}}+a_{\vec{k}}^{\ast }{u}_{\vec{k}}^{\ast })$ $\Rightarrow$ $\hat{\varphi }$ with ${\hat{a}}_{\vec{k}},{\hat{a}}_{\vec{k}}^{\ast }$ , $\hat{\Pi }={\partial }_t\hat{\varphi }$ canonical commutation relation: $[\hat{\varphi }(x),\hat{\Pi }(x)]=?$ Hints: for multiple particle:

• $[{\hat{x}}_a(t),{\hat{p}}_b(t)]=i{\delta }_{ab}$ (kronecker delta)
• $[{\hat{x}}_a,{\hat{x}}_b]=0$ , $[{\hat{p}}_a,{\hat{p}}_b]=0$
• $\Rightarrow$
  • $[\hat{\varphi }(t,\vec{x}),\hat{\varphi }(t,{\vec{x}}^{\prime })]=0$
  • $[\hat{\Pi }(t,\vec{x}),\hat{\Pi }(t,{\vec{x}}^{\prime })]=0$
  • $[\hat{\varphi }(t,\vec{x}),\hat{\pi }(t,{\vec{x}}^{\prime })]=i{\delta }^{(3)}(\vec{x}-{\vec{x}}^{\prime })$ (dirac delta)
  • $\Rightarrow$
    • $[{\hat{a}}_{\vec{k}},{\hat{a}}_{{\vec{k}}^{\prime }}]=[{\hat{a}}_{\vec{k}}^{†},{\hat{a}}_{{\vec{k}}^{\prime }}^{†}]=0$
    • $[{\hat{a}}_{\vec{k}},{\hat{a}}_{{\vec{k}}^{\prime }}^{†}]=(2\pi {)}^3{\delta }_{(k)}^{(3)}(\vec{k}-{\vec{k}}^{\prime })$

$\hat{H}=\int {\mathrm{d}}^{3}x\mathcal{H}=\cdots =\frac{1}{2}\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}{\omega }_{\vec{k}}({\hat{a}}_{\vec{k}}^{†}{\hat{a}}_{\vec{k}}+{\hat{a}}_{\vec{k}}{\hat{a}}_{\vec{k}}^{†})=\frac{1}{2}\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}{\omega }_{\vec{k}}\left({\hat{a}}_{\vec{k}}^{†}{\hat{a}}_{\vec{k}}+\frac{1}{2}[{\hat{a}}_{\vec{k}},{\hat{a}}_{\vec{k}}^{†}]\right)$ zero-point energy: $E_0=\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}{\omega }_{\vec{k}}\frac{1}{2}[{\hat{a}}_{\vec{k}},{\hat{a}}_{\vec{k}}^{†}]=\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}{\omega }_{\vec{k}}\frac{1}{2}(2\pi {)}^3{\delta }^{(3)}(0)$ , which is divergent

Remarks:

1. no time-dependence since $H$ is conserved quantity
2. QFT of $\varphi$ reduces to a system of continum of h.o. labeled by $\vec{k}$ with frequency $w_{\vec{k}}=\sqrt{k^2+m^2}$

The other conserved quantities:

1. spatial translation: $p^i=-\int {\mathrm{d}}^{3}x\hat{\Pi }(t,\vec{x}){\partial }_i\hat{\varphi }(t,\vec{x})=\int \frac{{\mathrm{d}}^3\vec{k}}{(2\pi {)}^3}\vec{k}{\hat{a}}_{\vec{k}}^{†}{\hat{a}}_{\vec{k}}$
2. Lorentz symmetry: $M^{\mu \nu }$

$E_0=\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}{\omega }_{\vec{k}}\frac{1}{2}(2\pi {)}^3{\delta }^{(3)}(0)E_0=\frac{1}{2}(2\pi {)}^3{\delta }^{(3)}(0)\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}{\omega }_{\vec{k}}$

$(2\pi {)}^3{\delta }^{(3)}(\vec{k})=\int {\mathrm{d}}^3\vec{x}{e}^{i\vec{k}\cdot \vec{x}}$

$(2\pi {)}^3{\delta }^{(3)}(0)=\int {\mathrm{d}}^3\vec{x}=\text{Volume}$

$E_0=\mathrm{V}{ϵ}_0$

${ϵ}_0=\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}\sqrt{k^2+m^2}$

how to deal with it?

${ϵ}_0={\int }^{\Lambda }\frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}\sqrt{k^2+m^2}\left(\left|\vec{k}\right|\ll \Lambda \right)$

• $\Lambda$ : cut-off
• large overall number → just ignore $E_0$ (current purples)

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Hilbert space

ground state(vaccum): ${\hat{a}}_{\vec{k}}|0⟩=|0⟩$ general stete: $|{\hat{a}}_{\overrightarrow{k_1}}^{†},{\hat{a}}_{\overrightarrow{k_2}}^{†},\cdots ⟩={\left({\hat{a}}_{\overrightarrow{k_1}}^{†}\right)}^{n_{{\vec{k}}_1}}{\left({\hat{a}}_{\overrightarrow{k_2}}^{†}\right)}^{n_{{\vec{k}}_2}}\cdots |0⟩$

• ${\hat{a}}_{\vec{k}}^{†}|0⟩=|\vec{k}⟩$
• ${\hat{a}}_{{\vec{k}}_1}^{†}{\hat{a}}_{{\vec{k}}_2}^{†}|0⟩=|{\vec{k}}_1,{\vec{k}}_2⟩$

ground state:

• $\hat{H}|0⟩=0$
• $P^i|0⟩=0$ general state:
• $\hat{H}|\vec{k}⟩=w_{\vec{k}}|\vec{k}⟩$
• ${\hat{P}}^i|\vec{k}⟩=k_i|\vec{k}⟩$
• $|\vec{k}⟩$ has $P^{\mu }=(w_{\vec{k}},\vec{k})$ : momentum of a relativistic particle of mass $m$

$|{\vec{k}}_1,{\vec{k}}_2⟩$ : $E={\omega }_{{\vec{k}}_1}+{\omega }_{{\vec{k}}_2}$ , $\vec{k}={\vec{k}}_1+{\vec{k}}_2$ $\Rightarrow$ $(E,\vec{k})$ : two particles of momentum $({\omega }_{{\vec{k}}_1},{\vec{k}}_1)$ and $({\omega }_{{\vec{k}}_2},{\vec{k}}_2)$

$|n_{{\vec{k}}_1},n_{{\vec{k}}_2}⟩$ : $n_{{\vec{k}}_1}$ particles of momentum ${\vec{k}}_1$, $n_{{\vec{k}}_2}$ particles of momentum ${\vec{k}}_2$

Remarks:

1. this can describe any # of particles
2. $\left[a_{{\vec{k}}_1}^{†},a_{{\vec{k}}_2}^{†}\right]=0$ full symmetry in permuting particles in a general state $\Rightarrow$ bosons
3. all particles are positive energies: even $E^2={\vec{k}}^2+m^2$ has two solutions, when we look at the eigenstate, we only have positive energy
4. total energy of state = sum of energies of all particles (only for no interaction, a theory of free particle)
5. normalization:
   • $⟨\vec{k}||\vec{k}⟩=(2\pi {)}^3{\delta }^{(3)}(\vec{k}-{\vec{k}}^{\prime })$ is not nice
   • $|k⟩\equiv \sqrt{2{\omega }_{\vec{k}}}|\vec{k}⟩=\sqrt{2{\omega }_{\vec{k}}}{\hat{a}}_k^{†}|0⟩$
   • $⟨k||k^{\prime }⟩=2{\omega }_{\vec{k}}(2\pi {)}^3{\delta }^{(3)}(\vec{k}-{\vec{k}}^{\prime })$
     • ${\omega }_{\vec{k}}(2\pi {)}^3{\delta }^{(3)}(\vec{k}-{\vec{k}}^{\prime })$ transforms nicely under Lorentz transformation
     • can show $u_{\Lambda }|k⟩=|\Lambda k⟩$
       • $\Lambda$ : Lorentz transformation
       • $u_{\Lambda }$ : quantum operator
6. general single-particle states $|\psi ⟩=\int \frac{{\mathrm{d}}^3\vec{k}}{(2\pi {)}^3}f(\vec{k})|k⟩$
   • can shoose $f(\vec{k})$ to construct a localized (in space) wave packet
     • $\int \frac{{\mathrm{d}}^3{\vec{k}}_1}{(2\pi {)}^3}\frac{{\mathrm{d}}^3{\vec{k}}_2}{(2\pi {)}^3}f({\vec{k}}_1)f({\vec{k}}_2)|k_1,k_2⟩$
7. structure of Hilbert spcae $\mathcal{H}=|0⟩\oplus {\mathcal{H}}_1\oplus {\mathcal{H}}_2\cdots$
   • the set of states with a finite # of particle is called a Fock space
8. $|\vec{k}⟩,|k⟩,|k_1,k_2⟩$ are not normalizable, just as $\psi =e^{ikx}$ in NR QM
   • not general state
   • $⟨\psi ||\psi ⟩=\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}{\left|f(k)\right|}^2<\infty$
   • then $|\psi ⟩$ is normalizable state (with choosing a basis $\{f_i\}$ )

initially we only have field, but at last we get particle with we give the energy free sclar field finally create bosons

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1. 从经典场论出发，得到量子化之后的守恒荷算符，和它的对易关系（生成无限小变换）、对应的幺正变换算符（生成有限变换）
2. 对复标量场的相位U(1)对称性应用守恒荷算符，表达式分离之后得到了表示粒子和反粒子的算符

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1. Starting from classical field theory, we derive the quantized conserved charge operator, and its commutation relation (which generates infinitesimal transformations) and the corresponding unitary transformation operator (which generates finite transformations).
2. We apply the conserved charge operator to the U(1) phase symmetry to the complex scalar field, after separation, the resulting expression yields operators representing particles and antiparticles.

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