theories:quantum_field_theory:canonical
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theories:quantum_field_theory:canonical [2019/01/31 10:06] jakobadmin [History] |
theories:quantum_field_theory:canonical [2020/01/02 19:59] 62.151.109.225 |
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| \begin{equation}\label{KGsol} \Phi(x)= \int \mathrm{d }k^3 \frac{1}{(2\pi)^3 2\omega_k} \left( a(k){\mathrm{e }}^{ -i(k x)} + a^\dagger(k) {\mathrm{e }}^{ i(kx)}\right)\end{equation} | \begin{equation}\label{KGsol} \Phi(x)= \int \mathrm{d }k^3 \frac{1}{(2\pi)^3 2\omega_k} \left( a(k){\mathrm{e }}^{ -i(k x)} + a^\dagger(k) {\mathrm{e }}^{ i(kx)}\right)\end{equation} | ||
| - | Now, if we combine this solution with the canonical commutation relation we see that $a(k)$ and $a^\dagger(k)$ can‘t be numbers, but must be __operators__. Using $[\Phi(x), \pi(y)] = i \delta(x-y)$ we cam we can compute | + | Now, if we combine this solution with the canonical commutation relation we see that $a(k)$ and $a^\dagger(k)$ can‘t be numbers, but must be __operators__. Using $[\Phi(x), \pi(y)] = i \delta(x-y)$ we can compute |
| $$ [a^\dagger(k), a(k)] = i \delta(x-y) .$$ | $$ [a^\dagger(k), a(k)] = i \delta(x-y) .$$ | ||
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| So what are these operators? | So what are these operators? | ||
| - | To answer this question, we next look at something that understand: energy. | + | To answer this question, we next look at something that we understand: energy. |
| - | We can compute, using the Lagrangian, the corresponding Hamiltonian, which represents the energy and is the conserved quantity that follows from invariance under time-translation, if we use Noether‘s theorem. | + | Using the Lagrangian, we can compute the corresponding Hamiltonian, which represents the energy and is the conserved quantity that follows from invariance under time-translation, according to Noether‘s theorem. |
| For example, for spin $0$ fields we have\begin{equation} \label{hamil} H= \frac{1}{2} \Big( \big(\partial_0 \Phi \big)^2 + ( \partial_i \Phi )^2 + m \Phi^2 \Big). \end{equation}The fields are operators and therefore the Hamiltonian is an operator. For reasons explained above we call it the energy operator, which means if we act with the Hamiltonian on an abstract state $ | \Psi \rangle$ that describes the system in question, we get the energy of the system:$$ H | \Psi \rangle = E | \Psi \rangle $$ | For example, for spin $0$ fields we have\begin{equation} \label{hamil} H= \frac{1}{2} \Big( \big(\partial_0 \Phi \big)^2 + ( \partial_i \Phi )^2 + m \Phi^2 \Big). \end{equation}The fields are operators and therefore the Hamiltonian is an operator. For reasons explained above we call it the energy operator, which means if we act with the Hamiltonian on an abstract state $ | \Psi \rangle$ that describes the system in question, we get the energy of the system:$$ H | \Psi \rangle = E | \Psi \rangle $$ | ||
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| $$ H a^\dagger(k') a^\dagger(k') | 0\rangle = 2\omega_{k'} a^\dagger(k') a^\dagger(k') | 0\rangle $$ | $$ H a^\dagger(k') a^\dagger(k') | 0\rangle = 2\omega_{k'} a^\dagger(k') a^\dagger(k') | 0\rangle $$ | ||
| - | Recall that $a^\dagger(k') and $a(k') $ are the operator parts of our quantum fields. Here we learn that quantum fields create and annihilate particles! | + | Recall that $a^\dagger(k')$ and $a(k')$ are the operator parts of our quantum fields. Here we learn that quantum fields create and annihilate particles! |
| $| 0\rangle$ is an empty system | $| 0\rangle$ is an empty system | ||
theories/quantum_field_theory/canonical.txt · Last modified: 2020/01/03 02:54 by 110.70.51.250
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