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 [2020/01/03 02:54] 110.70.51.250 [Intuitive] |
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| Then what is about the other types of particles? The quarks, the gluons, the Higgs? Well, these belong just to other fields. That is, our universe is filled up with many fields, all existing simultaneously at every point in space and time. | Then what is about the other types of particles? The quarks, the gluons, the Higgs? Well, these belong just to other fields. That is, our universe is filled up with many fields, all existing simultaneously at every point in space and time. | ||
| - | You may be wondering how this should work, and if this is not a bit crowded. But you know already that fields are mathematical concepts. For example, you can associate with every point in space and time a temperature, and thus create a temperature field. At the same time, there is an atmospheric pressure field. Both can happily exist simultaneously. But they are not ignoring each other. As you know, both a related with each other: If either changes this indicates a change of the other as well. Though this analogy is not exactly the same as the particle physics fields, and there are more things involved, the basic idea is the same. | + | You may be wondering how this should work, and if this is not a bit crowded. But you know already that fields are mathematical concepts. For example, you can associate with every point in space and time a temperature, and thus create a temperature field. At the same time, there is an atmospheric pressure field. Both can happily exist simultaneously. But they are not ignoring each other. As you know, both are related with each other: If either changes this indicates a change of the other as well. Though this analogy is not exactly the same as the particle physics fields, and there are more things involved, the basic idea is the same. |
| Also the particle physics fields interact, and thus not ignore each other. [...] | Also the particle physics fields interact, and thus not ignore each other. [...] | ||
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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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