Question on Canonical Quantization

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SUMMARY

This discussion centers on the canonical quantization process in Quantum Field Theory (QFT), specifically the relationship between canonical coordinate operators, canonical momentum operators, and field operators. The canonical quantization imposes commutation relations between these operators, leading to the Hamiltonian being represented as a densely-defined self-adjoint linear operator. The participants emphasize the necessity of solving classical field equations and utilizing creation and annihilation operators to establish the connection between these operator types. Recommended readings include "An Introduction to Quantum Field Theory" by Peskin & Schroeder and "The Quantum Theory of Fields" by Steven Weinberg.

PREREQUISITES
  • Understanding of Lagrangian density in Quantum Field Theory
  • Familiarity with canonical quantization procedures
  • Knowledge of creation and annihilation operators in quantum mechanics
  • Basic principles of Hamiltonian mechanics
NEXT STEPS
  • Study the classical field equations in Hamiltonian mechanics
  • Learn about the quantization of fields and the role of creation and annihilation operators
  • Read "An Introduction to Quantum Field Theory" by Peskin & Schroeder
  • Explore the first volume of "The Quantum Theory of Fields" by Steven Weinberg
USEFUL FOR

This discussion is beneficial for students and researchers in theoretical physics, particularly those studying Quantum Field Theory and seeking to understand the intricacies of operator relationships in quantization processes.

Inquisitive_Mind
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This semester I took a course in Quantum Field Theory. It is difficult (the professor assumes you know everything) and I have so many questions...

Starting with a lagrangian density, I was told that canonical quantisation is a procedure where we impose the usual commutation relation between the cannonical coordinates and their conjugate momentum. In formulae,

[tex]S = \int d^3 x dt L(\phi_i, \partial_{\miu} \phi_i)[/tex]
[tex][\phi_{i}(x), \frac{\delta S}{\delta \partial_{0} \phi_j}(x^{'})] = i\hbar \delta(x-x^{'})\delta_{ij}[/tex]

Now the hamiltonian can be obtained from the lagrangian and this above procedure turns the hamiltonian into an operator (which is in terms of the "cannonical coordinate operators" and "cannonical momentum operators").

However, my biggest question is here: there is another representation of the lagrangian in the "field operators" (i.e. operators which increase the number of particles at point x by 1). How may one turn the hamiltonian in terms of the "cannonical coordinate operators" and "cannonical momentum operators" to one in terms of the "field operators"?

To be specific, I am asking the question is there a definite relationship between the "cannonical coordinate operators", the "cannonical momentum operators" and the "field operators"? I suspect that there may be a relationship like

[tex]\Psi (x) = \frac {1}{\sqrt{2}}(\alpha \phi (x) - i \frac{1}{\alpha} \pi (x))[/tex]

where [tex]\Psi(x)[/tex] is the field operator and [tex]\pi (x)[/tex] is the cannonical momentum operator. The usual commutation rules (for field operators) are then satisfied. However, the definition is ambuiguous up the the factor [tex]\alpha[/tex]? Am I completely wrong in doing the things that way?

I suspect that the relation I seek (between phi, pi and psi) can only be given if we can diagonalise the hamiltonian in the form of harmonic oscillators. Is this true?

I am really in stress...because I really don't understand and I can't even finish the assignment problems. Please help!
 
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Well i am sure you know how to write creation and annihilation operators in terms of q (position operator) and p (momentum operator) in the case of a harmonic oscillator. This is done in ordinary QM. Now all you have to do is convert these equations along with the commutation relations for these operators, in terms of field theory.

[tex]q --> \phi[/tex]
[tex]p --> \pi[/tex]

and so on

marlon
 
Inquisitive_Mind said:
This semester I took a course in Quantum Field Theory. It is difficult (the professor assumes you know everything) and I have so many questions...

Starting with a lagrangian density, I was told that canonical quantisation is a procedure where we impose the usual commutation relation between the cannonical coordinates and their conjugate momentum.

Yes.That's how we do in QM,but not in QFT.

Inquisitive_Mind said:
In formulae,

[tex]S = \int d^3 x dt L(\phi_i, \partial_{\miu} \phi_i)[/tex]
[tex][\phi_{i}(x), \frac{\delta S}{\delta \partial_{0} \phi_j}(x^{'})] = i\hbar \delta(x-x^{'})\delta_{ij}[/tex]

Now the hamiltonian can be obtained from the lagrangian and this above procedure turns the hamiltonian into an operator (which is in terms of the "cannonical coordinate operators" and "cannonical momentum operators").

Nope.The quantization of the classical hamiltionian of the field turns the Hamiltionian into a densly-defined self-adjoint linear operator.

Inquisitive_Mind said:
However, my biggest question is here: there is another representation of the lagrangian in the "field operators" (i.e. operators which increase the number of particles at point x by 1). How may one turn the hamiltonian in terms of the "cannonical coordinate operators" and "cannonical momentum operators" to one in terms of the "field operators"?

Yoy need to solve the classical field equations (of Hamilton) and then use the amplitudes which would become,by quantization creation & annihilation operators.

Inquisitive_Mind said:
To be specific, I am asking the question is there a definite relationship between the "cannonical coordinate operators", the "cannonical momentum operators" and the "field operators"? I suspect that there may be a relationship like

[tex]\Psi (x) = \frac {1}{\sqrt{2}}(\alpha \phi (x) - i \frac{1}{\alpha} \pi (x))[/tex]

where [tex]\Psi(x)[/tex] is the field operator and [tex]\pi (x)[/tex] is the cannonical momentum operator.


Trust me,this is QFT,there are ONLY FIELD OPERATORS...


Inquisitive_Mind said:
Please help!

Please,read from Peskin & Schroeder and Weinberg (first volume).

Daniel.
 

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