Diagonalizable Hamilton denisty?

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In summary, the conversation discusses working with a complex scalar field and its lagrange and hamilton densities. The hamilton density is shown to be diagonalizable using annihilation and creation operators, and it is demonstrated that the theory contains two particles with mass m. The process for finding the masses and calculating expressions for \phi^*(x), \nabla\phi(x), (\nabla\phi(x))^*, and H is also mentioned.
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>>S/Z<<
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Homework Statement


I'm working with a complex scalar field with the lagrange density [tex] L= \partial_{\mu} \phi^{\ast} \partial^{\mu} \phi - m^2 \phi^{\ast} \phi [/tex] And I've shown that's its hamilton density [tex] H= \int d^3 x ( \pi^{\ast} \pi + \nabla \phi^{\ast} \cdot \nabla \phi + m^2 \phi^{\ast} \phi ) [/tex]

Now the annihilation and creation operateurs are introduced. And I have to show that hamilton density H is diagonalizable by writing
[tex] \phi(x) = d^3 p \left( \frac{1}{(2 \pi)^3} \frac{1}{\sqrt{2E_p}} (a_{\vec{p}} e ^{-i p \cdot x} + b_{\vec{p}}^{\dagger} e^{ip \cdot x}) \right) [/tex]

I also have to show that the theory contains two particles with mass m. How do I do this? I don't really know how to get started? Neither with how to show it is diagonalizable or how to find the masses:(
 
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  • #2
>>S/Z<< said:

Homework Statement


I'm working with a complex scalar field with the lagrange density [tex] L= \partial_{\mu} \phi^{\ast} \partial^{\mu} \phi - m^2 \phi^{\ast} \phi [/tex] And I've shown that's its hamilton density [tex] H= \int d^3 x ( \pi^{\ast} \pi + \nabla \phi^{\ast} \cdot \nabla \phi + m^2 \phi^{\ast} \phi ) [/tex]

Now the annihilation and creation operateurs are introduced. And I have to show that hamilton density H is diagonalizable by writing
[tex] \phi(x) = \int d^3 p \left( \frac{1}{(2 \pi)^3} \frac{1}{\sqrt{2E_p}} (a_{\vec{p}} e ^{-i p \cdot x} + b_{\vec{p}}^{\dagger} e^{ip \cdot x}) \right) [/tex]

Use this to calculate an expression for [itex]\phi^*(x)[/itex] (Most physicists use [itex]{}^{\dagger}[/itex] instead of [itex]{}^{*}[/itex] for Hermitian conjugation ), [itex]\mathbf{\nabla}\phi(x)[/itex], [itex](\mathbf{\nabla}\phi(x))^*[/itex] and then [itex]H[/itex]. Carry out the integration over [tex]x[/itex], and you should get something like

[tex]H=\int\int d^3pd^3p' f(\textbf{p},\textbf{p}') \delta(\textbf{p}-\textbf{p}')[/tex]

The presence of the delta function means that all non-diagonal elements vanish (since it is zero when [itex]\textbf{p}\neq\textbf{p}'[/itex] ).
 

1. What is a diagonalizable Hamilton density?

A diagonalizable Hamilton density is a mathematical concept used in quantum mechanics to describe the energy states of a system. It is represented as a matrix that can be diagonalized, meaning that it can be simplified into a diagonal form, making it easier to calculate and interpret the energy levels of the system.

2. How is diagonalizable Hamilton density related to quantum mechanics?

Diagonalizable Hamilton density is a fundamental concept in quantum mechanics, as it helps to describe the possible energy states and transitions of a system. It is used to calculate the energy spectrum, wave functions, and other properties of quantum systems.

3. What are the benefits of using diagonalizable Hamilton density?

Diagonalizable Hamilton density offers several benefits, including simplifying calculations and providing a clear understanding of the energy levels and transitions of a quantum system. It also allows for the use of mathematical techniques such as diagonalization, which can provide further insights into the system.

4. How is diagonalizable Hamilton density determined experimentally?

Diagonalizable Hamilton density is typically determined experimentally by measuring the energy levels and transitions of a quantum system. These measurements are then used to construct a matrix representation of the Hamiltonian, which can then be diagonalized to determine the diagonalizable Hamilton density.

5. Can all Hamilton densities be diagonalized?

No, not all Hamilton densities can be diagonalized. In order for a Hamilton density to be diagonalizable, it must meet certain conditions, such as being a Hermitian matrix and having eigenvalues that are all real and distinct. If these conditions are not met, the Hamilton density cannot be diagonalized.

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