From Quantum Mechanics to Lagrangian Densities

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Presentation transcript:

From Quantum Mechanics to Lagrangian Densities Just as there is no “derivation” of quantum mechanics from classical mechanics, there is no derivation of relativistic field theory from quantum mechanics. The “route” from one to the other is based on physically reasonable postulates and the imposition of Lorentz invariance and relativistic kinematics. The final “theory” is a model whose survival depends absolutely on its success in producing “numbers” which agree with experiment.

Summary: Quantum Mechanics The ten minute course in QM. Momentum Becomes an operator and use The Hamiltonian becomes an operator.

Physical interpretation of the wave function. This condition places a strong mathematical condition on the wave function.

Note that the Schrodinger equation reflects this relationship

Quantization arises from placing boundary conditions on the wave function. It is a mathematical result! L = r x P = r x

A “toy” model postulate approach to quantum field theory

- p - - + - p

Note that *(r,t) (r,t) does not represent the probability per unit volume density of the particle being at (r,t).

- -

The resulting “wave equation”:

The “negative energy” states arose from This emerges from starting out with . We know the energy of a real particle can’t be < 0.

Suppose the mass, m, is zero: This is the same equation we derived from Maxwell’s equations for the A vector (except, of course, above we have a scalar, ).

Following the same derivation used for the A vector we have: Solving the wave equation Following the same derivation used for the A vector we have: This is just like the E & M equation – except for the mass term. We try a solution of the form

We can see that after taking the partial derivatives there is a condition on the components of k and k0. Note, now the 4-dimensional dot product cannot = 0. The k and k0 (with p= k and p0 =  k0 ) must satisfy the same conditions as a relativistic particle with rest mass, m.

We have the following two linearly independent solutions to the “wave equation”: The most general (complete) solution to the wave equation is

The field operator for a neutral, spin =0, particle is creates a single particle with momentum p= k and p0 =  k0 at (r,t) Destroys a single particle with momentum p= k and p0 =  k0 at (r,t)

and the Euler-Lagrange equations give F = ma Lagrangians and the Lagrangian Density Recall that, and the Euler-Lagrange equations give F = ma In quantum field theory, the Euler-Lagrange equations give the particle wave function!

This calls for a different kind of “Lagrangian” -- not like the one used in classical or quantum mechanics. So, we have another postulate, defining what is meant by a “Lagrangian” – called a Lagrangian density. d/dt in the classical theory

Note that the Lagrangian density is quadratic in (r,t) and the Lorentz invariance is satisfied by using µ and µ

We can apply the Euler Lagrange equations to the above L: This part has a very simple result but it is hard to carry out. This part is easy.

Finally, with this Lagrangian density The Euler-Lagrange Equations give the wave equation for the neutral spin = 0 particle.

Summary for neutral (Q=0) scalar (spin = 0) particle, , with mass, m. Lagrangian density wave equation field operator

Charged (q = ±e) scalar (spin =0) particle with mass, m

charged scalar particle 8 terms cancel charged scalar particle

From the Lagrangian density and the Euler-Lagrange equation we can derive the wave equation

creates positively charged particle with momentum p= k and p0 =  k0 at (r,t) destroys negatively charged particle with momentum p= k and p0 =  k0 at (r,t) destroys negatively charged particle with momentum p= k and p0 =  k0 at (r,t) creates negatively charged particle with momentum p= k and p0 =  k0 at (r,t)

The scalar field (which represents a boson) must also satisfy a special boson commutation property: Creation and annihilation operators with the same k don’t commute. Everything else commutes.

Example of how the commutation relation is used We will use this when we calculate the charge of a particle. Later we will find that fermions satisfy a different commutation relation.

What follows is for the graduate students.

For the graduate students:

For the undergraduates: You can just remember this

Before taking the partial derivative, it is helpful to rewrite the L. Let  =  in this summation, so it does not become confused with the index of /x -- used later.

For the undergraduates: You can just remember this