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 Posted: Sun Jun 19, 2005 5:33 am Post subject:
Hey Dirac! What about bosons?
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Dirac wrote:
"According to relativistic quantum mechanics,
the energy spectrum of a massive particle
includes negative energy states.
However, an electron cannot tunnel
into the negative energy region
because THE NEGATIVE ENERGY BANDS ARE ALREADY FULL.
And if an electron were to be knocked out
of a negative energy band, the resulting hole
would be an anti-electron, ie a positron."
Which is fine - for FERMIONS.
But bosons bands CANNOT be full,
because bosons don't mind sharing.
So what stops bosons from tunnelling
into negative energy states? |
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Frank Hellmann
science forum beginner
Joined: 16 May 2005
Posts: 26
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| Posted: Sun Jun 19, 2005 10:24 pm Post subject:
Re: Hey Dirac! What about bosons?
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Dirac builds a theory for the electron.
This theory is historic, it predicted the positron but it can't
describe Bosons and has several other problems as well.
That's why Quantum Field Theory was developed which works for Boson,
Fermions and all known forms of matter, and all forces except for
gravity. In QFT there is no negative energy see, the negative energy
see is best considered a modell/image/metapher historically used
productively at some point in physics, but eventually replaced by a
consistent theory.
In Quantum Field Theory the negative frequency(formerly energy) modes
appear as thecreation operators of a different sort of particle, the
anti particle. A state with several such negative frequency modes
occupied has neithertheless positive energy. The physical energy
operator is no longer directly equivalent to the frequency part of a
wavefunction, and is always positive.
Frank.
timrobinson@paradise.net.nz wrote:
| Quote: | Dirac wrote:
"According to relativistic quantum mechanics,
the energy spectrum of a massive particle
includes negative energy states.
However, an electron cannot tunnel
into the negative energy region
because THE NEGATIVE ENERGY BANDS ARE ALREADY FULL.
And if an electron were to be knocked out
of a negative energy band, the resulting hole
would be an anti-electron, ie a positron."
Which is fine - for FERMIONS.
But bosons bands CANNOT be full,
because bosons don't mind sharing.
So what stops bosons from tunnelling
into negative energy states? |
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Hendrik van Hees
science forum addict
Joined: 15 May 2005
Posts: 60
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| Posted: Sun Jun 19, 2005 10:24 pm Post subject:
Re: Hey Dirac! What about bosons?
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timrobinson@paradise.net.nz wrote:
| Quote: | Which is fine - for FERMIONS.
But bosons bands CANNOT be full,
because bosons don't mind sharing.
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That's why nowadays we use quantum field theory instead of Dirac's hole
model to describe relativistic particles.
Take free scalar bosons and use the momentum representation to quantize
the theory, i.e., you solve the Klein-Gordon equation (KGE) with a
plane-wave ansatz:
phi_{\vec{p},\pm}(x)=1/sqrt(2 \omega(\vec{p})) exp(-i p.x)
|_{p0=\pm \omega(\vec{p})
Here \vec{p} is the three-momentum vector and
p.x=p0 x0-\vec{p} \vec{x}
the Minkowski product. Further
\omega(\vec{p})=sqrt(\vec{p}^2+m^2) (m=invariant mass of the particles).
It is not possible to interpret phi as "a wave function" in the same
sense as you do for wave function of the non-relativistisich
Schrödinger equation, if you like to interpret p0 as energy of a
quantum with momentum p, because then you would have quanta with
negative energy, and since \omega(\vec{p}) is unbounded, i.e., it can
grow as big as you want, you'd never get a stable ground state.
Especially bosons would like to go in the state of minimal energy, but
there would be none, they would fall into "nothing" sotosay, and you'd
get an infinite amount of energy out of nothing, a phenomenon never
observed.
This problem is very easily solved by just writing a creation operator
instead of an annihilation operator in the composition of the field
*operator* out of the plane-wave modes:
\hat{phi}(x)=\int d^3 p/(2 pi)^{3/2} [phi_{\vec{p},+} a(\vec{p})
+phi_{\vec{p},-}
b^{\dagger}(-\vec{p})]
So you interpret the modes with negative frequencies as anti-particles
with positive energy, and *momentum* \vec{p}, because, you can write as
well
\hat{phi}(x)=\int d^3 p/(2 pi)^{3/2} [phi_{\vec{p},+ a(\vec{p})
+phi_{\vec{p},+}^*
b^{\dagger}(\vec{p})]
where we have substituted \vec{p}->-\vec{p} in the integral with the
creation operator.
This is the socalled Feynman-Stückelberg trick.
The same trick of course works for Dirac (or Majorana) fermion fields,
and there is no need for an unobservable "Dirac sea" at all.
For fermions, the Dirac-sea construction leads to the socalled hole
theory which is equivalent to the field-theoretical formulation (at
least in the case of QED), but it has the shortcoming to be even more
complicated than the QFT formulation of QED and that it cannot be
generalized to bosons.
| Quote: |
So what stops bosons from tunnelling
into negative energy states?
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The simple fact that, thanks to the Feynman-Stückelberg trick, by
definition there are no states with negative energy since all the
quantum field theories are constructed such that the energy is positive
semi-definite.
--
Hendrik van Hees Texas A&M University
Phone: +1 979/845-1411 Cyclotron Institute, MS-3366
Fax: +1 979/845-1899 College Station, TX 77843-3366
http://theory.gsi.de/~vanhees/ mailto:hees@comp.tamu.edu |
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Jack Tremarco
science forum beginner
Joined: 19 Jun 2005
Posts: 7
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| Posted: Sun Jun 19, 2005 10:24 pm Post subject:
Re: Hey Dirac! What about bosons?
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The existence of negative energy states is due to the fact that the
variational equations of motion for Fermions are first order
differential equations.
For bosons they are second order differential equations, which do not
admit negative energy solutions. The energy spectrum positive
semi-definite. |
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Igor Khavkine
science forum Guru
Joined: 01 May 2005
Posts: 607
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| Posted: Tue Jun 21, 2005 7:15 am Post subject:
Re: Hey Dirac! What about bosons?
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On 2005-06-19, timrobinson@paradise.net.nz <timrobinson@paradise.net.nz> wrote:
| Quote: | Dirac wrote:
"According to relativistic quantum mechanics,
the energy spectrum of a massive particle
includes negative energy states.
However, an electron cannot tunnel
into the negative energy region
because THE NEGATIVE ENERGY BANDS ARE ALREADY FULL.
And if an electron were to be knocked out
of a negative energy band, the resulting hole
would be an anti-electron, ie a positron."
Which is fine - for FERMIONS.
But bosons bands CANNOT be full,
because bosons don't mind sharing.
So what stops bosons from tunnelling
into negative energy states?
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In this picture? Nothing. However, we don't see this happening around
us. One can only conclude that this picture does not apply to bosons.
In other words the frequency associated with each solution is not the
true energy. Whenever it's negative, it has to be prefixed with a minus
sign and gives a positive energy. The details are the
quantum-field-theoretic definition of bosinic theories.
Hope this helps.
Igor |
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