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Local gauge transformation in Fock space of charged particles

Physics Asked by Thomas Fritsch on August 25, 2021

I’m currently fiddling around with gauge-phase-transformations in Fock space.
Especially, I’m trying to write a local gauge-phase-transformation as
an operator in a basis-independent way.

Here is what I have so far.
Consider a system of indistinguishable particles (each with a charge $q$).

Total charge

Let’s take the total charge operator $hat{Q}$. It can be defined
by its action on the $n$-particle states (using Fock states in the position basis):
$$begin{align}
&hat{Q} |rangle &= & 0
&hat{Q} |vec{x}_1rangle &= & q |vec{x}_1rangle
&hat{Q} |vec{x}_1vec{x}_2rangle &= & 2q |vec{x}_1vec{x}_2rangle
&…
end{align} tag{1}$$

The operator $hat{Q}$ can be written in a basis-independent way:
$$hat{Q} = qhat{N} = qint d^3x hat{psi}^dagger(vec{x})hat{psi}(vec{x}) tag{2}$$
where $hat{N}$ is the total number operator, and
$hat{psi}^dagger(vec{x})$ and $hat{psi}(vec{x})$ are the canonical
creation and annihilation operators at position $vec{x}$.
It is easy to check that this operator (2) satifies the definition (1).

Global gauge transformation

Now let’s consider a global gauge-phase-transformation $hat{U}(f)$
with a global constant $f$. $hat{U}(f)$ can be defined by its action
on the $n$-particle states:
$$begin{align}
&hat{U}(f) |rangle &= & |rangle
&hat{U}(f) |vec{x}_1rangle &= & e^{iqf} |vec{x}_1rangle
&hat{U}(f) |vec{x}_1vec{x}_2rangle &= & e^{2iqf} |vec{x}_1vec{x}_2rangle
&…
end{align} tag{3}$$

It is easy to guess that $hat{U}(f)$ can be written
in a basis-independent way:
$$hat{U}(f) = e^{ihat{Q}f} tag{4}$$
And indeed, by using $hat{Q}$ from above
it can be verified that (4) satisfies the definition (3).

So far no problem.

Local gauge transformation

And now for the local gauge-phase-transformation $hat{U}(f)$
with a position-dependent function $f(vec{x})$.
Again $hat{U}(f)$ can be defined by its action on the $n$-particle states
(by generalizing the definition (3)):
$$begin{align}
&hat{U}(f) |rangle &= & |rangle
&hat{U}(f) |vec{x}_1rangle
&= & e^{iqf(vec{x}_1)} |vec{x}_1rangle
&hat{U}(f) |vec{x}_1vec{x}_2rangle
&= & e^{iqf(vec{x}_1)} e^{iqf(vec{x}_2)} |vec{x}_1vec{x}_2rangle

end{align} tag{5}$$

I was not able to write $hat{U}(f)$ in a basis-independent way
so that it will satisfy the definition (5).

  • $hat{U}(f) = int d^3x e^{ihat{Q}f(vec{x})}$
    is obviously wrong, because $hat{U}$ has the dimension of a volume,
    instead of being dimensionless.
  • $hat{U}(f) = e^{iint d^3x hat{Q}f(vec{x})}$
    is also wrong, because the exponent has the dimension of a volume,
    instead of being dimensionless.
  • $hat{U}(f) = int d^3x hat{psi}^dagger(vec{x})
    e^{ihat{Q}f(vec{x})} hat{psi}(vec{x})$

    is wrong, because when acting on the vacuum state
    the result is $hat{U}|rangle=0$ instead of $hat{U}|rangle=|rangle$.

Any ideas? Is it even possible?

One Answer

I'm quite sure the answer is $$hat{U}(f) = e^{iqint d^3x f(vec{x})hat{psi}^dagger(vec{x})hat{psi}(vec{x})}$$ But I wasn't able to prove it. So it is only a conjecture.

For the special case of $f(vec{x})=f=text{const}$, the above reduces to $$begin{align} hat{U}(f) &= e^{iqint d^3x f hat{psi}^dagger(vec{x})hat{psi}(vec{x})} &= e^{iqfint d^3x hat{psi}^dagger(vec{x})hat{psi}(vec{x})} &= e^{iqfhat{N}} &= e^{ihat{Q}f} end{align}$$ which is just the global gauge transformation from equation (4) in the question.


@ChiralAnomaly in his comment has already sketched an elegant proof using operator algebra.

Here is another proof on a more elementary level.

Let's use the abbreviation $$hat{Q}(f)=int d^3x f(vec{x})hat{psi}^dagger(vec{x})hat{psi}(vec{x}).$$

By applying $hat{Q}(f)$ to an $n$-particle state we get $$begin{align} & hat{Q}(f) |vec{x}_1 ... vec{x}_nrangle =& int d^3x f(vec{x})hat{psi}^dagger(vec{x})hat{psi}(vec{x}) |vec{x}_1 ... vec{x}_nrangle =& int d^3x f(vec{x})sum_{k=1}^n delta(vec{x}-vec{x}_k) |vec{x}_1 ... vec{x}_nrangle =& sum_{k=1}^n f(vec{x}_k) |vec{x}_1 ... vec{x}_nrangle end{align}$$

By applying $hat{Q}(f)$ again and again we get (for $j=1,2,3,...$) $$left(hat{Q}(f)right)^j |vec{x}_1 ... vec{x}_nrangle = left(sum_{k=1}^n f(vec{x}_k)right)^j |vec{x}_1 ... vec{x}_nrangle$$

By applying $sum_{j=0}^infty frac{1}{j!}(iq)^j$ to both sides of this equation we get the Taylor-series of the exponential function. $$e^{iqhat{Q}(f)} |vec{x}_1 ... vec{x}_nrangle = e^{iqsum_{k=1}^n f(vec{x}_k)} |vec{x}_1 ... vec{x}_nrangle$$

Now it is easy to prove equation (5) of the question: $$begin{align} & hat{U}(f) |vec{x}_1 ... vec{x}_nrangle =& e^{iqhat{Q}(f)} |vec{x}_1 ... vec{x}_nrangle =& e^{iqsum_{k=1}^n f(vec{x}_k)} |vec{x}_1 ... vec{x}_nrangle =& prod_{k=1}^n e^{iqf(vec{x}_k)} |vec{x}_1 ... vec{x}_nrangle end{align}$$

Correct answer by Thomas Fritsch on August 25, 2021

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