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): $$\begin{align} &\hat{Q}\ |\rangle &=\ & 0 \\ &\hat{Q}\ |\vec{x}_1\rangle &=\ & q\ |\vec{x}_1\rangle \\ &\hat{Q}\ |\vec{x}_1\vec{x}_2\rangle &=\ & 2q\ |\vec{x}_1\vec{x}_2\rangle \\ &... \end{align} \tag{1}$$ The operator $\hat{Q}$ can be written in a basis-independent way: $$\hat{Q} = q\hat{N} = q\int d^3x\ \hat{\psi}^\dagger(\vec{x})\hat{\psi}(\vec{x}) \tag{2}$$ 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: $$\begin{align} &\hat{U}(f)\ |\rangle &=\ & |\rangle \\ &\hat{U}(f)\ |\vec{x}_1\rangle &=\ & e^{iqf}\ |\vec{x}_1\rangle \\ &\hat{U}(f)\ |\vec{x}_1\vec{x}_2\rangle &=\ & e^{2iqf}\ |\vec{x}_1\vec{x}_2\rangle \\ &... \end{align} \tag{3}$$ It is easy to guess that $\hat{U}(f)$ can be written in a basis-independent way: $$\hat{U}(f) = e^{i\hat{Q}f} \tag{4}$$ 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)): $$\begin{align} &\hat{U}(f)\ |\rangle &=\ & |\rangle \\ &\hat{U}(f)\ |\vec{x}_1\rangle &=\ & e^{iqf(\vec{x}_1)}\ |\vec{x}_1\rangle \\ &\hat{U}(f)\ |\vec{x}_1\vec{x}_2\rangle &=\ & e^{iqf(\vec{x}_1)}\ e^{iqf(\vec{x}_2)}\ |\vec{x}_1\vec{x}_2\rangle \\ ... \end{align} \tag{5}$$

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

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

  • 1
    $\begingroup$ One thought is that, with a discrete number of points $N$, $\prod_{j=1}^{N} e^{i \hat{Q} f(\vec{x}_j)} = {\rm exp}[ \sum_{j=1}^N i \hat{Q} f(\vec{x}_j)]$ would work. In the limit $N\rightarrow \infty$ you'd get something like your second bullet point. But I guess the continuum limit is tricky, maybe one should replace $f$ or $\hat{Q}$ with a density. Another thought, is that the Hilbert space you get when quantizing a gauge theory depends on the gauge you choose, and in the end you want to restrict to a physical subspace. So, the states you are constructing should be identified. $\endgroup$
    – Andrew
    Commented Jul 22, 2020 at 18:39

1 Answer 1


I'm quite sure the answer is $$\hat{U}(f) = e^{iq\int d^3x\ f(\vec{x})\hat{\psi}^\dagger(\vec{x})\hat{\psi}(\vec{x})}$$ 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 $$\begin{align} \hat{U}(f) &= e^{iq\int d^3x\ f\ \hat{\psi}^\dagger(\vec{x})\hat{\psi}(\vec{x})} \\ &= e^{iqf\int d^3x\ \hat{\psi}^\dagger(\vec{x})\hat{\psi}(\vec{x})} \\ &= e^{iqf\hat{N}} \\ &= e^{i\hat{Q}f} \end{align}$$ 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 $$\begin{align} & \hat{Q}(f) |\vec{x}_1 ... \vec{x}_n\rangle \\ =& \int d^3x f(\vec{x})\hat{\psi}^\dagger(\vec{x})\hat{\psi}(\vec{x}) |\vec{x}_1 ... \vec{x}_n\rangle \\ =& \int d^3x f(\vec{x})\sum_{k=1}^n \delta(\vec{x}-\vec{x}_k) |\vec{x}_1 ... \vec{x}_n\rangle \\ =& \sum_{k=1}^n f(\vec{x}_k) |\vec{x}_1 ... \vec{x}_n\rangle \end{align}$$

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

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. $$e^{iq\hat{Q}(f)} |\vec{x}_1 ... \vec{x}_n\rangle = e^{iq\sum_{k=1}^n f(\vec{x}_k)} |\vec{x}_1 ... \vec{x}_n\rangle$$

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

  • 2
    $\begingroup$ Your conjecture is right, if I understand the notation in the question. I'm assuming that each $\vec x_n$ inside $|\cdots\rangle$ means that we've applied a factor of $\psi^\dagger(\vec x_n)$ to the vacuum state. In that case, your conjecture can be proven like this. Use the abbreviation $$Q(f)=\int f(x)\psi^\dagger(x)\psi(x).$$ Then $$[Q(f),\psi^\dagger(x)]=f(x)\psi^\dagger(x).$$ Equivalently, $$Q(f)\psi^\dagger(x)=\psi^\dagger(x)(f(x)+Q(f)).$$ This gives $$\exp(-iqQ(f))\psi^\dagger(x)\exp(iqQ(f))=\exp(iqf(x))\psi^\dagger(x).$$ To finish, use the fact that $Q(f)$ annihilates the vacuum state. $\endgroup$ Commented Jul 23, 2020 at 1:33
  • 1
    $\begingroup$ @ChiralAnomaly It took me a while to digest this sketched proof and fill the gaps. Very much appreciated. $\endgroup$ Commented Jul 25, 2020 at 18:03

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