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I am reading the Many-body quantum theory in condensed matter physics by Henrik Bruus. I am stuck at Eq. (4.51): The unitary transformation is given by $$U = \exp \left(i \int d\vec{r} \rho(\vec{r}) \phi(\vec{r}) \right)\quad ,$$ where $\rho(\vec{r}) = \Psi^\dagger(\vec{r})\Psi(\vec{r})$ and $$\tilde{\Psi}(\vec{r}) = U\Psi(\vec{r})U^{-1} \quad ,$$ where $\Psi(r)$ is the position operator. Then we have $$\tilde{\Psi}(\vec{r}) = \Psi(\vec{r})\exp\left(-i\phi(\vec{r})\right) \quad .$$ According to the book, this can be derived from a functional differential equation \begin{align}\frac{\delta}{\delta \phi(\vec{r'})}\tilde{\Psi}(\vec{r}) = i\tilde{\Psi}(\vec{r})\delta(\vec{r} - \vec{r'}) \quad .\end{align} I tried to multiply both side with $\phi(\vec{r})$ and integrate. but it just derives the above functional differential equation. I found an easy derivation from Deriving the unitary transformation for phase shifts. I understand this, but I want to understand Bruus' derivation.

How can I derive the third equation from the fourth one?

Tobias Fünke
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YJK
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  • Is there a sign error? In either the exponential of the suggested solution or in the on the RHS of the differential equation? – Tobias Fünke Dec 26 '22 at 19:24
  • Despite that, here is a guess/hint: Show that $\tilde\Psi$ is a solution to the differential equation subject to the initial value $\tilde\Psi[\phi=0] = \Psi$. It remains to show that this solution is unique. To do so, suppose there is another function $\psi$ obeying this initial value problem. Define $\varphi=\psi e^{i\phi}$ and show that the functional derivative is zero as well as $\varphi[\phi=0] = \psi[0]=\Psi$. This shows that $\tilde\Psi = \psi$, so the solution is unique. Again, this is a guess, but should be worth a trial. – Tobias Fünke Dec 26 '22 at 19:43

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