Complete proof of Von Neumann's uniqueness theorem I am looking for a complete detailed proof of Von Neumann's uniqueness theorem. I found two books. The first is a book written by Derezinski and Gerard ("Mathematics of quantization and quantum fields") and the second is 
"General principles of quantum field theory" of Bogoliubov et al.
Their proofs are detailed, but curiously they propose as an exercise nearly the same part of it.
Derezinski writes:
Proposition 4.33 Let us equip X with a Euclidean structure. Let P := $|\Psi_0\rangle \langle \Psi_0 |$ where $\langle x|\Psi_0\rangle \in L^2 (x)$ is given by
$$\Psi_0(x) = \pi^{-d/4} \exp(- x^2/2)$$
(It is a projection on the ground state of coherent states).
He writes that: 
1) $ P = P^2 = P^\dagger$ 
2) $ \pi^{d/2} P \exp(X^2) f(X) P = P \int f(x)~dx, ~ \mbox{for} ~f \in L^1 (x) \\ $. Here X is the position operator and x the position variable.
3) He proves that 
$$P = (2\pi)^{-d} \int e^{-\eta^2/4 - q^2/4 + iq\eta/2 } e^{i \eta X} e^{iq D}dq d\eta.$$
A function of X position operator and D momentum operator, the same formula that is given in the book of Bogoliubov to define P.
he writes that 1) is obvious, he explains how to get the formula in 3) from 1)
and 2) is left to the reader as an exercise.
I wrote the formula 3) a la Bogoliubov here https://math.stackexchange.com/q/3308183/ 
We have also like in 2) to prove that P exp(z* z) W(z) P is a projector but between the 2 P it is not an integrable function but the displacement operator that occurs in coherent states. and it is also left to the reader!
Could you prove these points? 
 A: Sorry, I will not address a formula which I have to go elsewhere to see. 
1) and 2) are textbook applications of Dirac's notation:
1) 
$$\langle \Psi_0 | \Psi_0\rangle = \int dx  \langle \Psi_0 | x\rangle\langle x | \Psi_0\rangle= \int \frac{dx}{\pi^{d/2}} e^{-x^2}=1.$$ Hence $ P=|\Psi_0\rangle\langle \Psi_0 |=P^\dagger $, while $P^2=    | \Psi_0\rangle\langle \Psi_0 | \Psi_0\rangle\langle \Psi_0 |= | \Psi_0\rangle\langle \Psi_0 |  =P $.
2)
$$ \pi^{d/2} | \Psi_0\rangle\langle \Psi_0 | 1\!\! 1   e^{X^2} f(X)      | \Psi_0\rangle\langle \Psi_0 |  = \\  
\pi^{d/2} \int dx | \Psi_0\rangle\langle \Psi_0    |x\rangle\langle x | e^{X^2} f(X)      | \Psi_0\rangle\langle \Psi_0 |  = \\
\pi^{d/2} \int dx | \Psi_0\rangle \frac{e^{-x^2/2}}{\pi^{\pi^{d/4}}} e^{x^2} f(x) \frac{e^{-x^2/2}}{\pi^{\pi^{d/4}}}      \langle \Psi_0 |  = \\  \therefore  ~~~~~~ 
  \int dx   f(x)   | \Psi_0\rangle       \langle \Psi_0 |  = \int dx   f(x)   P.
 $$
A: An opérator may have different définitions. Here Derezinski defines P as the orthogonal projection on the normed coherent state $\Phi_0$ centered on 0.
So $P = |\Phi_0 \rangle \langle \Phi_0|$
He proves in the link i gave above that P is equal to the integral
$$P = (2 \pi) ^{-d} \int e^{-\eta^2 /4} e^{-q^2 /4} e^{i q \eta /2 } e^{i \eta  x } e^{i q D } d\eta dq$$
$$P = (2 \pi) ^{-d} \int e^{-(\frac{\eta}{\sqrt{2}})^2 /2} e^{-(\frac{\eta}{\sqrt{2}})^2 /2} e^{i \frac{q}{\sqrt{2}} \frac{\eta}{\sqrt{2}} } e^{i \frac{\eta}{\sqrt{2}}  x\sqrt{2} } e^{i \frac{q}{\sqrt{2}} D\sqrt{2} } d\eta dq$$
In his book Bogoliubov gives the r.h.s of the equality as the the definition of the operator P.
After several changes of variables it reads
$$P = \int W(z) \pi^{-d} e^{-zz^* /2} d\eta dq$$
Here $z = q + i \eta$ and a is the annihilation operator with
$$W(z) = e^{-zz^* /2}e^{za^\dagger}e^{-z^* a}$$ 
it is then asked to prove that P is a projector and that $Pe^{zz^*/2}W(z)P = P$. This becomes straightforward if P = $|\Phi_0 \rangle \langle \Phi_0|$ (proof by Derezinski)
$$Pe^{zz^*/2}W(z)P =  |\Phi_0 \rangle \langle \Phi_0|e^{za^\dagger}e^{-z^* a}|\Phi_0 \rangle \langle \Phi_0|$$
As the operator a annihilates $\Phi_0$ all the only remaining terms in the exponentials are the identities and the result is P. (I would have preferred to use the only hints given by Bogoliubov!).
A: I found the solution of the execise proposed by Bogolov It only uses the properties of the CCR representations by the W(z).
we have to prove that
$\Pi = \int W(x+iy) \rho (x+iy) dx dy  $
with $\rho(z) = (1/\pi) exp(-|z|^2/2)$ is a projector.
Due to the properties of the W it is equal to
$\int exp (-(x^2 + y^2)/2) \int [W(Z) exp(z Z^*) \rho(Z) dX dY] \rho (x+iy) dx dy = $
$$(1/\pi)\iint exp(-(x^2 + y^2)) exp (x Z^* + y i Z^*)dx dy W(Z)\rho(Z) dX dY $$
We recognize in $\int exp(-(x^2 + y^2)) exp (x Z^* + y i Z^*)dx dy$
a gaussian integral 
I(A,a,b) = 
$$\iint e^{ - \frac{1}{2} \langle x,y|A|x,y\rangle} e^{\langle x,y|a,b\rangle} dx dy =$$
$$ \frac {2 \pi}{ \sqrt {det A}} e^{\langle a,b|A^{-1}|a,b\rangle}$$
where A is a 2 by 2 matrix with det A > 0 and where a and b can be complex numbers. Here A = 2 Id and $a = Z^*,b = iZ^* $
det A = 4 and the exponential of $ \langle Z^*, iZ^*|Id /|Z^*, iZ^* \rangle  $ is exp (0) = 1
this gaussian intégral is equal to $\pi$ ans so we get $$\Pi^2 = \Pi$$
