# Unitarily Inequivalent Representations

The definition of unitarily equivalent representations I am using is the one given here: https://en.wikipedia.org/wiki/Haag%27s_theorem.

Now in this text http://www.sa.infn.it/Massimo.Blasone/documents/cantrans.pdf we are given two sets of operators on some Hilbert space that both satisfy the canonical commutation relations from quantum field theory; $|0\rangle$ is the vacuum for one set of operators, and $|0(\theta)\rangle$ is the vacuum from the other. On page five of the author makes the claim that because $\langle 0|0(\theta)\rangle =0\,,$ we have two inequivalent representations. I am a little confused by why this is true.

I am also a little confused by why the author is claiming that acting by $U(\theta)$ on the vacuum leads out of the original Hilbert space when it seems everything there belongs to the same Hilbert space, the one originally defined, but maybe this is just a semantics issue.

• unfortunately the link to the lecture notes of Blasone is broken. Can you provide a new link or bibliographic details? Commented Sep 15, 2022 at 16:57

Let H be the prehilbertian vector space of the possible states of One particle. And F its associated Fock space. F has an orthonormal Hilbert occupancy number basis of vectors (each one is a sequence of entire numbers all equal to 0 except for a finite number of them). F is generated by finite linear combinations of these vectors AND by the converging $$\Sigma_{n_1,n_2,n_3,.}c(n_1,n_2,n_3,.) |n_1,n_2,n_3,..\rangle$$ where $$\Sigma_{n_1,n_2,n_3,.} |c(n_1,n_2,n_3,.)|^2$$ is finite. If dim(H) = 1 we have only one mode and the occupancy basis is (0) (1) (2) ... If dim(H) = 2 the sequences are (0,0) (0,1) (0,2) .... (1,0) (1,1) (1,2)....

A CCR representation is given by annihilation and création opérations $$a_k$$ and $$a^\dagger_k$$ (one for each mode).

Another representation is given by $$a_k + \theta_k$$ and $$a^\dagger_k + \theta^*_k$$. (here $$\theta$$ is a complex number. They are both canonical because $$[a_j,a^\dagger_k] = [a_j + \theta_j,a^\dagger_k + \theta_k^*]= \delta_{jk}$$ If these two representations are unitarily equivalent there is a unitary map U such that $$a_k U = U a_k + U \theta_k = U a_k + \theta_k U$$ for each mode. If $$|x\rangle = U |0\rangle$$ then $$\langle x|x \rangle = 1$$ and $$a_k |x\rangle = U (a_k + \theta_k) |0\rangle = \theta_k |x\rangle$$ this equality gives recurrence relations between the coordinates of x in the occupatin basis. One can easily find that if dim H = 1 (one mode) $$|x\rangle = c_0 |0\rangle + c_1 |1\rangle + .... = c_0 (1 + \theta_1 + \theta_1^2/2! ...)|0\rangle$$ so $$\langle x | x\rangle = |c_0|^2 exp (|\theta_1|^2)$$ there is no problem to find $$c_0$$ so that x is unitary. If dim H = 2 we find
$$\langle x | x\rangle = |c_{00}|^2 exp (|\theta_1|^2 + |\theta_2|^2)$$ if the set of modes is enumerable we have $$\langle x | x\rangle = |c_{00...}|^2 exp (\Sigma_k |\theta_k|^2)$$ this shows that when the sum does not converge, we cannot find $$c_{00...}$$ so thar $$\langle x | x\rangle = 1$$ and it is the case when all the $$\theta_k$$ are equal. In this case we get a contradiction with the existence of unitary equivalentrepresentations.

There is no need to consider that U | 0> is outside the Fock space (and orthogonal to it) because the unitary map U does not exist.

• That is the correct answer in my view. Commented Jul 18 at 6:55

The claim is that $$|0(\theta)\rangle$$ lies outside the Hilbert space built on the original vacuum $$|0\rangle$$.

To check that this true, consider the overlap of the new vacuum $$|0(\theta)\rangle$$ and the (unnormalized) basis states $$(a_k^\dagger)^n|0\rangle$$ generated from $$|0\rangle$$, taking into account that $$a_k(\theta) = a_k + \theta_k$$, $$a_k = a_k(\theta) - \theta_k$$, and $$a_k(\theta)\;|0(\theta)\rangle = 0$$. We have successively:

$$\begin{eqnarray} \langle 0|a_k^n|0(\theta)\rangle &=& \langle0|a_k^{n-1}[a_k(\theta)-\theta_k]|0(\theta)\rangle\\ &=& \langle 0|a_k^{n-1}a_k(\theta)|0(\theta)\rangle - \theta_k \langle 0|a_k^{n-1}|0(\theta)\rangle \\ &=& -\theta_k \langle 0|a_k^{n-1}|0(\theta)\rangle \\ &=& (-\theta_k)^2 \langle 0|a_k^{n-2}|0(\theta)\rangle \\ &=& \ldots\\ &=& (-\theta_k)^n \langle 0|0(\theta) \rangle\\ &=& 0 \end{eqnarray}$$

A similar result holds for basis states involving multiple modes $$k$$. In other words, the new vacuum $$|0(\theta) \rangle$$ is orthogonal not only on the original vacuum, but also on each of the canonical basis states generated from it. Hence it must lie outside the corresponding Hilbert space, despite being generated using a unitary $$U(\theta)$$ that is formulated in terms of $$a_k$$-s, $$a_k^\dagger$$-s.

• This is interesting. I find it a bit weird that you say $|0(\theta)\rangle$ lies outside the Hilbert space built on the original vacuum $|0\rangle;$ if it is true how you could even take the inner product of states living in different Hilbert spaces? I guess you mean the original Hilbert space, whatever it is, decomposes into a direct sum of Hilbert spaces, one which is generated by $|0\rangle$ and the other by $|0(\theta)\rangle\,...$ I am still unsure why they aren't unitarily equivalent though. Why couldn't there be a unitary transformation $U$ such that $a_k=U^{\dagger}a_k(\theta)U\,?$
– JLA
Commented Aug 14, 2015 at 17:18
• In fact haven't they really just shown that there is no nonzero state $|0(\theta)\rangle$ such that $a_k(\theta)|0(\theta)\rangle=0\,?$ If the two Hilbert spaces are different then how could you even write $a_k(\theta)=a_k+\theta_k\,?$
– JLA
Commented Aug 14, 2015 at 18:10
• @JLA: "the original Hilbert space, whatever it is, decomposes into a direct sum of Hilbert spaces, one which is generated by |0⟩ and the other by |0(θ)⟩" - this is correct. The reason it can happen has to do with the fact that the original Hilbert space is infinite dimensional. Otherwise you wouldn't be able to obtain <0|0(θ)>=0 from a finite superposition of orthogonal states.
– udrv
Commented Aug 14, 2015 at 23:59
• @JLA: "why they aren't unitarily equivalent" - the explanation of the term is given in the paper in the paragraph starting "Eq.(1.26) then.." following Eq.(1.33). It means "the unitary applied to |0> leads out of the original Hilbert (sub)space, so it is not a unitary on the original (sub)space, but between 2 orthogonal (sub)spaces/".
– udrv
Commented Aug 15, 2015 at 0:04
• As for "haven't they really just shown that there is no nonzero state |0(θ)⟩ such that ak(θ)|0(θ)⟩=0?", this is another good question. The answer is no. The explanation is given in the paragraph containing Eq.(1.34): the total number of particles in the vacuum |0(θ)⟩ is infinite in the infinite volume limit, but the density of particles per unit volume remains well-defined. This also means "the density of particles from each mode k is finite". Therefore the infinite volume state |0(θ)⟩ is not simply the null element of the original Hilbert space, which has 0 particle density for all modes.
– udrv
Commented Aug 15, 2015 at 0:31

For a succinct explanation of the orthogonality of the vaccua for the van Hove model (which seems to be what the initial pdf might have been referring to), see Section 2 of this paper.

Tne answer by Naima seems to be a bit of a roundabout way of saying that $$\lim \limits_{n \to \infty} \dfrac{n}{n}$$ does not exist because the numerator diverges.

$$\newcommand{\ket}[1]{|#1\rangle}$$ $$\newcommand{\braket}[2]{\langle #1|#2\rangle}$$

The $$\ket{x}$$ states in the example are known as coherent states (i.e., eigenvectors of $$a$$), and they are related to the single oscillator Hamiltonian eigenstates by

$$\ket{\alpha} = e^{-\frac{|\alpha|^2}{2}}\sum_{n = 0}^{\infty}\dfrac{\alpha^n}{\sqrt{n!}}\ket{n}, \tag{1}\label{cs}$$

which gives the following inner product

$$\braket{\alpha}{\alpha} = e^{-|\alpha|^2} e^{|\alpha|^2} = 1,$$

As one would expect. For a countable number of "modes", you would also get

$$\braket{\alpha_1 \alpha_2 \dots }{\alpha_1 \alpha_2 \ldots} = \braket{\alpha_1}{\alpha_1}\braket{\alpha_2}{\alpha_2}\ldots = \left(e^{-|\alpha_1|^2}e^{|\alpha_1|^2}\right)\left(e^{-|\alpha_2|^2}e^{|\alpha_2|^2}\right)\ldots = 1.$$

So anyway, $$|c_{00}|^2 = \exp(-\sum_k|\theta_k|^2)$$.

Now that I look at udrv's answer, I'm also unsure of that. Using the ansatz that coherent states are eigenstates of the anihilation operator, one can guess what $$\ket{0(\theta)}$$ is:

$$\ket{0(\theta)} = \ket{-\theta},$$

where $$\ket{-\theta}$$ is a coherent state. One can check that it is indeed anihilated by $$a(\theta)$$:

$$a(\theta)\ket{-\theta} = a\ket{-\theta} + \theta\ket{-\theta} = -\theta\ket{-\theta} + \theta\ket{-\theta} = 0.$$

But then we know the explicit form of $$\ket{-\theta}$$ in terms of QHO energy eigenstates (see $$\eqref{cs}$$), so

$$\braket{0}{0(\theta)} = e^{-\frac{|\theta|^2}{2}}.$$