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In our class the Fock Space $\mathcal{H}$ was defined as $\mathcal{H} := \mathcal{H}_0 \oplus \mathcal{H}_1 \oplus \dotsb$ where $\mathcal{H}_i$ denotes the $i$-Particle tensor hilbert space of $i$ particles. Further the creation operator $a_\varphi^\dagger : \mathcal{H} \rightarrow \mathcal{H}$ was introduced, which it acts as follows for any $\varphi \in \mathcal{H}_1$:

$$ a_\varphi^\dagger (\psi_0, \psi_1, \psi_2, ...) = (\psi'_0, \psi'_1, \psi'_2,...) $$ where

$$ \psi'_i(\vec{r}_1, ..., \vec{r}_i) = P_\sigma (\psi_{i-1} \otimes \varphi) (\vec{r}_1, ..., \vec{r}_i) $$ where $P_\sigma$ denotes the (anti)-symetrization operator.

Question:

Wouldn't these definitions for a state $\psi = (0,...,0,\psi_N,0,...) \in \mathcal{H}$ with a well defined particle imply, that after I create another particle with $a_\varphi^\dagger$ I end up with a state, which is not an eigen state of the particle number operator? I.e:

$$ a_\varphi^\dagger(0,...,0,\psi_N,0,...) = (\varphi, 0, ...,0,\psi'_{N+1}, 0, ...) $$

which as Entries as well in the $\mathcal{H}_1$ component as in the $\mathcal{H}_{N+1}$ component and therefore no well defined particle number.

Or does $a_\varphi^\dagger$ act differently on the $\mathcal{H}_0$ (vacuum) component depending, on whether there are already particles present in the other components of $\mathcal{H}$ or not?

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  • $\begingroup$ The creation operator essentially "shifts" all components that are different from zero at once. Let me also stress that the vacuum has first component one, and not zero ( and in general the state has a nonzero probability of having no particles if and only if the first component is different from zero). $\endgroup$ – yuggib Jan 19 '17 at 19:53
  • $\begingroup$ You should think of the "global" $a^\dagger$ as patching together all the $a^\dagger_n: H_n\to H_{n+1}$, since every $n$-particle state is turned into an $n+1$-particle state. In particular, you always end up with an eigen-state of the number operator in this special case. $\endgroup$ – Phoenix87 Jan 19 '17 at 20:21
  • $\begingroup$ @JanLukasBosse Regarding the statement in your title, "Creation operators creating non-pure states from pure states": since $a^\dagger$ is defined from the Fock space onto the Fock space the result of its action on a pure state will be a pure state. A non-pure state would be a mixed state described by a density matrix. $\endgroup$ – udrv Jan 19 '17 at 21:30
  • $\begingroup$ @yuggib So basically the entry in the first (vacuum) component tells me the probability of having no particle at all? And hence there is only a particle in the second (one particle component) created, if there was a finite probability of having no particle in the previous state? $\endgroup$ – Jan Lukas Bosse Jan 20 '17 at 9:29
  • $\begingroup$ @JanLukasBosse yes, the creation operator acting on the vacuum component (different from zero) gives a one-particle component. $\endgroup$ – yuggib Jan 20 '17 at 11:39

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