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I try to understand why the equality/inequality you can see below holds.

Let $\mathfrak{h}$ be a separable Hilbert space and define the Fock space $\mathcal{F}:= \oplus_{N=0}^\infty\otimes^N\mathfrak{h}$. For $\varphi\in\mathfrak{h}$ let $(a(\varphi), D(\sqrt{\mathcal{N}}))$ and $(a^\dagger(\varphi), D(\sqrt{\mathcal{N}}))$ be the annihilation and creation operator with $\mathcal{N}$ the number operator. Then for all $\Psi \in D(\sqrt{\mathcal{N}}))$ is holds that

\begin{align} \vert\vert a^\dagger(\varphi)\Psi\vert\vert_\mathcal F &= \vert\vert\varphi\vert\vert_\mathfrak h \cdot \vert\vert \sqrt{\mathcal N + 1}\Psi \vert\vert_\mathcal F\\ \vert\vert a(\varphi)\Psi\vert\vert_\mathcal F &\leq \vert\vert\varphi\vert\vert_\mathfrak h \cdot \vert\vert\sqrt{\mathcal N + 1}\Psi\vert\vert_\mathcal F \end{align}

For the first i tried to use the definition of the operator \begin{align} \vert\vert a^\dagger(\varphi)\Psi\vert\vert_\mathcal F = \vert\vert (0,\varphi\cdot \psi_0, \sqrt{2}\varphi\otimes\psi_1, \sqrt{3}\varphi\otimes\psi_2,\dots)\vert\vert \end{align} but i am note sure how to proceed and how to deal with the tensor product inside the norm.

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1 Answer 1

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When you write $\mathcal F = \bigoplus_{n=0}^\infty \otimes^n \mathfrak h$, the $\oplus$ and $\otimes$ are understood as direct sum and tensor product of Hilbert spaces. They carry a hermitian product and a norm built from that of $\mathfrak h$.

This is the definition of the norm in the case of binary $\oplus$ and $\otimes$. They generalize easily to infinite operations.

For the direct sum, the different summand are taken to be orthogonal : if $x\in \mathfrak h,y \in \mathfrak f$, then $(x,y)\in \mathfrak h\oplus \mathfrak f$ has norm : $$\|(x,y)\|_{\mathfrak h\oplus \mathfrak f}^2 = \|x\|_{\mathfrak{h}}^2 + \| y\|_{\mathfrak f}^2$$

For the tensor product, the hermitian product is defined by , if $x,x' \in \mathfrak h$, $y,y'\in \mathfrak f$ : $$ (x\otimes y , x'\otimes y')_{\mathfrak h\otimes \mathfrak f} = (x,x')_{\mathfrak h} (y,y')_{\mathfrak f}$$ In particular, $\|x\otimes y\|_{\mathfrak{f}\otimes \mathfrak f} = \|x \|_{\mathfrak h} \| y\|_{\mathfrak f}$

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  • $\begingroup$ thank you very much! $\endgroup$
    – uzizi_1
    May 4, 2021 at 9:15

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