Manuscript XVII-A@1 · XVII-A.tex · lemma · 711d7ee4

Lemma: Central smearing identity

Exact reconstruction of the named observable › The boundary smearing · lines 539–571 · proof included or linked

XVII-A@1#lem:smearing, XVII-A@1#eq:Q-identity

The boundary smearing satisfies

Q_{B} = U Q_{bulk} U^{†} .

Proof. For every positive integer $n$ ,

\frac{1}{2} \int_{- π /2}^{π /2} e^{\pm in t} d t = \frac{sin ( nπ /2 )}{n} .

Substituting (eq:boundary-field) into (eq:boundary-representative) gives

Q_{B} = n = 1 \sum^{K} \frac{( - 1 ) ^{n + 1} sin ( nπ /2 )}{π n} (b_{n} + b_{n}^{†}) .

If $n$ is even, the sine coefficient vanishes. If $n$ is odd, $(- 1)^{n + 1} = 1$ . Thus

Q_{B} = n = 1 \sum^{K} \frac{sin ( nπ /2 )}{π n} (b_{n} + b_{n}^{†}) = U Φ (0, 0) U^{†} .

Displayed formulas in this passage

XVII-A@1#eq:Q-identity · line 542
$Q_{B} = U Q_{bulk} U^{†} .$
unlabeled · line 550
$\frac{1}{2} \int_{- π /2}^{π /2} e^{\pm in t} d t = \frac{sin ( nπ /2 )}{n} .$
unlabeled · line 556
$Q_{B} = n = 1 \sum^{K} \frac{( - 1 ) ^{n + 1} sin ( nπ /2 )}{π n} (b_{n} + b_{n}^{†}) .$
unlabeled · line 564
$Q_{B} = n = 1 \sum^{K} \frac{sin ( nπ /2 )}{π n} (b_{n} + b_{n}^{†}) = U Φ (0, 0) U^{†} .$