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1

Regarding the first part of your question,they have just inserted a complete set of basis because $|\phi>$ is a basis in some infinite dimensional Hilbert space (in your case), therefore sum (integral) of all such bases is identity on the Hilbert space. Note that in second part $\langle r|\phi\rangle$=$\phi(r)$.

1

So first of all, the first equation you gave is only correct, if the $|ø\rangle$ form a basis. It has nothing to do with "in which basis they are". The easiest way to understand this is probably with a 3D vector-analogy. So if $b_i$, $i=1\dots3$ form a basis, for any vector $v$ it is legitimate to write $$v=\sum_{i=1}^3 b_i (b_i\cdot v)$$ There, the ...

2

The formula given is horribly unenlightening because it does not seem to use the fundamental fact about differential forms that they are alternating and thus adds $r$ equal terms, it also does not provide the connection to index notation that it supposedly tries to. Let us first understand the idea of the interior product. An $r$-form is something with $r$ ...

0

\begin{aligned} H|ψ\rangle =...=a ((c1+c2)|1\rangle+(c1-c2)|2\rangle)=& E|ψ\rangle. \end{aligned} Now you just have to remember how you have defined $|ψ\rangle$. Apply this definition in the equation above and knowing that $|1\rangle$ and $|2\rangle$ are independent, you can easily find $c_i$.

0

Sometimes it's helpful to consider finite dimensional special case to get insights on a problems like these. Let $P$ and $L$ be in $\mathbb{C}^{n \times n}$. If $P^2 = P$, then $Pv = \lambda v$ leads to $\lambda^2 = \lambda$, i.e., the eigenvalues are in $\{0, 1\}$. In the context of quantum, we consider Hermitian operators. In this case, we can find some ...

1

Without further constraints you cannot say anything about $f$. Observe that the identity is idempotent and $id \circ f = f$ holds for any function $f$ on any set. Alright. If $P$ is not the identity, it cannot be invertible, thus $\ker(P) \neq 0$. Idempotent operators are basically projection operators, they are diagonalizable and only have eigenvalues $0$ ...

5

Isn't your Ex 3 such an example? In any event, you might have trouble producing a pure photon state $\left|\,001\right>$, but you should have no trouble creating a coherent state. Just turn on your laser pointer. The coherent state is a superposition of single photon states, summing over all single photon number states. You can't impose a cutoff. ...

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