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1

Yes, your instructor is correct. If you simply perform the substitution they suggest, you should see fairly quickly that the two integrals are equal. The first equality holds by trivially substituting $-\vec y$ for $\vec y$ in the preceding expression. The second is generically not true; however, you can do the two integrals separately, and since they ...


2

Technically speaking, there is no eigenvector $\lvert N_i\rangle$ for continuous spectrum, so the expression $\int \lvert N_i\rangle\langle N_i\rvert \mathrm{d}i$ does not make sense. There is a notion of generalized eigenvector as a distribution, but apart from extremely simple cases (position or momentum operator) it is not very useful, at least in my ...


0

Imagine that the observable you want to measure is eg. position $\hat{x}$. When you measure it in a real-life experiment, the result will be an interval of certainty – something like $x = x_0 \pm \varepsilon$. You won't be able to project the original state onto one pure eigenvalue $\left| x_0\right>$, but instead you'll project it to a subspace $\left\{ ...


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