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In discrete bais, we can express a vector as $$ |\psi\rangle=\sum_{i} c_i|e_i\rangle $$ with orthonormality $$ \langle e_i|e_j\rangle=\delta_{ij}.$$ $\delta_{ij}$ is usual kronecker delta. If we promote the basis to continuum basis, we use dirac delta function instead, i.e., $$ \langle x|x'\rangle=\delta (x-x').$$ My question is that, if the definition of delta is $$ \delta(x-x')= \left\{\begin{array} 1\infty, & x=x' \\ 0, & x\ne x' \end{array} \right.$$ Does orthonormality hold? since $\langle x|x\rangle=\infty$, it is not 1 as same as discrete basis.

Qmechanic
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B. Silasan
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    You definition of dirac delta is wrong the true definision is that $\int_{-\infty}^{+\infty}\delta(x-x')dx=1$ and $\delta(x-x')=0; if \space x \neq x'$ – Sancol. Feb 15 '24 at 04:51
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    It is well-established that the correct continuum counterpart of the Kronecker delta function is the Dirac delta distribution. However, the fact that these entities live in Rigged Hilbert space rather than Hilbert space is a nuisance that we cannot avoid. Instead, the rigorous treatment of these issues is to heavily work on the standard Hilbert space methods, and show that the limit getting these Dirac delta distributions are all correct. – naturallyInconsistent Feb 15 '24 at 05:01
  • Not Strictly Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis, ADDENDUM. – Frobenius Feb 15 '24 at 07:04
  • Possible duplicates: https://physics.stackexchange.com/q/89958/2451, https://physics.stackexchange.com/q/273423/2451 and links therein. – Qmechanic Feb 15 '24 at 07:29

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