I was confused with the resolution of the identity, it states that if $ | \psi_i \rangle $ is a basis, then
$ \sum\limits_{i=0}^{\infty}{|\psi\rangle\langle\psi_i|} = 1 $.
After reading Quantum mechanics demystified, finally understood the meaning of the above.
As $ \psi $ is an abstract vector, we should NOT think of it as a wavefunction, or a tuple, or whatever, instead, we define $|a\rangle\langle b| $ as an operator to mean $ |a\rangle\langle b|( |c\rangle) = |a\rangle(\langle b|c\rangle) $,
Now it suddenly become clear! The 1 in the above identity is not the real/complex number 1, but the identity operator that takes a vector to itself!
$ \sum\limits_{i=0}^{\infty}{|\psi\rangle\langle\psi_i|} = 1 $.
After reading Quantum mechanics demystified, finally understood the meaning of the above.
As $ \psi $ is an abstract vector, we should NOT think of it as a wavefunction, or a tuple, or whatever, instead, we define $|a\rangle\langle b| $ as an operator to mean $ |a\rangle\langle b|( |c\rangle) = |a\rangle(\langle b|c\rangle) $,
Now it suddenly become clear! The 1 in the above identity is not the real/complex number 1, but the identity operator that takes a vector to itself!
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