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Hello could someone please check if my understanding of 'collapsing into eigenfunctions' is correct?

Say we have an observable, given by a linear self-adjoint operator $A$ and then we have $\{\psi_n\}_{n\geq 1}$ as a basis of eigenfunctions, such that $A\psi_n=c_n\psi_n$, for $\mathcal{H}.$

Suppose I have a normalised wave function in $\mathcal{H}$ given by $\psi(x)=a_1\psi_1+a_2\psi_2+a_3\psi_3.$ I should say $a_i's$ are just the coefficients for these eigenfunctions. For the sake of argument suppose $c_1=c_3$, that is, the eigenvalues for $\psi_1$ and $\psi_3$ are the same. Then after measurement, suppose I obtain $c_1$, then am I in the position of claiming the following:

the wavefunction after the measurement is now $a_1\psi_1+a_3\psi_3.$ This wave function is no longer normalised unless $a_2=0.$

Many thanks in advance!

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Since you know the initial state, then you just project the original state onto the degenerate space to get the new state. This state can be normalized; normalization is more mathematical than physical.

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Wave function collapse is not unitary, and does not preserve normalization. Physically, this is not a problem because states of your system (unlike Hilbert space vectors themselves) are only well-defined up to an overall complex factor - see projective Hilbert spaces on Wikipedia.

Since it's mathematically convenient to work with normalized vectors, typically one would simply re-normalize $\psi$ after such a measurement, but again this is purely a matter of convenience.

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