Undergrad Confused about the spectrum of an observable

[nomadreid]

TL;DR

Eigenvalues of an observable are probability amplitudes but also stated to be measurement values. What am I missing?

This is a very elementary question, from the beginnings of quantum mechanics.

For simplicity, I refer to a finite case with pure states.

If I understand correctly, the spectrum of an observable is the collection of eigenvalues formed by the inner product of states and hence equal to probability amplitudes; they are then associated to the possible values of a measurement.

Hence I am confused by statements such as the following

“Eigenvalues of observables are real and in fact are possible outcomes of measurements of a given observable.” (https://www.quantiki.org/wiki/observables-and-measurements, but not the only example.)

How can the square roots of probabilities, which are less than one, be values of measurements (which can be greater than one)?

 

[hilbert2]

Eigenvalues are not the same as probability amplitudes. The wave function that gives the prob. amplitudes is one representation of the eigenstate, and it can be either a position or momentum representation.

 

[hutchphd]

Summary:: Eigenvalues of an observable are probability amplitudes but also stated to be measurement values. What am I missing?

If I understand correctly, the spectrum of an observable is the collection of eigenvalues formed by the inner product of states and hence equal to probability amplitudes

Only the green part of the sentence is required. The possible eigenvalues of the system are determined from the potential independent of the actual state of the system. This is the spectrum.
The result of any measurement will be determined by the actual state of the system. The measurement will yield an eigenvalue. The state vector of the system will predict the probability of that value being measured (i.e if you repeated the measurement on similarly prepared states blah blah) which is given by the inner product of the state vector of the system with the corresponding eigenstate.

.

 

[nomadreid]

Thank you very much, hutchphd and hilbert2. An explanation similar to the last two sentences of hutchphd's explanation apparently was picked up by me somewhere and morphed into the version I posted. This clears it up.

 

[hutchphd]

This clears it up.

Good.
May I congratulate you on your Gary Larson cartoon. Perhaps my all-time favorite, and that's a difficult choice.