On Hermitian Operators

The second postulate of quantum mechanics states: “To every observable in classical mechanics corresponds a linear and Hermitian operator in quantum mechanics“. The third then continues entering the details of the relationship between this operator and the associated observable, and finally on how the original wave function changes following the measurement of the observable,

The first time I came across this relationship, I had the feeling of being in front of something magical. How could observables be associated only with hermitian operators?

Certainly the Hermitian operators have only real eigenvalues ​​and this is necessary if we want these eigenvalues ​​to be the measurable values ​​for the observable. But how can you be sure that this operator exists for any observable?
Things became clearer to me by reading the first chapters of the book “Principles of Quantum Mechanics” by professor R. Shankar, in which the author deals with the case of a particle that can only move along an x ​​axis. At each position x we ​​naturally associate a ket |x> which (being in the continuous case) can be imagined as a Dirac pulse in x itself.

More precisely, the following relationships apply:

Any wave function can be expressed as a function of x and we know that the probability of finding the particle in the interval [x, x + dx] is equal to

We also know that

can be expressed as “superposition” of the | x>.
Is there a way to relate the measurable position x with the relative state | x> via a linear operator? The answer is yes and it is through an X operator such that

where X is such that

It is a clearly Hermitian operator whose eigenvalues ​​and associated eigenvectors are x and | x> very simply. Whatever the orthonormal basis of the Hilbert space in which we decide to express X and | x>, the relationship just seen will always be worth:

But what we have seen for position x can be repeated for any observable. No observable is “better” then the others!

The presence therefore of a Hermitian linear operator associated with an observable (whose eigenvalues ​​and eigenstates have the well-known meaning) is therefore not so “magic”.

The postulate that the wave function is the same for all observables and that associated eigenstates bases belong to the same Hilbert space, this is the real magic of Quantum Mechanics!

2 Replies to “On Hermitian Operators”

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