1
$\begingroup$

When we define a basis for the Hilbert Space for a spin half particle I understand it being done using the principle of mutual exclusivity that is if $S_z = +\hbar/2$ then it cannot be $S_z = -\hbar/2$ hence they are orthonormal state vectors. The $S_z$ operator is nice in the sense that it has the above same state vectors as its eigenvectors and therefore just gets scaled to produce an observable. So the choice of basis beforehand seems useless as the operator anyway will only produce an observable along its eigenvectors. So do we automatically make the observables of our experiment the basis of our Hilbert space by default since they are the only observed values when we apply a specific operator based on our experiment on it ?

$\endgroup$
1
  • 1
    $\begingroup$ The basis of eigenstates of the operator corresponding to the observable of interest is typically the most practical basis to use. You can evaluate the action of the operator trivially in this basis. Using another basis would make the calculation of expectation values harder. This of course only holds if you evaluate a single operator/commuting operators and it also requires that you know the eigenvectors/values of the operator, which is not always the case. Finding the eigenvectors and eigenvalues of an operator can be a very difficult problem by itself. $\endgroup$
    – Hans Wurst
    Nov 27 '21 at 9:50
2
$\begingroup$

Observables aren’t a basis of the Hilbert space: their eigenvectors form a basis. The strategy is to find the largest possible set of commuting operators, so that their common eigenvectors are uniquely labelled by the eigenvalues in the commuting set. In your example, the Hilbert space is 2-dimensional and the eigenvalues of $\hat S_z$ are $\pm \frac{1}{2}$, so that’s enough to uniquely label the basis of your Hilbert space, so you don’t need anything else.

In your example, you could choose your basis vector $(1,0)^\top$ and $(0,1)^\top$ to be the eigenvectors of $\hat S_x$, and this would be just as fine: the matrix representation of $\hat S_z$ and $\hat S_y$ would then be non-diagonal.

The dimension of the Hilbert space is tied to the number of distinct mutually exclusive outcomes: experiment shows there’s only 2 possible distinct outcomes to measuring the spin of a spin-1/2 particle, and since these outcomes do not depend on the direction, the basis states of any operator of the form $$ n_x\hat S_x+n_y\hat S_y+n_z\hat S_z\, \qquad n_x^2+n_2^2+n_z^2=1 $$ would could serve as a basis for the 2-dimensional Hilbert space. It is convention to chose a basis where $\hat S_z$ is diagonal, but that’s just convention.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.