2
$\begingroup$

Let us consider an electron in an infinite square well. As we know that the electron has a spin=$1/2$ . The spin operator ($z$-direction) has two eigenvectors which span the vector space. But if we solve for the eigenvectors of Hamiltonian, We get infinite basis vectors which span the space. But in linear algebra, we cannot have two basis sets with different number of elements.

$\endgroup$
  • 4
    $\begingroup$ Have you considered that you are looking at two different vector spaces? $\endgroup$ – Aaron Stevens Feb 10 at 4:47
  • $\begingroup$ I might be missing something elementary, but can't we write any vector as a linear combination of spin eigenstates, even if they are energy eigenstates, meaning that we can produce any vector which is produced by energy eigenstates. $\endgroup$ – Jay Feb 10 at 5:11
  • 1
    $\begingroup$ The full state space can be decomposed in two infinite dimensional orthogonal factors, one containing all spin 1/2 states, and one with all spin -1/2 states. $\endgroup$ – doetoe Feb 10 at 6:54
  • 1
    $\begingroup$ In other words, there are two eigenvalues, but each with an infinite dimensional eigenspace $\endgroup$ – doetoe Feb 10 at 6:57
  • 1
    $\begingroup$ You can't specify the state of the particle just with the energy OR the spin, you need both. If you're considering the 1D square well, you will have degeneracy of two states for each energy eigenvalue (the same energy for both values of spin) and if you think about the degeneracy for the spin eigenvalues that should be (countable) infinity. What you should be looking for is a Complete Set of Commuting Observables, just like the $n, l, m$ and $s$ quantum numbers in atomic states. en.wikipedia.org/wiki/Complete_set_of_commuting_observables $\endgroup$ – Salvador Villarreal Feb 10 at 7:42
2
$\begingroup$

The spin operator $S_z$ has two eigenvalues, and its eigenvectors span the whole state space, but that doesn't mean it has two eigenvectors.

In your case, the full state space is spanned by states of e.g. definite spin and position, or spin and momentum, or if you want something like spin, energy and sign of the momentum, etc. Since all these values can be assumed independently (i.e. all combinations give a valid and different state), the full state space is the tensor product of the individual state spaces, in your example the abstract two dimensional spin state space, and the infinite dimensional position space.

That means that, as you asked in your comment, indeed a particle that has a definite spin can be in any superposition of position eigenstates.

$\endgroup$
  • $\begingroup$ So can I say that the statement "The eigenfunctions of an observable operator are complete, Any function in Hilbert space can be expressed as a linear combination of them." is incomplete, when it's really the tensor product of the individual spaces is hilbert space. $\endgroup$ – Jay Feb 10 at 16:54
  • $\begingroup$ @Jay No, it is actually correct (mostly, not all elements of an infinite dimensional Hilbert state space are linear combinations of eigenfunctions, but rather they can be approximated by them to any precision). In your case you could say you are considering the tensor product of state spaces, but it holds in full generality: whatever the state space of your system (obtained as a tensor product or otherwise), and whichever observable you consider, every state can be approximated by a linear combination of eigenstates of that particular observable. Does that answer your comment? $\endgroup$ – doetoe Feb 10 at 17:11
  • 2
    $\begingroup$ @Jay It is postulated that the eigenvectors of every observable form a basis, so the situation you describe cannot occur. Probably what you are having in mind is what AaronStevens already remarked: you are considering operators on different spaces. If you just look at spin, the full state space is 2 dimensional, and the spin operator has two eigenvectors. If you look at the full state space of an electron's position and spin, the state space is infinite dimensional, the spin operator can be extended to the full space, and for both spin eigenvalues we get an infinite dimensional eigenspace. $\endgroup$ – doetoe Feb 10 at 17:56
  • 1
    $\begingroup$ @Jay This total space can be realized as the tensor product of the spin state space and the position state space. The spin eigenspaces can be written as $\mathbb{C}|\uparrow\rangle\otimes\mathcal H$ and $\mathbb{C}|\downarrow\rangle\otimes\mathcal H$. Each position eigenfunction (whatever that exactly is) when tensored with a spin eigenvector, is simultaneously an eigenvector of spin and position. $\endgroup$ – doetoe Feb 10 at 17:59
  • 1
    $\begingroup$ In this specific case certainly, but even in the general case you could say that in a way it is tautologically true: you start out with a state space, elements of which describe the full state of the system under consideration. In your case that is position and spin (one of the surprises of QM is that the position state uniquely determines the momentum state and vice versa) but in other contexts you could very well consider systems in which the full state space is 2D, generated by spin alone. $\endgroup$ – doetoe Feb 11 at 8:31

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.