Suppose I want to represent the quantum state of a spinless particle. To do so, I employ a Hilbert space $\mathcal{H}_X$, which is an infinite-dimensional Hilbert space equipped with a position operator $\hat{X}$. If I want to represent the quantum state of a spin-1/2 particle, I need to "add in" the spin degrees of freedom by taking the tensor product of $\mathcal{H}_X$ with a spin Hilbert space $\mathcal{H}_S$, where $\mathcal{H}_S$ is a two-dimensional Hilbert space equipped with spin operators.

My question is this: what are the conditions under which we expand the Hilbert space in this way to capture new degrees of freedom (or perhaps better, what are the conditions for something to count as a "new" degree of freedom)? For example, contrast the case of spin with that of momentum: to capture the momentum degrees of freedom, I don't take a tensor product $\mathcal{H}_X \otimes \mathcal{H}_P$ - rather, it's just that there's a momentum operator $\hat{P}$ available on $\mathcal{H}_X$.

The most obvious difference is that the momentum operator does not commute with the position operator (whereas all the spin operators do so). So is the following true: a necessary and sufficient condition for two operators to "live" on the same Hilbert space is that they fail to commute?

  • $\begingroup$ I'm not sure how to answer your first question but the second is definitely false. Consider an identity operator and any other operator :). There are many other examples. You may have the wrong idea here though. An operator is simply a linear map between vector spaces. This definition implies nothing about commutativity. $\endgroup$ Commented Sep 14, 2015 at 19:13
  • $\begingroup$ Wow, you seem to forget that the Hilbert space is defined first, then you add various operators to it. The confusion which you claim to see does not arise in reality. To convince you, let me just say this. $H_X$ is not just a letter, its the space of (L2 integrable) functions $\psi(x)$ with a position operator x, and maybe a momentum operator p=-id/dx, and maybe even some other operators as well... $\endgroup$
    – Kphysics
    Commented Dec 24, 2018 at 22:51

1 Answer 1


Just to define the concepts:

Given a Hilbert space ${\mathcal H}$, there always exists a complete set of distinct, but commuting observables $\{ {\hat O}_1, {\hat O}_2, ...,{\hat O}_m \}$, $\left[{\hat O}_j, {\hat O}_k \right] = 0$, whose common eigenstates define a basis of ${\mathcal H}$ labeled exhaustively by the eigenvalues of each ${\hat O}_k$. This means that to each unique m-tuple of eigenvalues of the ${\hat O}_k$-s there corresponds a unique common eigenstate (no degeneracies). We can identify the observables in the complete commuting set as the "degrees of freedom" of the system described by ${\mathcal H}$. Note however that the definition of the DOFs is not unique. There are many equivalent ways to choose a complete set of observables for the same system.

If an observable ${\hat A}$ can be expressed uniquely in terms of some complete set of commuting ${\hat O}_k$-s, or if each eigenstate of ${\hat A}$ can be expressed uniquely in terms of the common eigenstates of the ${\hat O}_k$-s, then ${\hat A}$ is completely described within the ${\mathcal H}$ of the ${\hat O}_k$-s. Otherwise, ${\hat A}$ is, or includes, a new DOF and we need to expand ${\mathcal H}$ to accommodate it.

For example, a 1D spinless particle has a single degree of freedom that can be represented by its position operator ${\hat X}$. Alternatively, we can choose to represent it by the momentum ${\hat P}$. But ${\hat P}$ still acts in the same Hilbert space as ${\hat X}$ because we know how to express any eigenstate of ${\hat P}$ in terms of the eigenstates of ${\hat X}$.

For a spin-1/2 1D particle, the spin cannot be expressed in terms of position, momentum, or any function of them. In fact the spin value can be defined independently of either position or momentum. Correspondingly, the eigenstates of the spin observable cannot be expressed in terms of eigenstates of ${\hat X}$ or ${\hat P}$ and this means that the spin is a new degree of freedom that must be accommodated by expanding ${\mathcal H}_X$.

Same works for extending the 1D particle to 2D. The values of the additional coordinate $Y$ can be defined independently of $X$ and the corresponding DOF must be accounted for separately in the total Hilbert space. In this case we can even establish an isomorphism between the eigenstates of observables ${\hat X}$ and ${\hat Y}$, but they still need to be accounted separately because the $X$ and $Y$ DOFs must have independent values.

Caution: The fact that 2 observables commute or not is not in itself sufficient to decide whether they represent independent degrees of freedom. For instance, ${\hat P}$ and ${\hat P}^2$ do commute, but they are not independent DOFs. On the other hand, if ${\hat \sigma}$ is the spin observable, ${\hat X}$ and ${\hat \sigma}{\hat P}$ do not commute, but may be considered as independent DOFs, while ${\hat P}$ and ${\hat \sigma}{\hat P}$ do commute and may also be considered as independent DOFs. The important distinction is, ${\hat P}^2$ can be completely described in terms of ${\hat P}$, but ${\hat \sigma}{\hat P}$ cannot.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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