# Do eigenvectors of quantum operators span the whole Hilbert Space?

I am trying to solve an exercise in Shankar's QM book (concretely 4.2.1), and I am asked the probability of each possible value for the operator $L_x$ when the particle is in a certain eigenstate of another operator $L_z$. I understand that I have to "expand" the state vector in the $L_z$ basis, which in this case is just the appropriate eigenvector, and then change the base to that of $L_x$, i.e to write the $L_z$ eigenstate in the $L_x$ base. The problem is that there is just no way of doing it, since the eigenvectors of $L_x$ don't span the whole Hilbert Space.

I thought all quantum operators had a set of eigenvectors that spanned the whole space, am I wrong on that or am I wrong on my algebra?

• Can you comment on how you know the eigenvectors of $L_x$ don't span the whole (spin) space?
– BMS
Jan 6, 2014 at 19:23
• Well, the three Lx eigenvectors I get are (0, 0, 0), (1, 1, 1) and (1, -1, 1). How can you compose, say, the vector (0, 0, 1) out of them? Jan 6, 2014 at 19:34
• I figure that your $L_z$ state has $l=1$ right? Well, you just have to diagonalize $\frac{\hbar}{\sqrt{2}} \left( \begin{array}{ccc} 0&1&0\\ 1&0&1\\ 0&1&0\\ \end{array} \right)$ and this gives you the right eigenvectors of $L_x$ in the $L_z$ basis. then get the $L_z$'s in the $L_x$ basis and voilà Jan 6, 2014 at 19:58
• Without doing any calculations it should be clear if the eigenvectors L$_{z}$ span the Hilbert space, then so must the eigenvectors of L$_{x}$ and L$_{y}$ by rotational symmetry. In another person's coordinate system you're L$_{x}$ is their L$_{z}$, so it wouldn't make sense that you're L$_{z}$ gives you a basis but theirs does not. Jan 6, 2014 at 20:08
• I don't think those eigenvectors you gave are indeed the eigenvectors of Lx in the Lz basis. By definition, the null vector cannot be an eigenvector! Jan 6, 2014 at 20:44

I believe you've just made an algebraic error. To find the normalised eigenstates of $L_x$ in the (eigenstates of) $L_z$ basis, note that the matrices as given you in exercise 4.2.1 are already (a representation of the operators $L_i$) in the (eigenstates of) $L_z$ basis. This can be appreciated by noting that $L_z$ is diagonal; acting with it on the vectors $(1,0,0)$, $(0,1,0)$ or $(0,0,1)$ returns the same vector, scaled by an eigenvalue.

$$L_x = \frac{1}{\sqrt{2}}\begin{pmatrix} 0&1&0 \\1&0&1 \\0&1&0\end{pmatrix} \qquad L_y = \frac{1}{\sqrt{2}}\begin{pmatrix} 0&-i&0 \\i&0&-i \\0&i&0\end{pmatrix}\qquad L_z = \frac{1}{\sqrt{2}}\begin{pmatrix} 1&0&0 \\0&0&0 \\0&0&-1\end{pmatrix}$$

As such, the problem of finding the normalised eigenstates of $L_x$ in the $L_z$ basis reduces merely to finding the normalised eigenvectors of the matrix $L_x$ as given you. These are

$$\frac{1}{2} \begin{pmatrix} 1 \\ \sqrt{2} \\1\end{pmatrix} \qquad\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0\\-1\end{pmatrix} \qquad \frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt{2} \\1\end{pmatrix}$$

You should be able to finish the problem from here.

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Some words on matrices. You can think of a matrix as a representation of an operator in some specified basis, just as you can think of a column vector as a representation of an (abstract/geometric) vector in some specified basis. For instance, say we're considering the vector $|\psi \rangle$, which is a co-ordinate free object. Then if we choose a basis $\{|i\rangle,|j\rangle,|k\rangle\}$, we can expand our vector in terms of it

$$|\psi\rangle = 4|i\rangle + 2|j\rangle + |k\rangle$$

for instance. What we're saying is

$$|\psi\rangle \qquad \mathrm{can\ be\ represented\ by} \qquad \begin{pmatrix} 4 \\ 2 \\ 1\end{pmatrix}$$

Now the same is true of matrices and operators. For orthonormal bases, which is what we deal with in the vast majority of cases in quantum mechanics, the rule is this: say we choose a basis labelled by the index $i$ --- that is, let's call the vectors in our basis $\{|1\rangle,|2\rangle,|3\rangle,\ldots\}$ --- then the operator $\hat{L}_z$ can be represented by the matrix $L_z$, where the matrix $L_z$ is given by

$$(L_z)_{ij} = \langle i|\hat{L}_z |j \rangle$$

It's somewhat an abuse of notation that we often drop the hats, denoting the matrix and the operator by the same symbol. It's also common to write the operator as equal to its matrix, when we should really say that the matrix merely provides a representation of the operator, in some basis.

Now suppose that the basis we chose was the basis of eigenvectors of $\hat{L}_z$. That is, suppose we chose our basis such that

$$\hat{L}_z |1\rangle = \lambda_1 |1\rangle\,,\qquad\hat{L}_z |2\rangle = \lambda_2 |2\rangle\,,\qquad \hat{L}_z |3\rangle = \lambda_3 |3\rangle\,,$$

Then what will our matrix look like? Well, using the formula above

$$(L_z)_{ij} =\langle i|\hat{L}_z |j \rangle = \langle i|\lambda_j |j \rangle = \lambda_j\langle i| j \rangle = \lambda_j \delta_{ij} \qquad \mathrm{(no\ sum)}$$

Using the orthonormal nature of the basis in the last step. In other words, the matrix will be

$$L_z = \begin{pmatrix} \lambda_1 &0&0 \\ 0& \lambda_2 & 0\\ 0&0&\lambda_3\end{pmatrix}$$

It's diagonal, with the eigenvalues along the diagonal. So we have the operator $\hat{L}_z$ in this basis of eigenvectors, now let's ask what our eigenvectors are in this basis of eigenvectors. This is trivial. Our eigenvector $|1\rangle$ can be written as 'one lot of $|1\rangle$ plus zero lots of $|2\rangle$ plus zero lots of $|3\rangle$', and so we simply have

$$|1\rangle \qquad \mathrm{can\ be\ represented\ by} \qquad \begin{pmatrix} 1\\0\\0\end{pmatrix}$$

Essentially this is what it means to say that we're in the basis $\{|1\rangle,|2\rangle,|3\rangle\}$, the basis of eigenvectors of $\hat{L}_z$.

To summarise. Being in the basis of eigenvectors of a given operator means that the matrix representation of that operator will be diagonal. The fact that the matrix representing $\hat{L}_z$ is (in this case) diagonal therefore tells you that you're in the basis of eigenvectors of $\hat{L}_z$. The column vector representation of these eigenvectors in this basis is simply

$$\begin{pmatrix} 1\\0\\0\end{pmatrix} \qquad \begin{pmatrix} 0\\1\\0\end{pmatrix} \qquad \begin{pmatrix} 0\\0\\1\end{pmatrix}$$

As a final word, and I hope this does not confuse you: if you have a matrix representation of an operator in one basis, you can transform it into the matrix representation of that operator in another basis by pre- and post-multiplying by appropriate transformation matrices. It turns out that the transformation matrix you want has as its columns the column vector representations of the new basis vectors in terms of the old basis vectors. What this means is that if you want to diagonalise a matrix --- which means you want to get the matrix representation of the underlying operator in terms of the eigenvectors of that operator --- then the transformation matrix you want has, as its columns, the eigenvectors of the operator written in terms of the current basis you're in. This might be what you're thinking of.

Hope this helps!

• Oh, right, that's the whole problem, I didn't get the right eigenvectors. I just took them to be the coumns of the matrix. Aren't they supposed to be the eigenvectors? Jan 7, 2014 at 13:20
• Perhaps you could expand a little further on what you mean by that? The eigenvectors of a matrix aren't just its columns, no. In fact, it doesn't appear you've done this. The eigenvectors you first found (according to your comment above) are the columns of no matrix I quoted. Jan 7, 2014 at 15:34
• I was taught that the columns of a diagonalised matrix are its eigenvectors. But now I feel like I'm confused about that. Jan 7, 2014 at 15:44
• Yeah, that's not true. Let me edit my post to explain. Jan 7, 2014 at 16:21

If the operator $A$ has pure point spectrum, its eigenvectors span the whole Hilbert space.

If I understand correctly you are dealing with the operators $L_i$, components of the angular momentum of a particle. Each of those operators has indeed pure point spectrum (essentially as a consequence of Peter-Weyl theorem $SO(3)$ being compact), so the eigenvectors of each $L_i$ separately span the whole Hilbert space. For instance ${\bf L}^2(R^3)$ if the particle has no spin or other internal degrees of freedom.

In the considered case (spinless particle) the Hilbert space ${\bf L}^2(R^3)$ turns out to be a direct orthogonal sum of subspaces $H_j$ eigenspaces of $L^2$ with eigenvalue $\hbar j(j+1)$, where $j=0,1,2, \ldots$.

Each $H_j$ is invariant under the action of each $L_i$, so that you can find a orthonormal basis of each $L_i$ in every $H_j$. The union of all these bases, for a given $L_i$, form a complete Hilbert basis of the whole ${\bf L}^2(R^3)$.

As a matter of fact one has to decompose ${\bf L}^2(R^3) = {\bf L}^2(S^2) \otimes {\bf L}^2(R_+, r^2 dr)$. Where $S^2$ is the unit $2D$ sphere equipped with the $SO(3)$ invariant measure with total volume $4\pi$. With this decomposition, for $L_z$, the said spaces $H_j$ have the form:

$$H_j = \Omega_j \otimes {\bf L}^2(R_+, r^2 dr)$$

where $\Omega_j$ is the subspace of ${\bf L}^2(S^2)$ spanned by the spherical harmonics $Y^j_m$ defined on $S^2$ with $m= -j, -j+1, \ldots, j$.

Fixing any Hilbert basis $\{\psi_n\}_{n=1,2,\ldots}$ in ${\bf L}^2(R_+, r^2 dr)$, each space $H_j$ has a Hilbert basis made by the products $\psi_n \otimes Y^j_m$. The values $\hbar m$ are the eigenvalues of $L_z$:

$$L_z \psi_n \otimes Y^j_m = \hbar m \: \psi_n \otimes Y^j_m$$

Therefore:

$$\{\psi_n \otimes Y^j_m \:\:|\:\: n= 1,2,\ldots \:\: j= 0,1,2, \ldots, m= 0,\pm1 \pm2, \ldots\}$$

is just a Hilbert basis of eigenvectors of $L_z$. As you probably know $L_x$ and $L_y$ have the same spectrum as $L_z$ (in view of rotational invariance), but the eigenvectors are different (since the operators do not commute pairwise). Again in view of rotational invariance, however, the same basis of the $\psi_n$ can be used to construct the bases of $L_x$ and $L_y$ and one just work in each $\Omega_j$.

• What does mean the notation ${\bf L}^2(G)$, seen in the Wiki article about the the Peter-Weyl theorem ? It is equilvalent to your notation ${\bf L}^2(R^3) ?$ Jan 7, 2014 at 10:01
• $L^2(G)$ means the $L^2$-Hilbert space of the complex valued defined on a Hausdorff locally compact group admitting a two-side Haar measure, that is used as the measure to integrate functions. My $L^2(S^2)$ indicates, instead, the analogous $L^2$ space on the $2$-sphere. The measure is uniquely determined by the Haar mesure $SO(3)$, as $S^2$ can be obatined from the latter taking a quotient. $L^2(R^3)$ is the standard Hilbert space on $R^3$ referred to Lebesgue's measure. Jan 7, 2014 at 10:09
• The fact that the $L_i$ have pure point spectrum follows form PW theorem that assures that the unitary representations of the compact group $SO(3)$ (or $U(2)$) always are the direct sum of finite dimensional ones. The $L_i$ are the self-adjoint generators of these reps. In each finite dim invariant subspace the $L_j$ are "matrices" and thus their spectrum is a pure point spectrum. The overall spectrum is the union of all spectra and thus it remains a point spectrum (since it has no limit points). Jan 7, 2014 at 10:16
• Reading the Wiki article, it seems that it speaks directly (including Hilbert spaces) about the group $G$ and its representations, while it seems to me that you are speaking about the objects upon which these representations act (the two are linked, of course, but different). Is this correct ? Jan 7, 2014 at 10:40
• Actually it seems that what I said is missed in that Wiki page! PW theorem also says that (a) every irreducible unitary rep of $G$ in any Hilbert space is unitarily isomorphic to one irreducible unitary rep of the group in $L^2(G)$; (b) every unitary rep of $G$ in any Hilbert space is direct tum of unitary irreducible reps. sect. 1,2 of ch. 7 of Barut-Raczka textbook "theory of group representations and applications" Jan 7, 2014 at 10:54