4
$\begingroup$

I am seeking to extend the following theorem to the case of infinite dimensional Hilbert space:

Suppose we have two Hermitian operators $X$ and $Z$ in a finite dimensional Hilbert space $\mathcal H$. And they satisfy the following relation:

$$X^2=Z^2=I, \quad \{X,Z\}\equiv XZ+ZX= 0.$$

Then it is not very hard to prove that, up to a unitary change of basis,

$$X=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\otimes I,\qquad Z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\otimes I\tag{*}.$$

Indeed, we can find out all the vector pairs $v$ and $Xv$ such that $Zv=v,ZXv=-XZv=-Xv$. Under the the orthonormal basis made of these pairs, $X$ and $Z$ have the matrix representation given above.

I am just wondering if this theorem can fit into infinite dimension case. It may need rigorous definition of $X$ and $Z$, and reference to the spectral theory. How exactly should I redefine the problem in order to obtain result similar to $(*)$?

$\endgroup$
  • 2
    $\begingroup$ 1. That's not the spectral theorem. 2. This doesn't seem to be a physics question, as posed, it is a pure math question about operator theory in Hilbert spaces. $\endgroup$ – ACuriousMind Feb 28 '16 at 10:31
  • 2
    $\begingroup$ Actually it includes the theory of angular momentum...at least in my answer. I do not think that in math stackexchance this question would be easily answered... $\endgroup$ – V. Moretti Feb 28 '16 at 17:08
  • $\begingroup$ This question arises in my research on a problem in quantum computing, whose potential readers are computer scientists. So I am looking for "easy" proof, circumventing the need to introduce heavy math definitions or physics theorems. $\endgroup$ – Rui Mar 2 '16 at 21:00
  • $\begingroup$ I believe the correctness of the answer from @V.Moretti. But can we just appeal merely to the separability of the space and\or spectral theorem and\or Gram-Schmidt process? $\endgroup$ – Rui Mar 2 '16 at 21:03
  • $\begingroup$ Separability does not enter the problem: both the theorem of Nelson and the one of Peter-Weyl does not need it. $\endgroup$ – V. Moretti Mar 3 '16 at 7:54
5
$\begingroup$

I assume that your operators are bounded in a Hilbert space $H$, otherwise I am not sure that the result I am proving is still true in the absence of other assumptions like the essential self-adjointness of the operator $\sum_{a=1}^3 X_a^2$.

Define $iY=ZX$ and next $X_1:=X$, $X_2:=Y$, $X_3:= Z$.

With this definition and your hypotheses, you easily see that $X,Y,Z$ are bounded self-adjoint operators such that $$\{X_a,X_b\}= 2\delta_{ab}I\tag{1}$$ $$[X_a,X_b] = 2 i\sum_c \epsilon_{abc} X_c\:.\tag{2}$$ (2) are the commutation relations of $su(2)$.

As the operators are bounded and self-adjoint, $\sum_{a=1}^3 X_a^2$ is bounded and self-adjoint as well, so in particular it is essentially self-adjoint. Nelson's theorem implies that the Hilbert spece supports a strongly-continuous unitary representation of $SU(2)$, whose Lie-algebra is represented by operators $-iX_a$.

At this point, since $SU(2)$ is compact, Peter-Weyl theorem says that $H$ decomposes into an orthogonal direct sum of finite dimensional irreducible subrepresentatons $H = \oplus_{j}H_j$. Above $j=0,1/2,1,3/2,2,\ldots$. The generators of the $j$-th subrepresentation are the restrictions of the $X_a$ to $H_j$.

Let us focus on these generators $X_{aj}$. As well known $H_j$ is the eigenspace of $X_j^2= \sum_{a=1}^3 X_{aj}^2$ with eigenvalue $4j(j+1)$. So $$X_j^2 = 4j(j+1)I_j$$ but the constraint (1) implies $$3 I_j= 4j(j+1)I_j\:,$$ thus $j=1/2$. The only possible representation appearing in the decomposition of $H$ is the one with $j=1/2$. It may appear infinitely times if $H$ is infinite dimensional. Thus $$H = H_{1/2}\otimes K$$ where $K$ is infinite dimensional if $H$ is. The representation of $X_a$ in $H_{1/2}$ are the ones given in terms of Pauli matrices. We end up with $$X_1 = \sigma_1 \otimes I\:, \quad X_2 = \sigma_2 \otimes I \:, \quad X_3 = \sigma_3 \otimes I\:,$$ where $I$ is the identity operator in $K$.

$\endgroup$
  • $\begingroup$ Do you mean that the operator $-iX_a$ is a representation of su(2)? Could you give a reference (maybe paper) to Nelson Thm? Also, is it obvious that $H_j$ respects to eigenvalue $4j(j+1)$? $\endgroup$ – Rui Feb 29 '16 at 0:19
  • $\begingroup$ (a) Yes I do. (b) Nelson, E.: Analytic Vectors. Ann. Math. 70, 572-614 (1959). (c) It is obvious because it is an irreducible thus finite-dimensional representation of $SU(2)$ and we are dealing with its Casimir operator $J^2$ up to a factor... $\endgroup$ – V. Moretti Feb 29 '16 at 7:56
1
$\begingroup$

The following argument works both in finite and infinite dimension. The context in infinite dimension is that $X $ and $Z $ are bounded operators acting on a separable Hilbert space.

From $Z^2=I$, you get that the spectrum of $Z$ is contained in $\{1,-1\}$. From $ZX+XZ=0$ we get that $Z\ne \pm I$, so the spectrum of $Z$ is $\{1,-1\}$. It is then immediate (here is a proof) that $Z$ is unitarily equivalent to $$ Z=\begin{bmatrix}I_n&0\\0& -I_m\end{bmatrix}, $$ where the blocks are given, respectively, by the projections onto $\ker(Z-I)$ and $\ker(Z+I)$. If we represent $X$ as a block matrix with respect to the same basis, we have $$ X=\begin{bmatrix}A&B\\ B^*&C\end{bmatrix}. $$ But then $$ \begin{bmatrix}0&0\\0&0\end{bmatrix}=XZ+ZX=\begin{bmatrix}2A&0\\0&-2C\end{bmatrix}.$$ It follows that $A=C=0$. From $X^2=I$, we now get that $BB^*=I_n$, $B^*B=I_m$. If $n$ or $m$ is finite, taking the trace we see that $n=m$. So $n=m$ (whether they are finite or infinite) and $$ X=\begin{bmatrix}B&0\\0&I_n\end{bmatrix}\,\begin{bmatrix}0&I_n\\ I_n&0\end{bmatrix}\,\begin{bmatrix}B&0\\0&I_n\end{bmatrix}^*. $$ Now a straightforward computation shows (using that $BB^*=I_n$) that $$ \begin{bmatrix}B&0\\0&I_n\end{bmatrix}\,\begin{bmatrix}I_n&0\\0&- I_n\end{bmatrix}\,\begin{bmatrix}B&0\\0&I_n\end{bmatrix}^*=\begin{bmatrix}BB^*&0\\0&- I_n\end{bmatrix}=\begin{bmatrix}I_n&0\\0&- I_n\end{bmatrix}=Z. $$ Thus, writing $U=\begin{bmatrix}B&0\\0& I_n\end{bmatrix}$, we have $$ X=U(\sigma_x\otimes I_n)U^*,\ \ \ Z=U(\sigma_z\otimes I_n)U^*. $$

$\endgroup$
0
$\begingroup$

We have two self-adjoint operators $X$ and $Z$ satisfying (I assume that you mean with $\{X,Z\}$ the commutator $[X,Z]$)

\begin{eqnarray*} X^{2} &=&Z^{2}=I \\ \lbrack X,Z] &=&0 \end{eqnarray*} Then they share the same spectral measure $\{E(d\lambda ),\lambda \in \mathbb{R}\}$ and we can write \begin{eqnarray*} X &=&\int f(\lambda )E(d\lambda )\Rightarrow X^{2}=\int f(\lambda )^{2}E(d\lambda )=I\Rightarrow f(\lambda )^{2}=1\;\mathrm{a.e.} \\ Z &=&\int g(\lambda )E(d\lambda )\Rightarrow Z^{2}=\int g(\lambda )^{2}E(d\lambda )=I\Rightarrow g(\lambda )^{2}=1\;\mathrm{a.e.} \end{eqnarray*} Note that $f(\lambda )$ and $g$($\lambda )$ are real but in general not positive. We can decompose \begin{eqnarray*} f(\lambda ) &=&\chi _{A}(\lambda )-\chi _{B}(\lambda ) \\ g(\lambda ) &=&\chi _{C}(\lambda )-\chi _{D}(\lambda ) \end{eqnarray*} where \begin{equation*} A=\{\lambda |f(\lambda )\geqslant 0\},\;B=\{\lambda |f(\lambda )<0\} \end{equation*} and similar for $C$ and $D$. Here a.e. is understood.

Substraction gives \begin{equation*} \int \{f(\lambda )^{2}-g(\lambda )^{2}\}E(d\lambda )=0 \end{equation*} so \begin{equation*} f(\lambda )^{2}=g(\lambda )^{2}\;\mathrm{a.e.} \end{equation*} Thus \begin{eqnarray*} f(\lambda ) &=&g(\lambda ),\;\lambda \in \{A\cap C\}\cup \{B\cap D\} \\ f(\lambda ) &=&-g(\lambda ),\;\lambda \in \{A\cap D\}\cup \{B\cap C\} \end{eqnarray*} We note that in general there are many $X$ and $Z$ satisfying the requirements.

$\endgroup$
  • 5
    $\begingroup$ I believe OP's $\{\cdot,\cdot\}$ means anticommutator. $\endgroup$ – AccidentalFourierTransform Feb 28 '16 at 12:08
  • $\begingroup$ In that case my response is irrelevant. Sorry. $\endgroup$ – Urgje Feb 28 '16 at 20:10

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.