We start with some definitions. The symbol \({{\mathcal {P}}}(H)\) will stand for the set of all projections (idempotent and self-adjoint operators) on H, and \({{\mathcal {P}}}_1(H)\) will denote the set of all rank-one projections. The commutant of an effect A intersected with \({{\mathcal {E}}}(H)\) will be denoted by
$$\begin{aligned} A^c := \{ C\in {{\mathcal {E}}}(H) :CA=AC\}, \end{aligned}$$
and more generally, for a subset \({\mathcal {M}}\subset {{\mathcal {E}}}(H)\) we will use the notation \({\mathcal {M}}^c := \cap \{ A^c :A\in {\mathcal {M}} \}\). Also, we set \(A^{cc} := (A^c)^c\) and \({\mathcal {M}}^{cc} := ({\mathcal {M}}^c)^c\).
We continue with three known lemmas on the structure of coexistent pairs of effects that can all be found in [27]. The first two have been proved earlier, see [4, 21].
Lemma 2.1
For any \(A\in {{\mathcal {E}}}(H)\) and \(P \in {{\mathcal {P}}}(H)\) the following statements hold:
-
(a)
\(A^\sim = {{\mathcal {E}}}(H)\) if and only if \(A \in \mathcal {SC}(H)\),
-
(b)
\(P^\sim = P^c\),
-
(c)
\(A^c \subseteq A^\sim \).
Lemma 2.2
Let \(A,B \in {{\mathcal {E}}}(H)\) so that their matrices are diagonal with respect to some orthogonal decomposition \(H = \oplus _{i\in \mathcal {I}} H_i\), i.e. \(A = \oplus _{i\in \mathcal {I}} A_i\) and \(B = \oplus _{i\in \mathcal {I}} B_i \in {{\mathcal {E}}}(\oplus _{i\in \mathcal {I}} H_i)\). Then \(A\sim B\) if and only if \(A_i\sim B_i\) for all \(i\in \mathcal {I}\).
Lemma 2.3
Let \(A,B \in {{\mathcal {E}}}(H)\). Then the following are equivalent:
-
(i)
\(A \sim B\),
-
(ii)
there exist effects \(M,N \in {{\mathcal {E}}}(H)\) such that \(M \le A\), \(N \le I-A\), and \(M+N = B\).
We continue with a corollary of Lemma 2.1.
Corollary 2.4
For any effect A and projection \(P\in A^\sim \) we have \(P\in A^c\). In particular, we have \(A^\sim \cap {{\mathcal {P}}}(H) = A^c\cap {{\mathcal {P}}}(H)\).
Proof
Since coexistence is a symmetric relation, we obtain \(A\in P^\sim \), which implies \(AP=PA\). \(\square \)
The next four statements are easy consequences of Lemma 2.3, we only prove two of them.
Corollary 2.5
For any effect A we have \(A^\sim = \left( A^\perp \right) ^\sim \).
Corollary 2.6
Let \(A\in {{\mathcal {E}}}(H)\) such that either \(0\notin \sigma (A)\), or \(1\notin \sigma (A)\). Then there exists an \(\varepsilon >0\) such that \(\{C\in {{\mathcal {E}}}(H):C\le \varepsilon I\} \subseteq A^\sim \).
We recall the definition of the strength function of \(A\in {{\mathcal {E}}}(H)\):
$$\begin{aligned} \Lambda (A,P) = \max \{ \lambda \ge 0 :\lambda P \le A \} \qquad (P\in {{\mathcal {P}}}_1(H)), \end{aligned}$$
see [2] for more details and properties.
Corollary 2.7
Assume that \(A\in {{\mathcal {E}}}(H)\), \(0 < t \le 1\), and \(P\in {{\mathcal {P}}}_1(H)\). Then the following conditions are equivalent:
-
(i)
\(A\sim tP\);
-
(ii)
$$\begin{aligned} t \le \Lambda (A,P) + \Lambda (A^\perp ,P). \end{aligned}$$
(4)
Proof
By (ii) of Lemma 2.3 we have \(A\sim tP\) if and only if there exist \(t_1, t_2 \ge 0\) such that \(t = t_1 + t_2\), \(t_1 P \le A\) and \(t_2 P \le A^\perp \), which is of course equivalent to (4). \(\square \)
Corollary 2.8
Let \(A,B \in {{\mathcal {E}}}(H)\) such that \(A^\sim \subseteq B^\sim \). Assume that with respect to the orthogonal decomposition \(H = H_1 \oplus H_2\) the two effects have the following block-diagonal matrix forms:
$$\begin{aligned} A = \left[ \begin{matrix} A_1 &{} 0 \\ 0 &{} A_2 \end{matrix}\right] \qquad \text {and} \qquad B = \left[ \begin{matrix} B_1 &{} 0 \\ 0 &{} B_2 \end{matrix}\right] \in {{\mathcal {E}}}(H_1 \oplus H_2). \end{aligned}$$
Then we also have
$$\begin{aligned} A_1^\sim \subseteq B_1^\sim \qquad \text {and} \qquad A_2^\sim \subseteq B_2^\sim . \end{aligned}$$
(5)
In particular, if \(A^\sim = B^\sim \), then \(A_1^\sim = B_1^\sim \) and \(A_2^\sim = B_2^\sim \).
Proof
Let \(P_1\) be the orthogonal projection onto \(H_1\). By Lemma 2.2 we observe that
$$\begin{aligned}&\Bigg \{ \left[ \begin{matrix} C &{} 0 \\ 0 &{} D \end{matrix}\right] \in {{\mathcal {E}}}\left( H_1\oplus H_2 \right) :C \sim A_1, D \sim A_2 \Bigg \} = P_1^c \cap A^\sim \\&\qquad \subseteq P_1^c \cap B^\sim = \Bigg \{ \left[ \begin{matrix} E &{} 0 \\ 0 &{} F \end{matrix}\right] \in {{\mathcal {E}}}\left( H_1\oplus H_2 \right) :E \sim B_1, F \sim B_2 \Bigg \}, \end{aligned}$$
which immediately implies (5). \(\square \)
Next, we recall the Busch–Gudder theorem about the explicit form of the strength function, which we shall use frequently here. We also adopt their notation, so whenever it is important to emphasise that the range of a rank-one projection P is \({\mathbb {C}}\cdot x\) with some \(x\in H\) such that \(\Vert x\Vert =1\), we write \(P_x\) instead. Furthermore, the symbol \(A^{-1/2}\) denotes the algebraic inverse of the bijective restriction \(A^{1/2}|_{(\mathrm{Im\,}A)^-}:(\mathrm{Im\,}A)^-\rightarrow \mathrm{Im\,}(A^{1/2})\), where \(\cdot ^-\) stands for the closure of a set. In particular, for all \(x \in \mathrm{Im\,}(A^{1/2})\) the vector \(A^{-1/2} x\) is the unique element in \((\mathrm{Im\,}A)^-\) which \(A^{1/2}\) maps to x.
Busch–Gudder Theorem
(1999, Theorem 4 in [2]) For every effect \(A \in {{\mathcal {E}}}(H)\) and unit vector \(x\in H\) we have
$$\begin{aligned} \Lambda (A,P_x) = \left\{ \begin{matrix} \Vert A^{-1/2} x \Vert ^{-2}, &{} \text {if} \; x \in \mathrm{Im\,}(A^{1/2}), \\ 0, &{} \text {otherwise.} \end{matrix} \right. \end{aligned}$$
(6)
We proceed with proving some new results which will be crucial in the proofs of our main theorems. The first lemma is probably well-known, but as we did not find it in the literature, we state and prove it here. Recall that WOT and SOT stand for the weak- and strong operator topologies, respectively.
Lemma 2.9
For any effect \(A\in {{\mathcal {E}}}(H)\), the set \(A^\sim \) is convex and WOT-compact, hence it is also SOT- and norm-closed. Moreover, if H is separable, then the subset \(A^\sim \cap {{\mathcal {F}}}(H)\) is SOT-dense, hence also WOT-dense, in \(A^\sim \).
Proof
Let \(t\in [0,1]\) and \(B_1,B_2\in A^\sim \). By Lemma 2.3 there are \(M_1,M_2,N_1,N_2\in {{\mathcal {E}}}(H)\) such that \(M_1+N_1 = B_1\), \(M_2+N_2 = B_2\), \(M_1\le A, N_1\le I-A\) and \(M_2\le A, N_2\le I-A\). Hence setting \(M = tM_1+(1-t)M_2\in {{\mathcal {E}}}(H)\) and \(N = tN_1+(1-t)N_2\in {{\mathcal {E}}}(H)\) gives \(M+N = tB_1+(1-t)B_2\) and \(M\le A, N\le I-A\), thus \(tB_1+(1-t)B_2\sim A\), so \(A^\sim \) is indeed convex.
Next, we prove that \(A^\sim \) is WOT-compact. Clearly, \({{\mathcal {E}}}(H)\) is WOT-compact, as it is a bounded WOT-closed subset of \({{\mathcal {B}}}(H)\) (see [7, Proposition IX.5.5]), therefore it is enough to show that \(A^\sim \) is WOT-closed. Let \(\{B_\nu \}_\nu \subseteq A^\sim \) be an arbitrary net that WOT-converges to B, we shall show that \(B\sim A\) holds. For every \(\nu \) we can find two effects \(M_\nu \) and \(N_\nu \) such that \(M_\nu +N_\nu = B_\nu \), \(M_\nu \le A\) and \(N_\nu \le I-A\). By WOT-compactness of \({{\mathcal {E}}}(H)\), there exists a subnet \(\{B_\xi \}_\xi \) such that \(M_\xi \rightarrow M\) in WOT with some effect M. Again, by WOT-compactness of \({{\mathcal {E}}}(H)\), there exists a subnet \(\{B_\eta \}_\eta \) of the subnet \(\{B_\xi \}_\xi \) such that \(N_\eta \rightarrow N\) in WOT with some effect N. Obviously we also have \(B_\eta \rightarrow B\) and \(M_\eta \rightarrow M\) in WOT. Therefore we have \(M+N = B\) and by definition of WOT convergence we also obtain \(M\le A\), \(N\le I-A\), hence indeed \(B\sim A\). Closedness with respect to the other topologies is straightforward.
Concerning our last statement for separable spaces, first we point out that for every effect C there exists a net of finite rank effects \(\{C_\nu \}_\nu \) such that \(C_\nu \le C\) holds for all \(\nu \) and \(C_\nu \rightarrow C\) in SOT. Denote by \(E_C\) the projection-valued spectral measure of C, and set \(C_n = \sum _{j=0}^{n} \frac{j}{n} E_C\left( \left[ \frac{j}{n},\frac{j+1}{n}\right) \right) \) for every \(n\in {\mathbb {N}}\). Clearly, each \(C_n\) has finite spectrum, satisfies \(C_n\le C\), and \(\Vert C_n - C\Vert \rightarrow 0\) as \(n\rightarrow \infty \). For each spectral projection \(E_C\left( \left[ \frac{j}{n},\frac{j+1}{n}\right) \right) \) we can take a sequence of finite-rank projections \(\{P_k^{j,n}\}_{k=1}^\infty \) such that \(P_k^{j,n}\le E_C\left( \left[ \frac{j}{n},\frac{j+1}{n}\right) \right) \) for all k and \(P_k^{j,n} \rightarrow E_C\left( \left[ \frac{j}{n},\frac{j+1}{n}\right) \right) \) in SOT as \(k\rightarrow \infty \). Define \(C_{n,k} := \sum _{j=0}^{n} \frac{j}{n} P_k^{j,n}\). It is apparent that \(C_{n,k} \le C_n\) for all n and k, and that for each n we have \(C_{n,k} \rightarrow C_n\) in SOT as \(k\rightarrow \infty \). Therefore the SOT-closure of \(\{C_{n,k}:n,k\in {\mathbb {N}}\}\) contains each \(C_n\), hence also C, thus we can construct a net \(\{C_\nu \}_\nu \) with the required properties.
Now, let \(B\in A^\sim \) be arbitrary, and consider two other effects \(M,N\in {{\mathcal {E}}}(H)\) that satisfy the conditions of Lemma 2.3 (ii). Set \(C:= M\oplus N \in {{\mathcal {E}}}(H\oplus H)\), and denote by \(E_M\) and \(E_N\) the projection-valued spectral measures of M and N, respectively. Clearly, \(E_C\left( \left[ \frac{j}{n},\frac{j+1}{n}\right) \right) = E_M\left( \left[ \frac{j}{n},\frac{j+1}{n}\right) \right) \oplus E_N\left( \left[ \frac{j}{n},\frac{j+1}{n}\right) \right) \) for each j and n. In the above construction we can choose finite-rank projections of the form \(P_k^{j,n} = Q_k^{j,n}\oplus R_k^{j,n} \in {{\mathcal {P}}}(H\oplus H)\) where \(Q_k^{j,n}, R_k^{j,n} \in {{\mathcal {P}}}(H)\), \(Q_k^{j,n}\le E_M\left( \left[ \frac{j}{n},\frac{j+1}{n}\right) \right) \) and \(R_k^{j,n}\le E_N\left( \left[ \frac{j}{n},\frac{j+1}{n}\right) \right) \) holds for all k, n. Then each element \(C_\nu \) of the convergent net is an orthogonal sum of the form \(M_\nu \oplus N_\nu \in {{\mathcal {E}}}(H\oplus H))\). It is apparent that \(M_\nu , N_\nu \in {{\mathcal {F}}}(H)\), \(M_\nu \le M\) and \(N_\nu \le N\) for all \(\nu \), and that \(M_\nu \rightarrow M\), \(N_\nu \rightarrow N\) holds in SOT. Therefore \(M_\nu +N_\nu \in {{\mathcal {F}}}(H)\cap A^\sim \) and \(M_\nu +N_\nu \) converges to \(M+N = B\) in SOT, the proof is complete. \(\square \)
We proceed to investigate when do we have the equation \(A^\sim = B^\sim \) for two effects A and B, which will take several steps. We will denote the set of all rank-one effects by \({{\mathcal {F}}}_1(H) := \{tP :P\in {{\mathcal {P}}}_1(H), 0<t\le 1\}\).
Lemma 2.10
Let \(H = H_1\oplus H_2\) be an orthogonal decomposition and assume that \(A, B\in {{\mathcal {E}}}(H)\) have the following matrix decompositions:
$$\begin{aligned} A = \left[ \begin{matrix} \lambda _1 I_1 &{} 0 \\ 0 &{} \lambda _2 I_2 \end{matrix} \right] \quad \text {and} \quad B = \left[ \begin{matrix} \mu _1 I_1 &{} 0 \\ 0 &{} \mu _2 I_2 \end{matrix} \right] \; \in {{\mathcal {E}}}(H_1\oplus H_2) \end{aligned}$$
(7)
where \(\lambda _1, \lambda _2, \mu _1, \mu _2 \in [0,1]\), and \(I_1\) and \(I_2\) denote the identity operators on \(H_1\) and \(H_2\), respectively. Then the following are equivalent:
-
(i)
\(\Lambda (A,P) + \Lambda (A^\perp ,P) = \Lambda (B,P) + \Lambda (B^\perp ,P)\) holds for all \(P\in {{\mathcal {P}}}_1(H)\),
-
(ii)
\(A^\sim \cap {{\mathcal {F}}}_1(H) = B^\sim \cap {{\mathcal {F}}}_1(H)\),
-
(iii)
either \(\lambda _1=\lambda _2\) and \(\mu _1=\mu _2\), or \(\lambda _1=\mu _1\) and \(\lambda _2=\mu _2\), or \(\lambda _1 + \mu _1 = \lambda _2 + \mu _2 =1\).
Proof
The directions (iii)\(\Longrightarrow \)(ii)\(\iff \)(i) are trivial by Lemma 2.1 (a) and Corollaries 2.5, 2.7, so we shall only consider the direction (i)\(\Longrightarrow \)(iii). First, a straightforward calculation using the Busch–Gudder theorem gives the following for every \(x_1\in H_1, x_2\in H_2, \Vert x_1\Vert = \Vert x_2\Vert = 1\) and \(0\le \alpha \le \tfrac{\pi }{2}\):
$$\begin{aligned} \Lambda \left( A, P_{\cos \alpha x_1 + \sin \alpha x_2} \right) = \frac{1}{ \left( \tfrac{1}{\lambda _1}\right) \cdot \cos ^2\alpha + \left( \tfrac{1}{\lambda _2}\right) \cdot \sin ^2\alpha }, \end{aligned}$$
(8)
where we use the interpretations \(\tfrac{1}{0} = \infty \), \(\tfrac{1}{\infty } = 0\), \(\infty \cdot 0 = 0\), \(\infty + \infty = \infty \), and \(\infty + a = \infty \), \(\infty \cdot a = \infty \) (\(a>0\)), in order to make the formula valid also for the case when \(\lambda _1 = 0\) or \(\lambda _2 = 0\). Clearly, (8) depends only on \(\alpha \), but not on the specific choices of \(x_1\) and \(x_2\). We define the following two functions
$$\begin{aligned} T_A:\left[ 0,\tfrac{\pi }{2}\right] \rightarrow [0,1], \qquad T_A(\alpha ) = \Lambda \left( A, P_{\cos \alpha x_1 + \sin \alpha x_2} \right) + \Lambda \left( A^\perp , P_{\cos \alpha x_1 + \sin \alpha x_2} \right) \end{aligned}$$
and
$$\begin{aligned} T_B:\left[ 0,\tfrac{\pi }{2}\right] \rightarrow [0,1], \qquad T_B(\alpha ) = \Lambda \left( B, P_{\cos \alpha x_1 + \sin \alpha x_2} \right) + \Lambda \left( B^\perp , P_{\cos \alpha x_1 + \sin \alpha x_2} \right) , \end{aligned}$$
which are the same by our assumptions. By (8), for all \(0\le \alpha \le \tfrac{\pi }{2}\) we have
$$\begin{aligned} T_A(\alpha )&= \frac{1}{ \left( \tfrac{1}{\lambda _1}\right) \cdot \cos ^2\alpha + \left( \tfrac{1}{\lambda _2}\right) \cdot \sin ^2\alpha } + \frac{1}{ \left( \tfrac{1}{1-\lambda _1}\right) \cdot \cos ^2\alpha + \left( \tfrac{1}{1-\lambda _2}\right) \cdot \sin ^2\alpha } \nonumber \\&= \frac{1}{ \left( \tfrac{1}{\mu _1}\right) \cdot \cos ^2\alpha + \left( \tfrac{1}{\mu _2}\right) \cdot \sin ^2\alpha } + \frac{1}{ \left( \tfrac{1}{1-\mu _1}\right) \cdot \cos ^2\alpha + \left( \tfrac{1}{1-\mu _2}\right) \cdot \sin ^2\alpha } =T_B(\alpha ). \end{aligned}$$
(9)
Next, we observe the following implications:
-
if \(\lambda _1 = \lambda _2\), then \(T_A(\alpha )\) is the constant 1 function,
-
if \(\lambda _1 = 0\) and \(\lambda _2 = 1\), then \(T_A(\alpha )\) is the characteristic function \(\chi _{\{0,\pi /2\}}(\alpha )\),
-
if \(\lambda _1 = 0\) and \(0< \lambda _2 < 1\), then \(T_A(\alpha )\) is continuous on \(\left[ 0,\tfrac{\pi }{2}\right) \), but has a jump at \(\tfrac{\pi }{2}\), namely \(\lim _{\alpha \rightarrow \tfrac{\pi }{2}-} T_A(\alpha ) = 1-\lambda _2\) and \(T_A(\tfrac{\pi }{2}) = 1\),
-
if \(\lambda _1 = 1\) and \(0< \lambda _2 < 1\), then \(T_A(\alpha )\) is continuous on \(\left[ 0,\tfrac{\pi }{2}\right) \), but has a jump at \(\tfrac{\pi }{2}\), namely \(\lim _{\alpha \rightarrow \tfrac{\pi }{2}-} T_A(\alpha ) = \lambda _2\) and \(T_A(\tfrac{\pi }{2}) = 1\),
-
if \(\lambda _1,\lambda _2 \in (0,1)\), then \(T_A(\alpha )\) is continuous on \(\left[ 0,\tfrac{\pi }{2}\right] \),
-
if \(\lambda _1 \ne \lambda _2\), then we have \(T_A(0) = T_A(\tfrac{\pi }{2}) = 1\) and \(T_A(\alpha ) < 1\) for all \(0< \alpha < \tfrac{\pi }{2}\).
All of the above statements are rather straightforward computations using the formula (9), let us only show the last one here. Clearly, \(T_A(0) = T_A(\tfrac{\pi }{2}) = 1\) is obvious. As for the other assertion, if \(\lambda _1,\lambda _2 \in (0,1)\), then we can use the strict version of the weighted harmonic-arithmetic mean inequality:
$$\begin{aligned}&\frac{1}{ \tfrac{1}{\lambda _1} \cos ^2\alpha + \tfrac{1}{\lambda _2} \sin ^2\alpha } + \frac{1}{ \tfrac{1}{1-\lambda _1} \cos ^2\alpha + \tfrac{1}{1-\lambda _2} \sin ^2\alpha } \\&\quad< (\lambda _1 \cos ^2\alpha + \lambda _2 \sin ^2\alpha ) + ((1-\lambda _1) \cos ^2\alpha + (1-\lambda _2) \sin ^2\alpha ) = 1 \quad (0<\alpha <\tfrac{\pi }{2}). \end{aligned}$$
If \(\lambda _1 = 0< \lambda _2 < 1\), then we calculate in the following way:
$$\begin{aligned}&\frac{1}{ \left( \tfrac{1}{0}\right) \cos ^2\alpha + \tfrac{1}{\lambda _2} \sin ^2\alpha } + \frac{1}{ \cos ^2\alpha + \tfrac{1}{1-\lambda _2} \sin ^2\alpha }\\&\quad = \frac{1}{ 1-\sin ^2\alpha + \tfrac{1}{1-\lambda _2} \sin ^2\alpha }< 1 \quad (0<\alpha <\tfrac{\pi }{2}). \end{aligned}$$
The remaining cases are very similar.
The above observations together with Corollary 2.5 and (9) readily imply the following:
-
\(A\in \mathcal {SC}(H)\) if and only if \(B\in \mathcal {SC}(H)\),
-
\(A\in {{\mathcal {P}}}(H)\setminus \mathcal {SC}(H)\) if and only if \(B\in {{\mathcal {P}}}(H)\setminus \mathcal {SC}(H)\), in which case \(B \in \{A, A^\perp \}\),
-
there exists a \(P\in {{\mathcal {P}}}(H)\setminus \mathcal {SC}(H)\) and a \(t\in (0,1)\) with \(A\in \{tP, I-tP\}\) if and only if \(B\in \{tP, I-tP\}\),
-
\(\lambda _1,\lambda _2 \in (0,1)\) and \(\lambda _1\ne \lambda _2\) if and only if \(\mu _1, \mu _2 \in (0,1)\) and \(\mu _1\ne \mu _2\).
So what remained is to show that in the last case we further have \(B \in \{A, A^\perp \}\), which is what we shall do below.
Let us introduce the following functions:
$$\begin{aligned}&{\mathcal {T}}_A:[0,1]\rightarrow [0,1], \qquad {\mathcal {T}}_A(s) := T_A(\arcsin \sqrt{s}) = \frac{\lambda _1\lambda _2}{\lambda _1 s + \lambda _2 (1-s)} \\&\quad + \frac{(1-\lambda _1)(1-\lambda _2)}{(1-\lambda _1) s + (1-\lambda _2) (1-s)} \end{aligned}$$
and
$$\begin{aligned}&{\mathcal {T}}_B:[0,1]\rightarrow [0,1], \qquad {\mathcal {T}}_B(s) := T_B(\arcsin \sqrt{s}) = \frac{\mu _1\mu _2}{\mu _1 s + \mu _2 (1-s)}\\&\quad + \frac{(1-\mu _1)(1-\mu _2)}{(1-\mu _1) s + (1-\mu _2) (1-s)}. \end{aligned}$$
Our aim is to prove that \({\mathcal {T}}_A(s) = {\mathcal {T}}_B(s)\) (\(s\in [0,1]\)) implies either \(\lambda _1=\mu _1\) and \(\lambda _2=\mu _2\), or \(\lambda _1 + \mu _1 = \lambda _2 + \mu _2 =1\). The derivative of \({\mathcal {T}}_A\) is
$$\begin{aligned} \tfrac{d {\mathcal {T}}_A}{ds}(s) = (\lambda _1-\lambda _2)\left( \frac{-\lambda _1\lambda _2}{(\lambda _1 s + \lambda _2 (1-s))^2} + \frac{(1-\lambda _1)(1-\lambda _2)}{((1-\lambda _1) s + (1-\lambda _2) (1-s))^2}\right) , \end{aligned}$$
from which we calculate
$$\begin{aligned} \tfrac{d {\mathcal {T}}_A}{ds}(0) = -\frac{(\lambda _1-\lambda _2)^2}{(1-\lambda _2)\lambda _2} \quad \text {and} \quad \tfrac{d {\mathcal {T}}_A}{ds}(1) = \frac{(\lambda _1-\lambda _2)^2}{(1-\lambda _1)\lambda _1}. \end{aligned}$$
Therefore, if we managed to show that the function
$$\begin{aligned} F:(0,1)^2 \rightarrow {\mathbb {R}}^{2}, \quad F(x,y) = \left( \tfrac{(x-y)^2}{(1-x)x}, \tfrac{(x-y)^2}{(1-y)y} \right) \end{aligned}$$
is injective on the set \(\Delta := \{(x,y)\in {\mathbb {R}}^{2}:0<y<x<1 \}\), then we are done (note that \(F(x,y) = F(1-x,1-y)\)). For this assume that with some \(c,d>0\) we have
$$\begin{aligned} \frac{(x-y)^2}{(1-x)x} = \frac{1}{c} \quad \text {and} \quad \frac{(x-y)^2}{(1-y)y} = \frac{1}{d}, \end{aligned}$$
or equivalently,
$$\begin{aligned} (1-x)x = c(x-y)^2 \quad \text {and} \quad (1-y)y = d(x-y)^2. \end{aligned}$$
If we substitute \(u = \tfrac{x-y}{2}\) and \(v = \tfrac{x+y}{2}\), then we get
$$\begin{aligned} (u+v)^2-(u+v) = -4cu^2 \quad \text {and} \quad (v-u)^2-(v-u) = -4du^2. \end{aligned}$$
Now, considering the sum and difference of these two equations and manipulate them a bit gives
$$\begin{aligned} v^2-v = -(2c+2d+1)u^2 \quad \text {and} \quad v = (d-c) u + \tfrac{1}{2}. \end{aligned}$$
From these latter equations we conclude
$$\begin{aligned} x-y = 2u = \sqrt{\tfrac{1}{(d-c)^2+2c+2d+1}} \quad \text {and} \quad x+y = 2v = 2(d-c) u + 1, \end{aligned}$$
which clearly implies that F is globally injective on \(\Delta \), and the proof is complete. \(\square \)
We have an interesting consequence in finite dimensions.
Corollary 2.11
Assume that \(2\le \dim H < \infty \) and \(A,B \in {{\mathcal {E}}}(H)\). Then the following are equivalent:
-
(i)
\(\Lambda (A, P) + \Lambda (A^\perp , P) = \Lambda (B, P) + \Lambda (B^\perp , P)\) for all \(P\in P_1(H)\),
-
(ii)
\(A^\sim \cap {{\mathcal {F}}}_1(H) = B^\sim \cap {{\mathcal {F}}}_1(H)\),
-
(iii)
either \(A,B \in \mathcal {SC}(H)\), or \(A=B\), or \(A=B^\perp \).
Proof
The directions (i)\(\iff \)(ii)\(\Longleftarrow \)(iii) are trivial, so we shall only prove the (ii)\(\Longrightarrow \)(iii) direction. First, let us consider the two-dimensional case. As we saw in the proof of Lemma 2.10, we have \(A^\sim \cap {{\mathcal {F}}}_1(H) = {{\mathcal {F}}}_1(H)\) if and only if A is a scalar effect (see the first set of bullet points there). Therefore, without loss of generality we may assume that none of A and B are scalar effects. Notice that by Lemma 2.1, A and B commute with exactly the same rank-one projections, hence A and B possess the forms in (7) with some one-dimensional subspaces \(H_1\) and \(H_2\), and an easy application of Lemma 2.10 gives (iii).
As for the general case, since again A and B commute with exactly the same rank-one projections, we can jointly diagonalise them with respect to some orthonormal basis \(\{e_j\}_{j=1}^n\), where \(n = \dim H\):
$$\begin{aligned} A = \left[ \begin{matrix} \lambda _1 &{} 0 &{} \dots &{} 0 &{} 0 \\ 0 &{} \lambda _2 &{} \dots &{} 0 &{} 0 \\ \vdots &{} &{} \ddots &{} &{} \vdots \\ 0 &{} 0 &{} \dots &{} \lambda _{n-1} &{} 0 \\ 0 &{} 0 &{} \dots &{} 0 &{} \lambda _n \\ \end{matrix}\right] \quad \text {and} \quad B = \left[ \begin{matrix} \mu _1 &{} 0 &{} \dots &{} 0 &{} 0 \\ 0 &{} \mu _2 &{} \dots &{} 0 &{} 0 \\ \vdots &{} &{} \ddots &{} &{} \vdots \\ 0 &{} 0 &{} \dots &{} \mu _{n-1} &{} 0 \\ 0 &{} 0 &{} \dots &{} 0 &{} \mu _n \\ \end{matrix}\right] . \end{aligned}$$
Of course, for any two distinct \(i, j \in \{1,\dots , n\}\) we have the following equation for the strength functions:
$$\begin{aligned} \Lambda (A, P) + \Lambda (A^\perp , P) = \Lambda (B, P) + \Lambda (B^\perp , P) \qquad (P \in P_1({\mathbb {C}}\cdot e_i + {\mathbb {C}}\cdot e_j)), \end{aligned}$$
which instantly implies
$$\begin{aligned} \left[ \begin{matrix} \mu _i &{} 0 \\ 0 &{} \mu _j \end{matrix}\right] ^\sim \cap {{\mathcal {F}}}_1({\mathbb {C}}\cdot e_i + {\mathbb {C}}\cdot e_j) = \left[ \begin{matrix} \lambda _i &{} 0 \\ 0 &{} \lambda _j \end{matrix}\right] ^\sim \cap {{\mathcal {F}}}_1({\mathbb {C}}\cdot e_i + {\mathbb {C}}\cdot e_j). \end{aligned}$$
By the two-dimensional case this means that we have one of the following cases:
-
\(\lambda _i = \lambda _j\) and \(\mu _i = \mu _j\),
-
\(\lambda _i \ne \lambda _j\) and either \(\mu _i = \lambda _i\) and \(\mu _j = \lambda _j\), or \(\mu _i = 1-\lambda _i\) and \(\mu _j = 1-\lambda _j\).
From here it is easy to conclude (iii). \(\square \)
The commutant of an operator \(T\in {{\mathcal {B}}}(H)\) will be denoted by \(T' := \{ S\in {{\mathcal {B}}}(H) :ST=TS \}\), and more generally, if \({\mathcal {M}}\subseteq {{\mathcal {B}}}(H)\), then we set \({\mathcal {M}}' := \cap \{ T' :T\in {\mathcal {M}}\}\). We shall use the notations \(T'' := (T')'\) and \({\mathcal {M}}'' := ({\mathcal {M}}')'\) for the double commutants.
Lemma 2.12
For any \(A,B \in {{\mathcal {E}}}(H)\) the following three assertions hold:
-
(a)
If \(A^\sim \subseteq B^\sim \), then \(B\in A''\).
-
(b)
If \(\dim H \le \aleph _0\) and \(A^\sim \subseteq B^\sim \), then there exists a Borel function \(f:[0,1] \rightarrow [0,1]\) such that \(B = f(A)\).
-
(c)
If B is a convex combination of \(A, A^\perp , 0\) and I, then \(A^\sim \subseteq B^\sim \).
Proof
(a): Assume that \(C\in A'\). Our goal is to show \(B\in C'\). We express C in the following way:
$$\begin{aligned} C = C_{\mathfrak {R}} + i C_{\mathfrak {I}}, \; C_{\mathfrak {R}} = \frac{C+C^*}{2}, \; C_{\mathfrak {I}} = \frac{C-C^*}{2i} \end{aligned}$$
where \(C_{\mathfrak {R}}\) and \(C_{\mathfrak {I}}\) are self-adjoint (they are usually called the real and imaginary parts of C). Since A is self-adjoint, \(C^* \in A'\), hence \(C_{\mathfrak {R}}, C_{\mathfrak {I}} \in A'\). Let \(E_{\mathfrak {R}}\) and \(E_{\mathfrak {I}}\) denote the projection-valued spectral measures of \(C_{\mathfrak {R}}\) and \(C_{\mathfrak {I}}\), respectively. By the spectral theorem ([7, Theorem IX.2.2]), Lemma 2.1 and Corollary 2.4, we have \(E_{\mathfrak {R}}(\Delta ), E_{\mathfrak {I}}(\Delta ) \in A^c \subseteq A^\sim \subseteq B^\sim \), therefore also \(E_{\mathfrak {R}}(\Delta ), E_{\mathfrak {I}}(\Delta ) \in B'\) for all \(\Delta \in {{\mathcal {B}}}_{\mathbb {R}}\), which gives \(C\in B'\).
(b): This is an easy consequence of [7, Proposition IX.8.1 and Lemma IX.8.7].
(c): If \(A\sim C\), then also \(A^\perp , 0\) and \(I\sim C\). Hence by the convexity of \(C^\sim \) we obtain \(B\sim C\). \(\square \)
Now, we are in the position to prove our first main result.
Proof of Theorem 1.1
If H is separable, then the equivalence (ii)\(\iff \)(iii) is straightforward by Lemma 2.9. For general H the direction (i)\(\Longrightarrow \)(ii) is obvious, therefore we shall only prove (ii)\(\Longrightarrow \)(i), first in the separable, and then in the general case. By Lemma 2.1, we may assume throughout the rest of the proof that A and B are non-scalar effects. We will denote the spectral subspace of a self-adjoint operator T associated to a Borel set \(\Delta \subseteq {\mathbb {R}}\) by \(H_T(\Delta )\).
(ii)\(\Longrightarrow \)(i) in the separable case: We split this part into two steps.
STEP 1: Here, we establish two estimations, (11) and (12), for the strength functions of A and B on certain subspaces of H. Let \(\lambda _1, \lambda _2 \in \sigma (A), \lambda _1 \ne \lambda _2\) and \(0< \varepsilon < \tfrac{1}{2}|\lambda _1-\lambda _2|\). Then the spectral subspaces \(H_1 = H_A\left( (\lambda _1-\varepsilon , \lambda _1+\varepsilon ) \right) \) and \(H_2 = H_A\left( (\lambda _2-\varepsilon , \lambda _2+\varepsilon ) \right) \) are non-trivial and orthogonal. Set \(H_3\) to be the orthogonal complement of \(H_1\oplus H_2\), then the matrix of A written in the orthogonal decomposition \(H = H_1\oplus H_2 \oplus H_3\) is diagonal:
$$\begin{aligned} A = \left[ \begin{matrix} A_1 &{} 0 &{} 0 \\ 0 &{} A_2 &{} 0 \\ 0 &{} 0 &{} A_3 \end{matrix}\right] \in {{\mathcal {B}}}(H_1\oplus H_2 \oplus H_3). \end{aligned}$$
Note that \(H_3\) might be a trivial subspace. Since by Corollary 2.4A and B commute with exactly the same projections, the matrix of B in \(H = H_1\oplus H_2 \oplus H_3\) is also diagonal:
$$\begin{aligned} B = \left[ \begin{matrix} B_1 &{} 0 &{} 0 \\ 0 &{} B_2 &{} 0 \\ 0 &{} 0 &{} B_3 \end{matrix}\right] \in {{\mathcal {B}}}(H_1\oplus H_2 \oplus H_3). \end{aligned}$$
At this point, let us emphasise that of course \(H_j, A_j\) and \(B_j\) (\(j=1,2,3\)) all depend on \(\lambda _1, \lambda _2\) and \(\varepsilon \), but in order to keep our notation as simple as possible, we will stick with these symbols. However, if at any point it becomes important to point out this dependence, we shall use for instance \(B_j^{(\lambda _1, \lambda _2, \varepsilon )}\) instead of \(B_j\). Similar conventions apply later on.
Observe that by Corollary 2.8 we have
$$\begin{aligned} \left[ \begin{matrix} A_1 &{} 0 \\ 0 &{} A_2 \end{matrix}\right] ^\sim = \left[ \begin{matrix} B_1 &{} 0 \\ 0 &{} B_2 \end{matrix}\right] ^\sim . \end{aligned}$$
Now, we pick two arbitrary points \(\mu _1\in \sigma (B_1)\) and \(\mu _2\in \sigma (B_2)\). Then obviously, the following two subspaces are non-zero subspaces of \(H_1\) and \(H_2\), respectively:
$$\begin{aligned} {\widehat{H}}_1 := (H_1)_{B_1}\big ( (\mu _1-\varepsilon , \mu _1+\varepsilon ) \big ), \;\; {\widehat{H}}_2 := (H_2)_{B_2}\big ( (\mu _2-\varepsilon , \mu _2+\varepsilon ) \big ). \end{aligned}$$
Similarly as above, we have the following matrix forms where \({\check{H}}_j = H_j \ominus {\widehat{H}}_j\) \((j=1,2)\):
$$\begin{aligned} B_1 = \left[ \begin{matrix} {\widehat{B}}_1 &{} 0 \\ 0 &{} {\check{B}}_1 \end{matrix}\right] \in {{\mathcal {B}}}({\widehat{H}}_1 \oplus {\check{H}}_1) \;\; \text {and} \;\; B_2 = \left[ \begin{matrix} {\widehat{B}}_2 &{} 0 \\ 0 &{} {\check{B}}_2 \end{matrix}\right] \in {{\mathcal {B}}}({\widehat{H}}_2 \oplus {\check{H}}_2) \end{aligned}$$
and
$$\begin{aligned} A_1 = \left[ \begin{matrix} {\widehat{A}}_1 &{} 0 \\ 0 &{} {\check{A}}_1 \end{matrix}\right] \in {{\mathcal {B}}}({\widehat{H}}_1 \oplus {\check{H}}_1) \;\; \text {and} \;\; A_2 = \left[ \begin{matrix} {\widehat{A}}_2 &{} 0 \\ 0 &{} {\check{A}}_2 \end{matrix}\right] \in {{\mathcal {B}}}({\widehat{H}}_2 \oplus {\check{H}}_2). \end{aligned}$$
Note that \({\check{H}}_1\) or \({\check{H}}_2\) might be trivial subspaces. Again by Corollary 2.8, we have
$$\begin{aligned} \left[ \begin{matrix} {\widehat{A}}_1 &{} 0 \\ 0 &{} {\widehat{A}}_2 \end{matrix}\right] ^\sim = \left[ \begin{matrix} {\widehat{B}}_1 &{} 0 \\ 0 &{} {\widehat{B}}_2 \end{matrix}\right] ^\sim . \end{aligned}$$
Let us point out that by construction \(\sigma ({\widehat{A}}_j) \subseteq [\lambda _j-\varepsilon ,\lambda _j+\varepsilon ]\) and \(\sigma ({\widehat{B}}_j) \subseteq [\mu _j-\varepsilon ,\mu _j+\varepsilon ]\). Corollary 2.7 gives the following identity for the strength functions, where \({\widehat{I}}_j\) denotes the identity on \({\widehat{H}}_j\) \((j=1,2)\):
$$\begin{aligned}&\Lambda \left( \left[ \begin{matrix} {\widehat{A}}_1 &{} 0 \\ 0 &{} {\widehat{A}}_2 \end{matrix}\right] , P \right) + \Lambda \left( \left[ \begin{matrix} {\widehat{I}}_1 - {\widehat{A}}_1 &{} 0 \\ 0 &{} {\widehat{I}}_2 - {\widehat{A}}_2 \end{matrix}\right] , P \right) \nonumber \\&\quad = \Lambda \left( \left[ \begin{matrix} {\widehat{B}}_1 &{} 0 \\ 0 &{} {\widehat{B}}_2 \end{matrix}\right] , P \right) + \Lambda \left( \left[ \begin{matrix} {\widehat{I}}_1 - {\widehat{B}}_1 &{} 0 \\ 0 &{} {\widehat{I}}_2 - {\widehat{B}}_2 \end{matrix}\right] , P \right) \quad \left( \forall \; P\in {{\mathcal {P}}}_1\left( {\widehat{H}}_1\oplus {\widehat{H}}_2\right) \right) . \end{aligned}$$
(10)
Define
$$\begin{aligned} \Theta :{\mathbb {R}}\rightarrow [0,1], \quad \Theta (t) = \left\{ \begin{matrix} 0 &{} \text {if} \; t< 0 \\ t &{} \text {if} \; 0 \le t \le 1 \\ 1&{} \text {if} \; 1 < t \end{matrix} \right. , \end{aligned}$$
and notice that we have the following two estimations for all rank-one projections P:
$$\begin{aligned}&\Lambda \left( \left[ \begin{matrix} \Theta (\lambda _1-\varepsilon ){\widehat{I}}_1 &{} 0 \\ 0 &{} \Theta (\lambda _2-\varepsilon ){\widehat{I}}_2 \end{matrix}\right] , P \right) \nonumber \\&\qquad + \Lambda \left( \left[ \begin{matrix} \Theta (1-\lambda _1-\varepsilon ){\widehat{I}}_1 &{} 0 \\ 0 &{} \Theta (1-\lambda _2-\varepsilon ){\widehat{I}}_2 \end{matrix}\right] , P \right) \nonumber \\&\quad \le \text {the expression in }(10) \nonumber \\&\quad \le \Lambda \left( \left[ \begin{matrix} \Theta (\lambda _1+\varepsilon ){\widehat{I}}_1 &{} 0 \\ 0 &{} \Theta (\lambda _2+\varepsilon ){\widehat{I}}_2 \end{matrix}\right] , P \right) \nonumber \\&\qquad + \Lambda \left( \left[ \begin{matrix} \Theta (1-\lambda _1+\varepsilon ){\widehat{I}}_1 &{} 0 \\ 0 &{} \Theta (1-\lambda _2+\varepsilon ){\widehat{I}}_2 \end{matrix}\right] , P \right) \end{aligned}$$
(11)
and
$$\begin{aligned}&\Lambda \left( \left[ \begin{matrix} \Theta (\mu _1-\varepsilon ){\widehat{I}}_1 &{} 0 \\ 0 &{} \Theta (\mu _2-\varepsilon ){\widehat{I}}_2 \end{matrix}\right] , P \right) \nonumber \\&\qquad + \Lambda \left( \left[ \begin{matrix} \Theta (1-\mu _1-\varepsilon ){\widehat{I}}_1 &{} 0 \\ 0 &{} \Theta (1-\mu _2-\varepsilon ){\widehat{I}}_2 \end{matrix}\right] , P \right) \nonumber \\&\quad \le \text {the expression in} (10) \nonumber \\&\quad \le \Lambda \left( \left[ \begin{matrix} \Theta (\mu _1+\varepsilon ){\widehat{I}}_1 &{} 0 \\ 0 &{} \Theta (\mu _2+\varepsilon ){\widehat{I}}_2 \end{matrix}\right] , P \right) \nonumber \\&\qquad + \Lambda \left( \left[ \begin{matrix} \Theta (1-\mu _1+\varepsilon ){\widehat{I}}_1 &{} 0 \\ 0 &{} \Theta (1-\mu _2+\varepsilon ){\widehat{I}}_2 \end{matrix}\right] , P \right) . \end{aligned}$$
(12)
Note that the above estimations hold for any arbitrarily small \(\varepsilon \) and for all suitable choices of \(\mu _1\) and \(\mu _2\) (which of course depend on \(\varepsilon \)).
STEP 2: Here we show that \(B\in \{A,A^\perp \}\). Let us define the following set that depends only on \(\lambda _j\):
$$\begin{aligned} {\mathcal {C}}_j= & {} {\mathcal {C}}_j^{(\lambda _j)} := \bigcap \left\{ \sigma \left( B_j^{(\lambda _1, \lambda _2, \varepsilon )} \right) :0< \varepsilon< \tfrac{1}{2}|\lambda _1 - \lambda _2| \right\} \\= & {} \bigcap \left\{ \sigma \left( B|_{H_A((\lambda _j-\varepsilon ,\lambda _j+\varepsilon ))} \right) :0 < \varepsilon \right\} \quad (j=1,2). \end{aligned}$$
Notice that as this set is an intersection of monotonically decreasing (as \(\varepsilon \searrow 0\)), compact, non-empty sets, it must contain at least one element. Also, observe that if \(\mu _1\in {\mathcal {C}}_1\) and \(\mu _2\in {\mathcal {C}}_2\), then (11) and (12) hold for all \(\varepsilon >0\).
We proceed with proving that either \({\mathcal {C}}_1 = \{\lambda _1\}\) and \({\mathcal {C}}_2 = \{\lambda _2\}\), or \({\mathcal {C}}_1 = \{1-\lambda _1\}\) and \({\mathcal {C}}_2 = \{1-\lambda _2\}\) hold. Fix two arbitrary elements \(\mu _1 \in {\mathcal {C}}_1\) and \(\mu _2 \in {\mathcal {C}}_2\), and assume that neither \(\lambda _1 = \mu _1\) and \(\lambda _2 = \mu _2\), nor \(\lambda _1 + \mu _1 = \lambda _2 + \mu _2 = 1\) hold. From here our aim is to get a contradiction. As we showed in the proof of Lemma 2.10, there exists an \(\alpha _0 \in \left( 0,\tfrac{\pi }{2}\right) \) such that we have
$$\begin{aligned}&\frac{1}{ \left( \tfrac{1}{\lambda _1}\right) \cdot \cos ^2\alpha _0 + \left( \tfrac{1}{\lambda _2}\right) \cdot \sin ^2\alpha _0 } + \frac{1}{ \left( \tfrac{1}{1-\lambda _1}\right) \cdot \cos ^2\alpha _0 + \left( \tfrac{1}{1-\lambda _2}\right) \cdot \sin ^2\alpha _0 } \\&\quad \ne \frac{1}{ \left( \tfrac{1}{\mu _1}\right) \cdot \cos ^2\alpha _0 + \left( \tfrac{1}{\mu _2}\right) \cdot \sin ^2\alpha _0 } + \frac{1}{ \left( \tfrac{1}{1-\mu _1}\right) \cdot \cos ^2\alpha _0 + \left( \tfrac{1}{1-\mu _2}\right) \cdot \sin ^2\alpha _0 } \end{aligned}$$
where we interpret both sides as in (8). Notice that both summands on both sides depend continuously on \(\lambda _1,\lambda _2,\mu _1\) and \(\mu _2\). Therefore there exists an \(\varepsilon >0\) small enough and a rank-one projection \(P = P_{\cos \alpha _0 {\widehat{x}}_1+\sin \alpha _0 {\widehat{x}}_2}\), with \({\widehat{x}}_1\in {\widehat{H}}_1, {\widehat{x}}_2\in {\widehat{H}}_2, \Vert {\widehat{x}}_1\Vert =\Vert {\widehat{x}}_2\Vert =1\), such that the closed intervals bounded by the right- and left-hand sides of (11), and those of (12) are disjoint—which is a contradiction.
Observe that as we can do the above for any two disjoint elements of the spectrum \(\sigma (A)\), we can conclude that one of the following possibilities occur:
$$\begin{aligned} \{\lambda \} = \bigcap \left\{ \sigma \left( B|_{H_A((\lambda -\varepsilon , \lambda +\varepsilon ))} \right) :\varepsilon > 0 \right\} \qquad (\lambda \in \sigma (A)) \end{aligned}$$
(13)
or
$$\begin{aligned} \{1-\lambda \} = \bigcap \left\{ \sigma \left( B|_{H_A((\lambda -\varepsilon , \lambda +\varepsilon ))} \right) :\varepsilon > 0 \right\} \qquad (\lambda \in \sigma (A)). \end{aligned}$$
(14)
From here, we show that (13) implies \(A=B\), and (14) implies \(B=A^\perp \). As the latter can be reduced to the case (13), by considering \(B^\perp \) instead of B, we may assume without loss of generality that (13) holds. By Lemma 2.12 and [7, Theorem IX.8.10], there exists a function \(f\in L^\infty (\mu )\), where \(\mu \) is a scalar-valued spectral measure of A, such that \(B = f(A)\). Moreover, we have \(B = A\) if and only if \(f(\lambda ) = \lambda \) \(\mu \)-a.e, so we only have to prove the latter equation. Let us fix an arbitrarily small number \(\delta > 0\). By the spectral mapping theorem ( [7, Theorem IX.8.11]) and (13) we notice that for every \(\lambda \in \sigma (A)\) there exists an \(0< \varepsilon _{\lambda } < \delta \) such that
$$\begin{aligned} \mu -\mathrm {essran} \left( f|_{(\lambda -\varepsilon _{\lambda },\lambda +\varepsilon _{\lambda })} \right) = \sigma \left( B|_{H_A((\lambda -\varepsilon _{\lambda },\lambda +\varepsilon _{\lambda }))} \right) \subseteq (\lambda -\delta ,\lambda +\delta ), \end{aligned}$$
(15)
where \(\mu -\mathrm {essran}\) denotes the essential range of a function with respect to \(\mu \) (see [7, Example IX.2.6]). Now, for every \(\lambda \in \sigma (A)\) we fix such an \(\varepsilon _{\lambda }\). Clearly, the intervals \(\{(\lambda -\varepsilon _{\lambda },\lambda +\varepsilon _{\lambda }) :\lambda \in \sigma (A)\}\) cover the whole spectrum \(\sigma (A)\), which is a compact set. Therefore we can find finitely many of them, let’s say \(\lambda _1, \dots , \lambda _n\) so that
$$\begin{aligned} \sigma (A) \subseteq \bigcup _{j=1}^{n} \left( \lambda _j-\varepsilon _{\lambda _j},\lambda _j+\varepsilon _{\lambda _j} \right) . \end{aligned}$$
Finally, we define the function
$$\begin{aligned} h(\lambda ) = \lambda _j, \text { where } |\lambda -\lambda _{i}| \ge \varepsilon _{\lambda _i} \text { for all } 1\le i< j \text { and } |\lambda -\lambda _{j}| < \varepsilon _{\lambda _j}. \end{aligned}$$
By definition we have \(\Vert h - \mathrm {id}_{\sigma (A)} \Vert _{\infty } \le \delta \) where the \(\infty \)-norm is taken with respect to \(\mu \) and \(\mathrm {id}_{\sigma (A)}(\lambda ) = \lambda \) \((\lambda \in \sigma (A))\). But notice that by (15) we also have \(\Vert h - f \Vert _{\infty } \le \delta \), and hence \(\Vert f - \mathrm {id}_{\sigma (A)} \Vert _\infty \le 2\delta \). As this inequality holds for all positive \(\delta \), we actually get that \(f(\lambda ) = \lambda \) for \(\mu \)-a.e. \(\lambda \).
(ii)\(\Longrightarrow \)(i) in the non-separable case: It is well-known that there exists an orthogonal decomposition \(H = \oplus _{i\in \mathcal {I}} H_i\) such that each \(H_i\) is a non-trivial, separable, invariant subspace of A, see for instance [7, Proposition IX.4.4]. Since A and B commute with exactly the same projections, both are diagonal with respect to the decomposition \(H = \oplus _{i\in \mathcal {I}} H_i\):
$$\begin{aligned} A = \oplus _{i\in \mathcal {I}} A_i \;\; \text {and} \;\; B = \oplus _{i\in \mathcal {I}} B_i \in {{\mathcal {E}}}(\oplus _{i\in \mathcal {I}} H_i). \end{aligned}$$
By Corollary 2.8 we have \(A_i^\sim = B_i^\sim \) for all \(i\in \mathcal {I}\), therefore the separable case implies
$$\begin{aligned} \text {either} \; A_i = B_i, \;\; \text {or} \; B_i = A_i^\perp , \;\; \text {or} \; A_i, B_i \in \mathcal {SC}(H_i) \qquad (i\in \mathcal {I}). \end{aligned}$$
Without loss of generality we may assume from now on that there exists an \(i_0\in \mathcal {I}\) so that \(A_{i_0}\) is not a scalar effect. (In case all of them are scalar, we simply combine two subspaces \(H_{i_1}\) and \(H_{i_2}\) so that \(\sigma (A_{i_1})\ne \sigma (A_{i_2})\)). This implies either \(A_{i_0} = B_{i_0}\), or \(B_{i_0} = A_{i_0}^\perp \). By considering \(B^\perp \) instead of B if necessary, we may assume from now on that \(A_{i_0} = B_{i_0}\) holds.
Finally, let \(i_1 \in \mathcal {I} \setminus \{i_0\}\) be arbitrary, and let us consider the orthogonal decomposition \(H = \oplus _{i\in \mathcal {I}\setminus \{i_0,i_1\}} H_i \oplus K\) where \(K = H_{i_0}\oplus H_{i_1}\). Similarly as above, we obtain either \(A_{i_0}\oplus A_{i_1} = B_{i_0} \oplus B_{i_1}\), or \(B_{i_0}\oplus B_{i_1} = A_{i_0}^\perp \oplus A_{i_1}^\perp \), but since \(A_{i_0} = B_{i_0}\), we must have \(A_{i_1} = B_{i_1}\). As this holds for arbitrary \(i_1\), the proof is complete. \(\square \)
Now, we are in the position to give an alternative proof of Molnár’s theorem which also extends to the two-dimensional case.
Proof of Theorem 1.2 and Molnár’s theorem
By (a) of Lemma 2.1 and (\(\sim \)) we obtain \(\phi (\mathcal {SC}(H)) = \mathcal {SC}(H)\), moreover, the property (\(\le \)) implies the existence of a strictly increasing bijection \(g:[0,1] \rightarrow [0,1]\) such that \(\phi (\lambda I) = g(\lambda ) I\) for every \(\lambda \in [0,1]\). By Theorem 1.1 we conclude
$$\begin{aligned} \phi (A^\perp ) = \phi (A)^\perp \qquad (A\in {{\mathcal {E}}}(H)\setminus \mathcal {SC}(H)). \end{aligned}$$
We only have to show that the same holds for scalar operators, because then the theorem is reduced to Ludwig’s theorem. For any effect A and any set of effects \({{\mathcal {S}}}\) let us define the following sets \(A^\le := \{B\in {{\mathcal {E}}}(H):A\le B\}\), \(A^\ge := \{B\in {{\mathcal {E}}}(H):A\ge B\}\) and \({{\mathcal {S}}}^\perp := \{B^\perp :B\in {{\mathcal {S}}}\}\). Observe that for any \(s,t\in [0,1]\) we have
$$\begin{aligned} \left( (sI)^\le \cap (tI)^\ge \setminus \mathcal {SC}(H)\right) ^\perp = (sI)^\le \cap (tI)^\ge \setminus \mathcal {SC}(H) \ne \emptyset \end{aligned}$$
(16)
if and only if \(t=1-s\) and \(s< \tfrac{1}{2}\). Thus for all \(s < \tfrac{1}{2}\) we obtain
$$\begin{aligned} \emptyset&\ne \left( (g(s)I)^\le \cap (g(1-s)I)^\ge \setminus \mathcal {SC}(H)\right) ^\perp = \phi \left( \left( (sI)^\le \cap ((1-s)I)^\ge \setminus \mathcal {SC}(H)\right) ^\perp \right) \\&= \phi \left( (sI)^\le \cap ((1-s)I)^\ge \setminus \mathcal {SC}(H)\right) = (g(s)I)^\le \cap (g(1-s)I)^\ge \setminus \mathcal {SC}(H), \end{aligned}$$
which by (16) implies \(g(1-s) = 1-g(s)\) and \(g(s)< \tfrac{1}{2}\), therefore we indeed have (\(\perp \)) for every effect. \(\square \)