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I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But notice that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.) Conversely, two commuting self-adjoint operators $B$ and $C$ can be packed into a normal operator (2). We stress that the commutativity of $B$ and $C$ precisely encodes the normality condition (1).

Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$.

We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a commuting pair of standard real-valued observables, i.e. self-adjoint operators. For this reason, the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

For more on real-valued observables, see e.g. thisthis Phys.SE post and links therein.

--

$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$.

I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But notice that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.) Conversely, two commuting self-adjoint operators $B$ and $C$ can be packed into a normal operator (2). We stress that the commutativity of $B$ and $C$ precisely encodes the normality condition (1).

Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$.

We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a commuting pair of standard real-valued observables, i.e. self-adjoint operators. For this reason, the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

For more on real-valued observables, see e.g. this Phys.SE post and links therein.

--

$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$.

I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But notice that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.) Conversely, two commuting self-adjoint operators $B$ and $C$ can be packed into a normal operator (2). We stress that the commutativity of $B$ and $C$ precisely encodes the normality condition (1).

Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$.

We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a commuting pair of standard real-valued observables, i.e. self-adjoint operators. For this reason, the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

For more on real-valued observables, see e.g. this Phys.SE post and links therein.

--

$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$.

6 added explanation
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I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But notice that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.) Conversely, two commuting self-adjoint operators $B$ and $C$ can be packed into a normal operator (2). We stress that the commutativity of $B$ and $C$ precisely encodes the normality condition (1).

Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$.

We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a commuting pair of standard real-valued observables, i.e. self-adjoint operators. HenceFor this reason, the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

For more on real-valued observables, see e.g. this Phys.SE post and links therein.

--

$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$.

I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But notice that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.)

Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$.

We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a pair of standard real-valued observables, i.e. self-adjoint operators. Hence the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

For more on real-valued observables, see e.g. this Phys.SE post and links therein.

--

$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$.

I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But notice that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.) Conversely, two commuting self-adjoint operators $B$ and $C$ can be packed into a normal operator (2). We stress that the commutativity of $B$ and $C$ precisely encodes the normality condition (1).

Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$.

We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a commuting pair of standard real-valued observables, i.e. self-adjoint operators. For this reason, the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

For more on real-valued observables, see e.g. this Phys.SE post and links therein.

--

$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$.

5 Added explanation
source | link

I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But notenotice that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

Thus($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.)

Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$.

We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a pair of standard real-valued observables, i.e. self-adjoint operators. Hence the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

For more on real-valued observables, see e.g. this Phys.SE post and links therein.

--

$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$.

I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But note that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

Thus a normal operator does not lead to anything fundamentally new which couldn't have been covered by standard real-valued observables, i.e. self-adjoint operators. Hence the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

--

$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$.

I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But notice that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.)

Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$.

We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a pair of standard real-valued observables, i.e. self-adjoint operators. Hence the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

For more on real-valued observables, see e.g. this Phys.SE post and links therein.

--

$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$.

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