7 replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/ edited Apr 13 '17 at 12:39 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 edited Nov 2 '13 at 23:34 Qmechanic♦ 112k1313 gold badges219219 silver badges13341334 bronze badges 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 edited Oct 29 '13 at 11:08 Qmechanic♦ 112k1313 gold badges219219 silver badges13341334 bronze badges 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 $$C=\frac{A-A^{\dagger}}{2i}$$. 4 added 100 characters in body edited Oct 29 '13 at 2:55 Qmechanic♦ 112k1313 gold badges219219 silver badges13341334 bronze badges 3 added 555 characters in body edited Oct 29 '13 at 0:12 Qmechanic♦ 112k1313 gold badges219219 silver badges13341334 bronze badges 2 added 145 characters in body edited Oct 28 '13 at 23:44 Qmechanic♦ 112k1313 gold badges219219 silver badges13341334 bronze badges 1 answered Oct 28 '13 at 23:35 Qmechanic♦ 112k1313 gold badges219219 silver badges13341334 bronze badges