I was wondering if it is important in Quantum Mechanics to deal with operators that have an orthonormal basis of eigenstates? Imagine that we would have an operator (finite-dimensional) acting on a spin system that has real eigenvalues, but its eigenvectors are not perpendicular to each other. Is there any reason why such an operator cannot describe an actual physical quantity?
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$\begingroup$ No. PT symmetric, but not Hermitian Hamiltonians are one example. $\endgroup$– ACuriousMind ♦Commented Jan 22, 2015 at 23:47
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$\begingroup$ non-normal operators don't have functional calculus, and this is something you want most of the times. $\endgroup$– Phoenix87Commented Jan 22, 2015 at 23:53
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$\begingroup$ @Phoenix87 the functional calculus is rather a mathematical construct, I was wondering more about why this is necessary from a physical perspective. $\endgroup$– Xin WangCommented Jan 22, 2015 at 23:54
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$\begingroup$ @ACuriousMind sorry, I don't understand what PT symmetric means. $\endgroup$– Xin WangCommented Jan 22, 2015 at 23:55
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1$\begingroup$ @XinWang Physically, orthogonality of the eigenvectors means that the states are distinguishable. Therefore, if one has an "observable" with non-orthogonal eigenvectors, it means there does not exist even in principle a measurement that allows you to determine the value of that observable with certainty. This is a very strange property for an observable to have, although I can't see why this should actually be forbidden. $\endgroup$– Mark MitchisonCommented Jan 22, 2015 at 23:57
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3 Answers
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If two states are orthogonal, this means that $\langle \psi | \phi \rangle = 0$. Physically this means that if a system is in state $|\psi\rangle$ then there is no possibility that we will find the system in state $|\phi\rangle$ on measurement, and vis versa. In other words the 2 states in some sense mutually exclusive. This is an important property for operators because it means that the results of a measurement are unambiguous. A state with a well defined momentum $p_1$, i.e. an eigenstate of the momentum operator, cannot also have a momentum $p_2 \ne p_1$. Observables having an orthogonal (and complete) set of eigenstates is therefore a requirement in order for the theory to make physical sense (or at least for repeated measurements to give consistent results, as is experimentally observed)
answered Jan 23, 2015 at 0:04
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$\begingroup$ Are not your conditions satisfied if there exists a basis $| \psi_n \rangle$ and a dual basis $\langle \phi_n |$ such that $\langle \phi_n | \psi_m \rangle = \delta_{nm}$? Contrary to your answer, this condition is much weaker than orthonormality, which requires additionally that the dual is provided by taking the Hermitian conjugate, i.e. $\langle \phi_n | = ( | \psi_n \rangle )^\dagger$. $\endgroup$ Commented Feb 3, 2021 at 21:54
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1$\begingroup$ @ComptonScattering the Hermitian conjugate is defined relative to some inner product. Given that we have particular inner product singled out (implicitly) by the axioms of QM, it makes sense to define hermitian conjugation relative to that. In other words $\langle \phi_n| = (|\psi_n\rangle)^\dagger$ by (the only sensible) definition n this context. $\endgroup$ Commented Feb 4, 2021 at 16:53
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If the eigenstates of your operator is not a orthogonal set, then your operator is not a hermitian operator, or in other words, is not an observable.
Actually, non-hermitian operators "appears" all the time, but if you investigate decoherence mechanism you may note that this operators don't affect directly the classical realm. This is because you can't construct consistent histories upon question about this non-hermitian operators. More concretely, if you have the coherent state $|\alpha \rangle$, i.e.
$$ a|\alpha \rangle=\alpha |\alpha \rangle $$
where $a$ is the annihilation operator, then asking the probability of the system have some value of $\alpha$ don't make sense because we always can represent one $\alpha$ state in superposition of others $\alpha's$.
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2$\begingroup$ (1) This is mistaken: "Non-hermitian operators is operators that have at least one complex eigenvalue." Its converse is true, but this statement is not. (2) This is not quite true: "We cannot measure quantities that could assume complex values." There's no problem with generalizing hermitian operators to normal operators, thus allowing complex-valued observables (spectral theorem works, etc.). It's just not conventionally done because it's redundant and not useful, being equivalent to a simultaneous measurement of two commuting real-valued observables. $\endgroup$ Commented Jan 23, 2015 at 6:07
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A measurement extracts information from a quantum system that can be copied. Not all of the information in a quantum system can be copied as a result of the no-cloning theorem. The information that can be copied has to be described by an operator that remains unchanged under the unitary operator representing the measurement. And the only operators that respect this requirement are normal operators.
A similar argument can be given in terms of state vectors showing that only outcomes in orthogonal state vectors can be recorded:
