# What is the meaning of “ Ψ is not a measurable quantity in itself”?

I want to know that why the wavefunction Ψ as a complex quantity (i.e $A+iB$ form) in quantum mechanics and somewhere I have studied that Ψ is not a measurable quantity in itself that's why we multiply it by a it's complex conjugate Ψ* to measure Ψ. What does it mean that "Ψ is not a measurable quantity in it self" ?

• – ACuriousMind Apr 25 '15 at 19:06
• possible duplicate of Why is wave function so important? – 299792458 Apr 25 '15 at 19:09
• ^ not an exact duplicate, but those answers answer this question too. Therefore, it is a subset, not an exact duplicate. – 299792458 Apr 25 '15 at 19:10
• I'm a little confused by the question, that sentence means exactly what it says: you can't measure $\psi$. Did you expect that it was hiding something more subtle? – dmckee Apr 25 '15 at 23:55
• Concerning the complex nature of the wavefunction, see also physics.stackexchange.com/q/8062/2451 and links therein. – Qmechanic Apr 26 '15 at 10:29

## 3 Answers

What does measurable mean ?

It means that one can do an experiment and get a value for a+ib , the complex number.

A complex number to be measurable one should be able to measure a value at the same time for a and b and put a point on the complex plane. This means two independent variables, a and b can be measured and a point defined.

In quantum mechanics we have real distribution which are probability distribution, i.e. one independent variable. The mathematics of quantum mechanics give us though a function Psi, which is a complex number. The mathematics is very successful in deriving the spectra of atoms ( for example), using this Psi formalism. An additional bonus comes from the same equations by identifying that the probability of finding the electron at (x,y,z) is given by the complex conjugate square of this Psi function, as follows:

the probability of measuring a given eigenvalue \lambda_i will equal to where P_i is the projection onto the eigenspace of A corresponding to \lambda_i.

A is the hermitian operator that has the eigenvelue \lambda_i , (in the above example the (x,y,z) point )

To get at the real measurable value \lambda_i the two variables of the Psi function have been folded with one real number measurable as a probability distribution after a lot of measurements, and the other being as an undefined and unmeasurable variable, identified with the phase in the complex plane. If one takes the identity operator as A one has a point in the probability density for the specific problem.

• $|\psi(\mathbf r)|^2$ gives probability density, not probability. Neither of these is possible to measure. What can be measured in this context are positions of particles $\mathbf r$. By making lots of such measurements $\mathbf r_k$, we obtain frequencies and can infer the best probability distribution and $|\psi|$ given these measurements. But we never measure $|\psi|^2$ directly; it is just a mathematical tool. – Ján Lalinský Apr 26 '15 at 10:35
• I am just describing the Born rule, simplified .en.wikipedia.org/wiki/Born_rule#The_rule, "the probability of measuring a given eigenvalue \lambda_i will equal ...." – anna v Apr 26 '15 at 10:51
• "this Psi function, i.e. its two variables are turned into one measurable as probability" here I see somewhat misleading statement - you are saying probability is measurable. That's not true. – Ján Lalinský Apr 26 '15 at 10:56
• @JánLalinský I will try an edit – anna v Apr 26 '15 at 10:57

Further to Anna V's answer, in the case of an electron there is an important physical meaning to the "lack" measurability of the phase of the electron's wavefunction.

This is because the electron is coupled to the electromagnetic field. And, if one models this by the Minimal Coupling between the electron and the electromagnetic field, one gets the Maxwell-Dirac Equation (see my answer here for further details). So, one can add a smooth phase as a function of space $\chi(x,\,y,\,z)$ to the electron's wavefunction and in turn that phase is absorbed by the familiar gauge transformation $\phi\mapsto\phi+\partial_t\chi$, $\vec{A}\mapsto\vec{A}+\nabla\chi$ wrought on the Maxwell equation in the nonlinear coupled set. So the electron-electromagnetic field system has a $U(1)$ gauge symmetry: it's like Hugh Lofting's Pushmepullyou character: the electron can "hide" its phase by pushing it into the electromagnetic field, which pulls away a compensating phase into a Maxwellian gauge transformation.

This is in contrast, the one photon "wavefunction's" phase, as I describe in my asnwer here (although, strictly speaking, we're dealing with a particular representation of the photon's quantum state that is not what most people would rigorously define as a wavefunction, which can't be defined for the photon). Although a global phase added to a one photon Fock state cannot be measured, spatially varying phase is indirectly meaningful because it alters the diffraction of the wavefunction: the one-photon state propagates following Maxwell's equations, and spatially varying phase added to the classical analogue - the Maxwellian EM field - alters how that classical field diffracts with time.

The foregoing paragraph has the following further physical meaning: photons are Bosons, which means, in contrast with the Fermionic electron, any number of them can be in the same state. So, for every one photon state, we can imagine squillions of photons in the same state. These will work together to yield a classical EM field, whose phase is most certainly meaningful as the simple notion of a lens or phasemask dramatically illustrates.

Ψ is supposedly a probability density amplitude. ΨΨ* is the probability density which in theory can be measured. For example in electron diffraction through a crystal a statistical measure of the electrons in, divided into the electrons out in a small region divided by the volume would allow ΨΨ* to be approximately measured. Phase information is lost when ΨΨ* is calculated. Only phase difference is needed in diffraction experiments. The phase of Ψ cannot be measured.

• What do you mean that $\psi$ is supposedly a probability density amplitude? Isn't that what it is? – Kyle Kanos Apr 26 '15 at 1:45