The book "Introduction to quantum mechanics" by Griffiths starts by introducing the wave function. The squared of the integral of the wave function gives you the probability of measuring the position of a particle between a and b. Later it talks about eigenstates and eigenvalues.The eigenstate describes the determinate state of the system after a measurement of an observable with operator $Q.$ The eigenvalues of $Q$ is what we can measure from $Q,$ i.e. $L_zf = hm_lf$ where $L_z$ is the angular momentum operator, f is the eigenstates and hm is the eigenvalues. An arbitrary state is ie $ \psi = c_1f_1 + c_2f_2$ where $c_1^2,c_2^2$ is the probability of measuring the eigenvalues corresponding to the given eigenstates.

My question is the following: why is the square of the wave function in the beginning giving us the probability of measuring the particle between a and b? Is it just that the wave function is a eigenstate of the position operator? If we take the squared of the eigenstates we get from measuring $L_z$ will we get the probability of measuring a particle between a and b?I know that the wave function just describes a system's state like the eigenstate.

  • $\begingroup$ $L_z$ is the angular momentum operator in the z direction. Why do you thinks it tell you anything about the position operator? The wavefunction is a vector, not an operator. Answer to (1) it is a postulate $\endgroup$ – user126422 Jun 11 '17 at 0:50
  • $\begingroup$ To be sure, it's the integral of the absolute square of the wavefunction that gives the probability. $\endgroup$ – Alfred Centauri Jun 11 '17 at 2:21
  • $\begingroup$ But both the wavefunction and eigenstates of a an operator describes the state of a system. Isn't a wavefunction and eigenstate the same? $\endgroup$ – N.J Abel Jun 11 '17 at 9:45

This link might help in the definition of the wavefunction.

My question is the following: why is the square of the wave function in the beginning giving us the probability of measuring the particle between a and b?

As said in the comments, it is part of the postulates of quantum mechanics, which allow calculations and the theory is well validated.

Is it just that the wave function is a eigenstate of the position operator?

No. One has to use the position operator:

To relate a quantum mechanical calculation to something you can observe in the laboratory, the "expectation value" of the measurable parameter is calculated. For the position x, the expectation value is defined as


This integral can be interpreted as the average value of x that we would expect to obtain from a large number of measurements. Alternatively it could be viewed as the average value of position for a large number of particles which are described by the same wavefunction.

Continuing from the same site:

The wavefunction for a given physical system contains the measurable information about the system. To obtain specific values for physical parameters, for example energy, you operate on the wavefunction with the quantum mechanical operator associated with that parameter. The operator associated with energy is the Hamiltonian, and the operation on the wavefunction is the Schrodinger equation. Solutions exist for the time independent Schrodinger equation only for certain values of energy, and these values are called "eigenvalues*" of energy.

The Schrodinger equation is part of the postulates of quantum mechanics, i.e. the extra axioms imposed on the mathematics in order to connect with measurents/data.

So the general solution of the shrodinger equation for a specific potential will give specific wavefunctions on which the expectation value of the appropriate operator will give a number to compare with measurement.

For a position operator, look at page 3 here .

Correspondingly there will exist a $L_z$ operator to operate on the wavefunction. In general if the operators commute, then there could be simultaneous measurement of both, this is not the case for angular momentum, and position.

| cite | improve this answer | |
  • $\begingroup$ But are generally wavefunctions and eigenstates the same? $\endgroup$ – N.J Abel Jun 11 '17 at 9:47
  • $\begingroup$ Not the same, a wavefunction is a general solution of the Schrodinger equation. An eigenstate is a specific wavefunction that will give a specific eigenvalue for a given operator. $\endgroup$ – anna v Jun 11 '17 at 10:12
  • $\begingroup$ Thank you! Can they both describe a state?Can a wavefunction be a linear combination of eigenstates? $\endgroup$ – N.J Abel Jun 11 '17 at 20:25
  • $\begingroup$ look at page three on ocw.mit.edu/courses/materials-science-and-engineering/… . The eigenfunction solutions are complete sets and any function can be expanded in a complete set, so the answer is yes. $\endgroup$ – anna v Jun 12 '17 at 4:25

Why the probability density is proportional to the squared module of the wave function is actually a rule of Quantum Mechanics, known as the Born Rule, certainly it has been experimentally tested and up to this day it holds true. Whether you accept this rule as a fundamental building block of the theory depends on the flavor of the theory, so far I know you can actually build a self consistent theory without this rule, but then you need to add other elements, see for instance this article.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.