- #1

- 151

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter metroplex021
- Start date

- #1

- 151

- 0

- #2

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 413

- #3

A. Neumaier

Science Advisor

- 7,568

- 3,418

The |n> must be eigenstates of H in order that the equation H|n>=E_n|n> is valid.

- #4

- 128

- 0

Quantum mechanics answers the questions: If we measure the observable [tex]\hat A[/tex] for a specified experimental configuration, 1) what are the possible results of a measurement? And 2) what is the probability of obtaining each result? Answer 1) The possible results of a measurement are eigenvalues of [tex]\hat A[/tex]. Answer 2) The probability of obtaining each result is [tex]\left| {\left\langle {{a_k }}

\mathrel{\left | {\vphantom {{a_k } \psi }}

\right. \kern-\nulldelimiterspace}

{\psi } \right\rangle } \right|^2 [/tex], where the [tex]\left| {a_k } \right\rangle [/tex] are the eigenvectors of [tex]\hat A[/tex]. [tex]\left| \psi \right\rangle [/tex] is the state vector, which is determined by the experimental configuration. Thus, we need to know the eigenvalues and eigenvectors, which are obtained by solving the eigenvalue equation of the measured observable. Solving eigenvalue equations is part of doing quantum mechanics. If we are going to measure the energy then we must solve the energy eigenvalue equation [tex]H\left| n \right\rangle = E_n \left| n \right\rangle [/tex], where [tex]\left| n \right\rangle [/tex]

is the eigenvector corresponding to the eigenvalue [tex]E_n [/tex]. [tex]H[/tex] is the Hamiltonian operator.

If the state vector is [tex]\left| n \right\rangle [/tex], an eigenvector of the Hamiltonian, then a measurement of the energy always yields the value [tex]E_n [/tex]. There is no uncertainty in energy when the particle is in an energy eigenstate. Generally, repeated measurements yield the entire eigenvalue spectrum. By blocking out all results except [tex]E_n [/tex], the particle will be in eigenstate [tex]\left| n \right\rangle [/tex].

Best wishes.

- #5

- 151

- 0

Basically what I'm wondering is whether the applicability of eigenvalue equations in physics is a peculiarity of QM, or a feature that we should expect to crop up in theories of physics quite generally. Any thoughts (or references) would be much appreciated!

- #6

A. Neumaier

Science Advisor

- 7,568

- 3,418

whether it is a peculiarity of quantum mechanics that we can represent the values that physical systems have for various properties this way,

It is the essence of quantum mechanics. The established view is that all observables are represented by operators whose eigenvalues define their possible measurement values,

which are predictable with certainty if the system is in the corresponding eigenstate.

For example, the eigenvectors of the z-component J_3 of the angular momentum operator

give the states where the measurement of J-1 has a definite value.

This has no analogue in classical mechanics. It surely will survive coming changes in our theories, at least in a good approximation.

Share:

- Replies
- 5

- Views
- 859

- Replies
- 0

- Views
- 1K