- #1

- 33

- 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 sruthisupriya
- Start date

- #1

- 33

- 0

- #2

George Jones

Staff Emeritus

Science Advisor

Gold Member

- 7,423

- 1,063

Here's one thing. Suppose that two orthogonal eigstates of the observable A correspond to distinct eigenvalues a1 and a2. Perform a meausurement of A and assume that the result is a1. Perform a second measurement of A immediately after the first measurement. The probabilty that the result is a2 is zero.

- #3

- 223

- 2

If I can measure certain eigenstates, the probability of measuring a particle in the state [tex]|\Psi\rangle[/tex] and obtaining it being in the state [tex]|\phi\rangle[/tex] after measurement is

[tex] P(\phi) = |\langle \phi | \Psi \rangle|^2 [/tex]

So what if these two states are orthogonal to each other?

- #4

reilly

Science Advisor

- 1,075

- 1

Regards,

Reilly Atkinson

- #5

- 41

- 0

In the case of degeneracy, such as with unpaired electron spin not in a magnetic field, the situation is a little more complicated, because the degenerate orthogonal solutions can mix.

-Jim

Share:

- Replies
- 1

- Views
- 953