Solving Projector Question on Mixed States

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In summary, the conversation discusses an arbitrary mixed state represented by the equation ( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |) \rho ( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |), which can be simplified using the distributive law to ( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |) (\rho | 0 \rangle \langle 0 | +\rho | 1 \rangle \langle 1 |). The simplified form is then expanded to show the individual components. The conversation also mentions a fast online latex equation editor for
  • #1
raisin_raisin
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A few months ago I wrote this line down, but it does not seem to follow any more. Am I mistaking a mistake now or when I first wrote it down? thanks
[tex] ( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |) \rho ( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |) [/tex]


[tex]= | 0 \rangle \langle 0 | \rho | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | \rho | 1 \rangle \langle 1 | [/tex]

where [tex]\rho[/tex] is an arbitrary mixed state.

(It is not letting me preview my latex so fingers crossed this works as expected)
 
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  • #2
You have to use the distributive law
1. a*(b+c) = ab + ac
2. (a+b)*(c+d) = (ac + ad + bc + bd)[tex]
( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |) \rho ( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |)
[/tex]

[tex]
=( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |) (\rho | 0 \rangle \langle 0 | +\rho | 1 \rangle \langle 1 |)
[/tex]

[tex]
= \langle 0 | \rho | 0 \rangle \cdot | 0 \rangle \langle 0 | + \langle 0 | \rho | 1 \rangle \cdot | 0 \rangle \langle 1 | + \langle 1 | \rho | 0 \rangle \cdot | 1 \rangle \langle 0 | + \langle 1 | \rho | 1 \rangle \cdot | 1 \rangle \langle 1 |
[/tex]A fast online latex equation editor is here:
http://www.codecogs.com/latex/eqneditor.php
 
Last edited:
  • #3
To workaround the preview problem, hit your browser's refresh button when the incorrect image comes up, and then confirm that you want to send the information one more time. You can also edit your post up to 24 hours after you posted it.
 
  • #4
Thanks for this trick, didn't know about it.
 

1. What is a mixed state in quantum mechanics?

A mixed state in quantum mechanics is a statistical combination of pure states. It represents a system that is in a state of uncertainty, where the exact state of the system cannot be determined.

2. How do you solve a projector question on mixed states?

To solve a projector question on mixed states, you first need to determine the eigenstates of the given mixed state. Then, you can use the projection operator to calculate the probabilities of each eigenstate being measured. Finally, you can use these probabilities to find the expected value of the observable in the mixed state.

3. What is the projection operator in quantum mechanics?

The projection operator in quantum mechanics is a mathematical operator that projects a state onto a subspace of the total state space. It is used to calculate the probability of obtaining a certain measurement result in a given state.

4. Can mixed states be entangled in quantum mechanics?

Yes, mixed states can be entangled in quantum mechanics. Entanglement occurs when two or more particles become correlated in a way that their properties cannot be described independently. This can happen in both pure and mixed states.

5. How are mixed states different from pure states in quantum mechanics?

Mixed states and pure states are different in that pure states represent a system that is in a definite state, while mixed states represent a system that is in a state of uncertainty. Pure states can be described by a single wave function, while mixed states require a statistical combination of pure states to be described.

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