Mathematical meaning of measurement

[nateHI]

I pasted a paragraph from http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-845-quantum-complexity-theory-fall-2010/lecture-notes/MIT6_845F10_lec02.pdf" below.

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Consider the following matrix which represents a 45◦ counter clock wise rotation in the plane

| 1/√2 −1/√2 |
| 1/√2 1/√2 |

Suppose you are in the state |0> and you apply this operation twice in succession and measure, you would get |1>. However if you apply it once, then measure, then apply it again you would get |0> half of the time and |1> half of the time.
This is analog to interference in the double slit experiment.
-----------------------------------

I understand how you would mathematically apply the 45 degree rotation matrix "twice in succession" it would simply be R*R*|0> = |1>

But, how do you *mathematically* "apply it once, then measure, then apply it again" ?

Thanks
-Nate

 

[SpectraCat]

I pasted a paragraph from http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-845-quantum-complexity-theory-fall-2010/lecture-notes/MIT6_845F10_lec02.pdf" below.

------------------------------
Consider the following matrix which represents a 45◦ counter clock wise rotation in the plane

| 1/√2 −1/√2 |
| 1/√2 1/√2 |

Suppose you are in the state |0> and you apply this operation twice in succession and measure, you would get |1>. However if you apply it once, then measure, then apply it again you would get |0> half of the time and |1> half of the time.
This is analog to interference in the double slit experiment.
-----------------------------------

I understand how you would mathematically apply the 45 degree rotation matrix "twice in succession" it would simply be R*R*|0> = |1>

But, how do you *mathematically* "apply it once, then measure, then apply it again" ?

Thanks
-Nate

The only thing I can think of is that the author of the quote was trying to emphasize that applying the rotation twice with a measurement in between is effectively the same as applying it once. I say this because the first rotation will put the system in a 50-50 superposition of |0> and |1>, so a measurement of that system will "collapse" the wavefunction to EITHER |0> OR |1>. Then you are back where you started and the 45º rotation on the resulting state will again put you in the 50-50 superposition.

As far as the mathematical representation of the entire process is concerned ... that's hard ... there is currently no known way to represent the collapse of the wavefunction. It's even hard to conceive of how to do this, because the measurement will give EITHER |1> OR |0>, and you don't know which. That suggests you might write it as a superposition, but we know that's wrong, because after the measurement, it will be in ONLY one pure quantum state. However, we have a handy Q.M. postulate (the measurement postulate), which tells us that we are allowed to rationalize the way I have above.

 

[Hurkyl]

Suppose you are in the state |0> and you apply this operation twice in succession and measure, you would get |1>.

You are measuring in the {|0>, |1>} basis I presume.

However if you apply it once, then measure, then apply it again you would get |0> half of the time and |1> half of the time.

I think you meant "apply it once, then measure", without the additional "apply it again". Or maybe "apply it once, then measure, then apply it again, then measure again".

But, how do you *mathematically* "apply it once, then measure, then apply it again" ?

Applying the operator R to "50% |0> and 50% |1>" is simply "50% R|0> and 50% |1>"

 

[Hurkyl]

...
I say this because the first rotation will put the system in a 50-50 superposition of |0> and |1>
...

You seem to be mixing up the ideas of "superposition of kets" and "statistical mixture of states".

 

[SpectraCat]

Applying the operator R to "50% |0> and 50% |1>" is simply "50% R|0> and 50% |1>"

Of course ... I was being dense in my previous post ... guess it is past my bedtime.. You could just say the result of the first measurement is a mixed state, as opposed to the superposition, which is a pure state, right?

 

[Hurkyl]

Of course ... I was being dense in my previous post ... guess it is past my bedtime.. You could just say the result of the first measurement is a mixed state, as opposed to the superposition, which is a pure state, right?

Right.

And if you want, since we have a preferred basis and the mixture is a convex combination of them, we can think of the mixed state as being a probability distribution on the two possible outcomes |0> and |1>, and think of the statistics in a classical fashion.

 

[nateHI]

OK, I guess I don't know the difference between statistical mixture and superposition of two eigenstates. In fact, I always thought the superposition of two eigenstates where the norm is 1 was a statistical mixture. Thanks though, now I know where the hole in my basic knowledge is and what I need to study.

 

[nateHI]

I'm trying to test my understanding between a mixed state and a pure state.

Suppose you have two electrons in a quantum system. If I want to calculate the expectation value of the operator, A, of the system I have to use a density matrix in the calculation.

Now suppose the two electrons become entangled, I can now use simply use <s|A|s>

Is that true?