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Qubits probability

  • Thread starter t_n_p
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1. Homework Statement
http://img249.imageshack.us/img249/4476/46715318ov5.jpg [Broken]

3. The Attempt at a Solution

I know to calculate the pr we square the number in front of the state.

e.g) for state psi 00, the pr is (1/root(2))^2

But for state psi 01, do I include the imaginary number part? That is, do I calculate the square of (1+i/2root(2))?

Thanks
 
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Answers and Replies

nrqed
Science Advisor
Homework Helper
Gold Member
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1. Homework Statement
http://img249.imageshack.us/img249/4476/46715318ov5.jpg [Broken]

3. The Attempt at a Solution

I know to calculate the pr we square the number in front of the state.

e.g) for state psi 00, the pr is (1/root(2))^2

But for state psi 01, do I include the imaginary number part? That is, do I calculate the square of (1+i/2root(2))?

Thanks
The probability is the square of the magnitude of the coefficient (if the wavefunction is normalized). So the formula is that the probability of measuring E_{01} for example is

[tex] \vert C_{01} \vert^2 = C_{01} \times C_{01}^* [/tex]

which is obviousy a rel number.
This should be familiar to you. Of course, when the coefficient is real, this becomes simply the ordinary square of the coefficient.
 
Last edited by a moderator:
595
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Ah so its the complex conjugate. Makes sense!
 

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