Quantum Mechanics: Probability of the result h-bar/2 if Sy is measured

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SUMMARY

The probability of measuring h-bar/2 when measuring Sy is directly linked to the state ψ+. The discussion confirms that the solution provided in the problem statement is incorrect due to the inclusion of the matrix Sy in the probability calculation. The correct approach should yield a dimensionless probability, whereas the original calculation resulted in a dimensional value of h², indicating a fundamental error in the problem setup.

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  • Knowledge of the mathematical representation of quantum states, such as ψ+.
  • Basic understanding of dimensional analysis in physics.

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  • Study the role of operators in quantum mechanics, focusing on angular momentum operators like Sy.
  • Learn about the mathematical framework of quantum states and their probabilities.
  • Explore dimensional analysis techniques in physics to ensure correct interpretations of physical quantities.
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bbnl1990
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Homework Statement



http://dl.dropbox.com/u/24280471/Question.PNG
http://dl.dropbox.com/u/24280471/Answer.PNG

Homework Equations



Isn't the probability of measuring h-bar/2 the same as the probability of finding the particle in the state ψ+? Since Syψ+ = (h-bar/2)Syψ+.

The Attempt at a Solution



(Sorry, don't know how to use LaTex.)
http://dl.dropbox.com/u/24280471/Attempt.PNG
 
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bbnl1990 said:
Isn't the probability of measuring h-bar/2 the same as the probability of finding the particle in the state ψ+?
Yes, your solution is correct. The answer in the problem statement is incorrect. They should have left out the matrix Sy in the calculation of the probability. Note that the probability must be a dimensionless number, but they are getting something that has the dimensions h2.
 


Yes, your solution is correct. The answer in the problem statement is incorrect. They should have left out the matrix Sy in the calculation of the probability. Note that the probability must be a dimensionless number, but they are getting something that has the dimensions h2.

Thank you for your reply!
 

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