Question
and
are the orthonormal basis states of a two-dimensional Hilbert space. A Hermitian operator
in this basis is given by the spectral representation:
. For a normalized state given by
Answer :
Word Count : 314
The given Hermitian operator is [ \hat{A} = 3|1\rangle\langle 1| - 1|2\rangle\langle 2| ] and the normalized state is [ |\psi\rangle = \frac{1}{2}|1\rangle + \frac{\sqrt{3}}{2}|2\rangle. ] First, the expectation value of (\hat{A}) in state (|\psi\rangle) is [ \langle \hat{A} \rangle = \langle \psi | \hat{A} | \psi \rangle. ] Expanding using the basis states: [ \langle \psi | \hat{A} | \psi \rangle = \left(\frac{1}{2}\langle 1| + \frac{\sqrt{3}}{2}\langle 2|\right) \left(3|1\rangle\langle 1| - 1|2\rangle\langle 2|\right) \left(\frac{1}{2}|1\rangle + \frac{\sqrt{3}}{2}|2\rangle\right). ] Compute _______ _______ ____ _______ __________ ______ _________.
___ _____ ______ _______ ___ _____.
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_________ ____ __________ _______ ______ __________.
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________ ________ ____ ___ _________ _________ ____ ______.
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______ ___ ______ ____ ________ __________ ____ ________.
________ _________ ________ _______ ___.
_____ _______ _________ _____ _____ __________ _______ _______ __________ __________.
______ _______ ___ _________ ______ _____.
______ ________ _________ __________ ___ _____ ________ ______ _________ ___ ______ __________.
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The given Hermitian operator is [ \hat{A} = 3|1\rangle\langle 1| - 1|2\rangle\langle 2| ] and the normalized state is [ |\psi\rangle = \frac{1}{2}|1\rangle + \frac{\sqrt{3}}{2}|2\rangle. ] First, the expectation value of (\hat{A}) in state (|\psi\rangle) is [ \langle \hat{A} \rangle = \langle \psi | \hat{A} | \psi \rangle. ] Expanding using the basis states: [ \langle \psi | \hat{A} | \psi \rangle = \left(\frac{1}{2}\langle 1| + \frac{\sqrt{3}}{2}\langle 2|\right) \left(3|1\rangle\langle 1| - 1|2\rangle\langle 2|\right) \left(\frac{1}{2}|1\rangle + \frac{\sqrt{3}}{2}|2\rangle\right). ] Compute _______ _______ ____ _______ __________ ______ _________.
___ _____ ______ _______ ___ _____.
_________ _____ ____ _________ ______ ______ _____ ____.
__________ ______ __________ ______ ___ ____.
___ ______ _____ ____ ________ ________ _________ ______ _________.
_____ ____ ____ _______ ___ _______ _____ ___ _____ ___ _________.
_______ _________ _____ _____ _____ ____ ________.
______ ________ ________ ___ _______ _______ _____.
_______ ____ __________ _________ _______ _____ __________ ____.
___ _________ __________ ______ ______ _____ ______ _______ ________.
__________ __________ _____ __________ __________ ____ ______ _________.
___ _________ ___ _______ ___ __________ ______ _____ ________ _________ __________.
________ ____ _______ _____ ___ _____ _____ _________ _________ __________ ___.
______ ______ ______ _________ _______ ___ _________ ___ _____.
____ _____ ____ _____ __________ __________ ______ ______ ______ ____ ______ __________.
____ _____ __________ _________ _______ ____ _________ _______ ________ _________ _________ ____.
_________ _________ _________ ___ ________ ________ _______.
_______ ________ _____ ____ _____ __________ ____ _________ ___.
_________ ____ __________ _______ ______ __________.
________ _____ ___ ______ ______.
___ _____ ____ ______ ______ ________ ___.
________ ________ ____ ___ _________ _________ ____ ______.
_________ ___ ___ _________ ___ ___ ______ __________ _________ ______ ______ _______.
______ ___ ______ ____ ________ __________ ____ ________.
________ _________ ________ _______ ___.
_____ _______ _________ _____ _____ __________ _______ _______ __________ __________.
______ _______ ___ _________ ______ _____.
______ ________ _________ __________ ___ _____ ________ ______ _________ ___ ______ __________.
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