squelch
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Homework Statement
A particle is in the state |\psi \rangle = \frac{1}{{\sqrt 3 }}|U\rangle + \frac{{a\sqrt {(2)} }}{{\sqrt {(3)} }}i|D\rangle. The up state |U\rangle = \left( {\begin{array}{*{20}{c}}<br /> 1\\<br /> 0<br /> \end{array}} \right) and the down state |D\rangle = \left( {\begin{array}{*{20}{c}}<br /> 0\\<br /> 1<br /> \end{array}} \right) correspond to the z-basis vectors. Detirmine:
a. The value of "a" such that the state is normalized.
b. \langle U|D\rangle
c. The probability of measuring down. |\langle D|\psi \rangle {|^2}
d. The duel vector \langle \psi |
e. The probability amplitude of measuring up.
f. Write out the state |\psi \rangle in terms of right |R\rangle and left |L\rangle.
g. What is the probability of measuring right?
h. A measurement is made on our initial |\psi \rangle with our apparatus oriented in the x-direction and |R\rangle was the outcome. Determine the probability of measuring |U\rangle if our apparatus is oriented back in the z-direction.
Homework Equations
None, but as a reference I am using http://ocw.mit.edu/courses/physics/...all-2013/lecture-notes/MIT8_05F13_Chap_04.pdf to guide me through the notation.
The Attempt at a Solution
I'm going to attempt to break this question down best I can as far as I can understand it, but I'm very uncertain (lol) on the details of what I'm doing. I don't want to do cargo-cult physics.
a. I understand that this is the value of a for while the probability \langle \psi |\psi \rangle = 1. It appears that to get there, a should simply equal 1, but it appears there's a possibility that a= 1/i. So, right now my answer is a=1, because I sort of understand the wave function is necessarily a complex function.
b. The probability of measuring up when the state is down should be 0 (zero).
c. It seems the probability of measuring down is a*(2/3) ... or, normalized, just 2/3.
d. The duel vector \langle \psi | is the complex conjugate of |\psi \rangle. So, I might say that \langle \psi | = \frac{1}{{\sqrt 3 }}|U\rangle - \frac{{a\sqrt 2 }}{{\sqrt 3 }}i|D\rangle
e. As a guess, I'd say \frac{1}{{\sqrt 3 }}, the coefficient on |U\rangle
f. I feel like I have to make an assumption here: that it's fifty-fifty either way. If so, I might write: |\psi \rangle = \frac{1}{{\sqrt 2 }}|R\rangle - \frac{1}{{\sqrt 2 }}i|L\rangle
g. From that state equation, 1/2.
h. The previous state is lost, so it becomes 1/3.