- #1

squelch

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## Homework Statement

A particle is in the state [itex]|\psi \rangle = \frac{1}{{\sqrt 3 }}|U\rangle + \frac{{a\sqrt {(2)} }}{{\sqrt {(3)} }}i|D\rangle [/itex]. The up state [itex]|U\rangle = \left( {\begin{array}{*{20}{c}}

1\\

0

\end{array}} \right)[/itex] and the down state [itex]|D\rangle = \left( {\begin{array}{*{20}{c}}

0\\

1

\end{array}} \right)[/itex] correspond to the z-basis vectors. Detirmine:

a. The value of "a" such that the state is normalized.

b. [itex]\langle U|D\rangle [/itex]

c. The probability of measuring down. [itex]|\langle D|\psi \rangle {|^2}[/itex]

d. The duel vector [itex]\langle \psi |[/itex]

e. The probability amplitude of measuring up.

f. Write out the state [itex]|\psi \rangle [/itex] in terms of right [itex]|R\rangle [/itex] and left [itex]|L\rangle [/itex].

g. What is the probability of measuring right?

h. A measurement is made on our initial [itex]|\psi \rangle [/itex] with our apparatus oriented in the x-direction and [itex]|R\rangle [/itex] was the outcome. Determine the probability of measuring [itex]|U\rangle [/itex] 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 [itex]\langle \psi |\psi \rangle = 1[/itex]. 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 [itex]\langle \psi |[/itex] is the complex conjugate of [itex]|\psi \rangle [/itex]. So, I might say that [itex]\langle \psi | = \frac{1}{{\sqrt 3 }}|U\rangle - \frac{{a\sqrt 2 }}{{\sqrt 3 }}i|D\rangle [/itex]

e. As a guess, I'd say [itex]\frac{1}{{\sqrt 3 }}[/itex], the coefficient on [itex]|U\rangle[/itex]

f. I feel like I have to make an assumption here: that it's fifty-fifty either way. If so, I might write: [itex]|\psi \rangle = \frac{1}{{\sqrt 2 }}|R\rangle - \frac{1}{{\sqrt 2 }}i|L\rangle [/itex]

g. From that state equation, 1/2.

h. The previous state is lost, so it becomes 1/3.