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Homework Help: Bra-Ket Notation, Wave Equation, Particle States

  1. May 3, 2015 #1


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    1. The problem statement, all variables and given/known data

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

    2. Relevant 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.

    3. 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.
  2. jcsd
  3. May 3, 2015 #2


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    The condition on a is ## |a|^2=a a^*=1 ## which is a circle in the complex plane. So there are an infinite number of solutions all equally valid. Two of them are 1 and 1/i.
    Actually ## \langle \psi |= \frac{1}{\sqrt 3} \langle U |-\frac{a\sqrt 2}{\sqrt 3}i\langle D |##.
    You should write up and down states in terms of right and left states and then substitute in ## |\psi\rangle ##. Its usually the case that right and left states are defined in terms of up and down states. So you should only solve two simultaneous algebraic equations to invert the definitions.
    All other answers seem fine to me(except g and h which should change according to f).
  4. May 3, 2015 #3


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    Thanks. I'll work on unpacking the correction you've given me for f.

    Was the link to the lecture notes I provided a good reference to these sorts of problems? The professor threw us some stuff that wasn't in the textbook and I feel very much like I'm groping in the dark at my notes with this.
  5. May 3, 2015 #4


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    I didn't check it carefully but it seemed good. Also I think you can trust MIT in such things! But I suggest you take a look at other notes of the same series too, just in case!
  6. May 4, 2015 #5


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    I'm having a lot of trouble unpacking your last statement :
    It seems like you're suggesting I say [itex]|U\rangle = \frac{1}{{\sqrt 2 }}|L\rangle + \frac{1}{{\sqrt 2 }}|R\rangle [/itex] and [itex]|D\rangle = \frac{1}{{\sqrt 2 }}|L\rangle + \frac{1}{{\sqrt 2 }}|R\rangle [/itex] and say:

    [tex]|\psi \rangle = \frac{1}{{\sqrt 3 }}(\frac{1}{{\sqrt 2 }}|L\rangle + \frac{1}{{\sqrt 2 }}|R\rangle ) + \frac{2ai}{{\sqrt 3 }}(\frac{1}{{\sqrt 2 }}|L\rangle + \frac{1}{{\sqrt 2 }}|R\rangle )[/tex]

    Simplified as:

    [tex]|\psi \rangle = \frac{1+2ai}{{\sqrt 6 }}|L\rangle + \frac{1+2ai}{{\sqrt 6 }}|R\rangle [/tex]

    edit: corrected some algebra
    Last edited: May 4, 2015
  7. May 4, 2015 #6


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    The way you wrote them, we have ## |U\rangle=|D\rangle ## which is wrong!
    We know that we can write up and down states as linear combinations of right and left states and now we want to determine the coefficients. Lets call them ## a_u, b_u,a_d,b_d##, with the conditions ## |a_u|^2=|b_u|^2=|a_d|^2=|b_d|^2=\frac 1 2##. This condition comes from the fact that when the system is in a up or down eigenstate, we have no knowledge about its right-left condition and so they should be equally probable. Another condition on the coefficients is that they should give us orthogonal up and down states, so we have ## a_u a_d^*+b_u b_d^*=0##.
    So we have four complex unknowns which means eight real unknowns. But the number of equations are six.(the norm equations are one equation each but the orthogonality equation has a real and a complex part.) So we have an underdetermined system which means we have some amount of freedom(two arbitrary real unknowns). What you should do, is first choosing two of the real unknowns and then use the above equations to determine the rest.
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