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Homework Help: Scaling of a quantum state

  1. Oct 24, 2012 #1
    Hi everyone,

    I'm new to quantum mechanics, so bear with me o:)

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

    I'm not sure if scaling is the right word here, but my problem is about the absolut value of a quantum mechanics state to be one. I have the state [tex] | \phi>[/tex] which is a linear combination of the states [tex] |+>[/tex] and [tex] |->[/tex]. The first task is about scaling my phi.

    2. Relevant equations


    3. The attempt at a solution

    I know [tex] | \phi> = \lambda_1 |+> + \lambda_2 |->[/tex] whereas the lambdas are complex numbers.

    Afterwards I used the definition of the scalar product to get the norm.
    [tex] <\phi|\phi> = (\lambda_{1}' <+| + \lambda_{2}'<-|)(\lambda_1 |+> + \lambda_2 |->) [/tex]

    lambda' is the complex conjugated
    If I solve this I get an expression like : sqrt(a+b) wheres my a equals the realpart of my lambdas1 and b equals the real part of my lambdas2

    Is that solution right or totally wrong?

    Thanks for your help
  2. jcsd
  3. Oct 24, 2012 #2


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    I'm not sure what you mean when you say "If I solve this I get an expression like : sqrt(a+b) " What are you saying sqrt(a+b) should equal?

    Anyway, I think the imaginary parts of each λ will be just as important as the real parts.

    Note ##(\lambda_{1}' <+| + \lambda_{2}'<-|)(\lambda_1 |+> + \lambda_2 |->)## will expand to

    ##\lambda_{1}'\lambda_{1} <+| +> \;+ \;\lambda_{1}'\lambda_{2}<+|-> \;+ \; \lambda_{2}'\lambda_{1}<-|+> \;+ \;\lambda_{2}'\lambda_{2}<-|->##

    Try to simplify this.
  4. Oct 24, 2012 #3
    first of all: thanks for the help

    Is my attempt at a solution even right? I'm uncertain if that's the right approach.

    Yes, I had the expression you mentioned already, where I think I can say the following:

    <+|+> and <-|-> equals 1 and <+|-> equals zero

    And while I'm writing this down. I realised I forgot to say in my beginning post that + and - are orthonormal quantum states :/. Sorry about that.

    edit: I got to the point where I have the followng left:
    [tex] <\phi|\phi> = \lambda_{1}'\lambda_{1} +\lambda_{2}'\lambda_{2}[/tex]

    And a complex number times its conjugated version gives me a real part only
  5. Oct 24, 2012 #4


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    That looks good. Yes, a complex number times it's conjugate gives a real number in the sense that if z = a + ib then z'z = a^2 + b^2, which includes both the real and imaginary parts of z. (I thought you might have been saying that only the real part of z would contribute to z'z.)

    When you talk about "scaling" ##|\phi>##, I am guessing you are talking about what is usually called "normalizing" ##|\phi>##. So, you want to find a number, ##A##, such that if you multiply ##|\phi>## by ##A##, the resultant state vector has a norm of 1. That is, if you let ##|\psi> = A|\phi>## then ##<\psi|\psi> = 1##.
  6. Oct 25, 2012 #5
    yes, normalizing is the word I'm looking for. I'm not a native speaker and my translation program didn't give me "to normalize" :) .

    So, my attempt is right?

    the normalized vector would then be:

    [tex] \frac {1}{\lambda_1' \lambda_1 \lambda_2' \lambda_2} \mid \phi \rangle[/tex]

    Is that correct?

    Furthermore I want to calculate the possibility of |+>. I've done it with

    [tex] p_n = \frac {|a_n|^2} {\sum |a_n|^2}[/tex]

    where my a_n's are the coefficients.
    So finally I'd get

    [tex] p_+=\frac {|\lambda_1|^2}{|\lambda_1|^2+|\lambda_2|^2}[/tex]

    is that correct?

    Thanks for the help
  7. Oct 25, 2012 #6


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    That's not quite correct. You want to find ##A## such that ##A|\phi>## is normalized. Which means

    ##<\phi|A'A|\phi> = A'A<\phi|\phi> = 1##. So, ##A'A = \frac{1}{<\phi|\phi>}## You are free to choose ##A## to be a real number, so you can choose ##A^2= \frac{1}{<\phi|\phi>}##
    That's correct.
  8. Oct 25, 2012 #7
    Hm I don't understand that.

    Let's take a look at "normal" vectors. When I want to normalize a vector its:

    [tex] \vec v' = \frac {1}{\sqrt{a²+b²+c²}} \cdot \vec v[/tex]

    doesn't that also work now for quantum states?

    like in my example

    [tex] \mid \phi \rangle' = \frac {1}{\sqrt{\lambda_1' \lambda_1 + \lambda_2' \lambda_2}} \cdot \mid \phi \rangle[/tex]
  9. Oct 25, 2012 #8


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    Yes, that's correct. Good.

    Earlier you had
  10. Oct 25, 2012 #9
    Ah ok, I forgot the square root aswell as the plus. My fault.
    Thank your very much for your help o:)
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