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Quantum Mechanics: a non-normalizable state

  1. Jul 18, 2008 #1
    1. The problem statement, all variables and given/known data
    At a given moment, the wave function of a particle is in a non-normalizable state [tex]\Psi[/tex](x) = 1 + sin²(kx). By measuring its kinetic energy, what values are possible and with what probability?.


    2. Relevant equations


    3. The attempt at a solution
    I can't calculate the probability because the integral that define it is unbounded.
    ([tex]\Psi[/tex], [tex]\widehat{K}\Psi[/tex]) = [tex]\infty[/tex]
     
  2. jcsd
  3. Jul 18, 2008 #2

    Dick

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    You can express that state as a sum of states with definite momentum and hence definite energy. What are their relative amplitudes? Hint,
    sin(kx)=(exp(ikx)-exp(-ikx))/2i.
     
  4. Jul 20, 2008 #3
    Thanks Dick for your hint.

    Here is what I did following your advice:

    [tex]\Psi[/tex](x) = exp{i(2k)x} - [tex]\frac{1}{4}[/tex]exp{i(-2k)x} + (1+i)exp{i(0)x},

    that is, I have expressed the wave function as a linear combination of self-states of the linear momentum.

    So, we have:

    [tex]\left\|\Psi\right\|[/tex]² = 1 + [tex]\frac{1}{16}[/tex]+2 [tex]\Rightarrow\left\|\Psi\right\|[/tex] = [tex]\frac{7}{4}[/tex]

    and, using the same notation, the normalized wave function is

    [tex]\Psi[/tex](x) = [tex]\frac{4}{7}[/tex] exp{i(2k)x} - [tex]\frac{1}{7}[/tex] exp{i(-2k)x} + [tex]\left(\frac{4}{7} + \frac{4}{7} i \right)[/tex] exp{i(0)x}.

    Hence,


    Pr([tex]p_{x}[/tex] = 2k) = [tex]\frac{16}{49}[/tex],

    Pr([tex]p_{x}[/tex] = -2k) = [tex]\frac{1}{49}[/tex],

    Pr([tex]p_{x}[/tex] = 0) = [tex]\frac{32}{49}[/tex],

    and therefore, denoting the kinetic energy by T, we finally have:

    Pr(T = 2k²/2m) = [tex]\frac{17}{49}[/tex],

    Pr(T = 0) = [tex]\frac{32}{49}[/tex],

    as the kinetic energy is related to linear momentum through the expression

    T = [tex]\frac{p^{2}_{x}}{2m}[/tex].

    Is it right?.
     
  5. Jul 20, 2008 #4

    Dick

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    Homework Helper

    Something went wrong with the very first line. I don't see how e^(2ikx) and e^(-2ikx) can have different coefficients. Did you miss a bracket in
    sin(kx)=[exp(ikx)-exp(-ikx)]/(2i)? Your general approach is right, though.
     
  6. Jul 20, 2008 #5
    Oh, yes, I missed a bracket in sin(kx)=[exp(ikx)-exp(-ikx)]/(2i). Thank you very much.

    Let's see if now is entirely correct.

    Taking into account the identity sin(kx) = [tex]\frac{exp(ikx)-exp(-ikx)}{2i}[/tex],

    the non-normalizable wave function [tex]\Psi (x)[/tex] = 1 + sin²(x) can be expressed as a linear combination of self-states of the linear momentum,

    [tex]\Psi (x)[/tex] = [tex]\frac{3}{2}[/tex] exp{i(0)x} - [tex]\frac{1}{4}[/tex] exp{i(2k)x} - [tex]\frac{1}{4}[/tex] exp{i(-2k)x},

    So, we have:

    [tex]\left\|\Psi\right\|[/tex]² = [tex]\frac{19}{8}\Rightarrow\left\|\Psi\right\|[/tex] = [tex]\frac{\sqrt{38}}{4}[/tex]

    and the normalized wave function, using the same notation, is

    [tex]\Psi [/tex](x) = [tex]\frac{6}{\sqrt{38}}[/tex] exp{i(0k)x} - [tex]\frac{1}{\sqrt{38}}[/tex] exp{i(2k)x} - [tex]\frac{1}{\sqrt{38}}[/tex] exp{i(-2k)x}

    Thus, by measuring the observable linear momentum, the only possible values are those relating to these three self-states of the linear momentum, because in the process of measuring the wave function is projected onto one of these (collapse or reduction of the state), and the probabilities for each value of the linear momentum is given by the square of the amplitude of the corresponding self-functions.

    Hence,

    Pr([tex]p_{x} = 0)[/tex] = [tex]\frac{36}{38}[/tex],

    Pr([tex]p_{x} = 2k)[/tex] = [tex]\frac{1}{38}[/tex],

    Pr([tex]p_{x} = -2k)[/tex] = [tex]\frac{1}{38}[/tex],

    and therefore, denoting the kinetic energy by T, we finally have:

    Pr(T = 0) = [tex]\frac{18}{19}[/tex],

    Pr(T = 2k²/m) = [tex]\frac{1}{19}[/tex],

    as the kinetic energy is related to linear momentum through the expression

    T = [tex]\frac{p^{2}_{x}}{2m}[/tex].

    Thanks in advance.
     
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