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Probability of observing E for particle in a box

  1. Mar 14, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-3-15_0-41-42.png

    2. Relevant equations


    3. The attempt at a solution
    Since this is discrete, I know the probability is equal to ##|a_{n}|^{2}##, hence the probability of observing ##E_{1}##, ##P(E_{1}) = \frac {9}{25}##, and ##P(E_{2}) = \frac {16}{25}##. However, I just sort of remembered this fact, I am not sure how to prove that the probability is equal to the square of the absolute value of the coefficient.

    For the second part, they pose a very good question, what in the heck is ##f##? I don't see any ##f## anywhere.

    For the third question, shouldn't it just be ##|q_{7}|^{2}##?
     
  2. jcsd
  3. Mar 14, 2015 #2

    DrClaude

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    There nothing to prove. This is simply the Born rule.

    I see two possibilities: a typo, and its supposed to be ##|\psi\rangle##. Or its any state ##|f\rangle##, and you have to prove a general statement.

    What is ##q_7## in the problem?
     
  4. Mar 14, 2015 #3

    Maylis

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    I'll just assume its for any state ##|f \rangle##,

    Then $$\langle f|f \rangle = \int_{-\infty}^{\infty} f^{*}f \hspace{0.02 in} dx $$

    And for question 3, I suppose ##q_{7}## is the operator ##\hat {Q}##
     
  5. Mar 14, 2015 #4

    DrClaude

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    I guess that the question is asking you to push this further, considering that you have a particle in a box.

    That's not correct. You have
    $$
    \hat{Q} | g_7 \rangle = q_7 | g_7 \rangle
    $$
    Can you identify all the elements in that equation?
     
  6. Mar 14, 2015 #5

    vela

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    Did you post the actual, complete problem statement as given to you? I'm thinking not because of the word eigenstuff. If you are, you should ask your instructor for clarification on what's being asked.

    The general idea is that if a system is in state ##\lvert \psi \rangle## and you want to calculate the probability you'll find it in the state ##\lvert \phi \rangle##, you calculate the overlap of the two states, ##\langle \phi \vert \psi \rangle##, which is called the probability amplitude. The probability is then given by the square of the modulus of the probability amplitude. Knowing this, you should be able to answer part (c).

    By the way, ##q_7## is a number. It's definitely not the operator ##\hat{Q}##. What is its relation to ##\hat{Q}##? That's what DrClaude wants you to understand.
     
  7. Mar 15, 2015 #6

    Maylis

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    This is the question statement, I just copied it off a youtube video.

    Assuming ##|g_{7} \rangle## is nonzero, they should cancel on both sides, hence ##q_{7} = \hat {Q}##. I know ##q_{7}## is an eigenvalue and ##\hat {Q}## is the operator, but I don't know what to make of it.

    For question 2, I will revert to @DrClaude other idea, to replace ##f## with ##\psi##

    Then
    $$\langle \psi | \psi \rangle = \int_{-\infty}^{\infty} (\frac {3}{5} | \psi_{1} \rangle + \frac {4i}{5} | \psi_{2} \rangle)(\frac {3}{5} | \psi_{1} \rangle + \frac {-4i}{5} | \psi_{2} \rangle) dx $$

    $$ = \int_{-\infty}^{\infty} \frac {9}{25} + \frac {16}{25} dx $$
    $$ = \int_{-\infty}^{\infty} 1 dx $$
    $$ = ??? $$
    Apparently this does not make sense.
     
    Last edited: Mar 15, 2015
  8. Mar 15, 2015 #7

    DrClaude

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    You can't do that. Since one is an operator and the other a scalar, they can't be equal. If you want to calculate things in QM, you need to understand how eigenequations work. With the equation ##\hat{Q} | g_7 \rangle = q_7 | g_7 \rangle##, you are being told that there is an operator ##\hat{Q}##, of which one eigenvector is ##| g_7 \rangle##, with eigenvalue ##q_7##.

    The question is not very well posed. Let me rephrase it: you have a system in state ##|\psi\rangle##. You measure Q, which corresponds to the observable (operator) ##\hat{Q}##. Write an expression for the probability of measuring ##Q = q_7##.

    You are mixing two notations. If you have functions ##\phi(x)## and ##\psi(x)##, then you can calculate the scalar product as
    $$
    \int_{-\infty}^\infty \phi(x)^* \psi(x) dx
    $$
    In the Dirac notation, that scalar product is simply written ##\langle \phi | \psi \rangle##. The product of a bra and a ket is already an integration, and you can't have a ket as a function of position. Bras and kets are abstract representation of states.
     
  9. Mar 15, 2015 #8

    Maylis

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    How can a ket not be a function of position? Certainly ##\psi = f(x)##, hence what Vela said the "probability amplitude" ##\langle \phi | \psi \rangle## would have a ket as a function of position. If what I wrote is not right, I have no clue what I am supposed to do to evaluate.
     
    Last edited: Mar 15, 2015
  10. Mar 15, 2015 #9

    DrClaude

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    Because it is an abstract state vector, that lives in a Hilbert space. To get a function of position, you need to calculate
    $$
    \psi(x) = \langle x | \psi \rangle
    $$

    If you know the position representation of the bras and kets, you can use that to calculate inner products, e.g., for the 1D harmonic oscillator eigenfunctions ##\phi_i##
    $$
    \langle \phi_i | \phi_j \rangle = \int_{-\infty}^\infty \phi_i^*(x) \phi_j(x) dx
    $$
    But you don't need to know that function representation, because since the eigenfunctions are othonormal, you get directly
    $$
    \langle \phi_i | \phi_j \rangle =\delta_{ij}
    $$
    In other words, you can (and should here) manipulate the kets as abstract state vectors.
     
  11. Mar 15, 2015 #10

    Maylis

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    So then the inner product ##\langle \psi | \psi \rangle## actually means ##\psi(\psi)##? Since you can't have a function with an argument of a function, so that makes no sense.
     
  12. Mar 15, 2015 #11

    DrClaude

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    The notation does not mean ##\langle a | b \rangle = b(a)##. I was referring to position eigenstates
    $$
    \hat{x} | x \rangle = x | x \rangle
    $$
    which have the property
    $$
    \langle x | x' \rangle = \delta(x - x')
    $$
    You need to understand Dirac notation better to progress with these exercises. The document http://www.hep.manchester.ac.uk/u/stevew/teaching/dirac.pdf might be a good start.
     
  13. Mar 15, 2015 #12

    Maylis

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    No wonder I have no idea what you are talking about, I'm still on section 3.5 of Griffiths, and "Dirac notation" does not appear until 3.6. However, I still don't see the consistency in what you are saying.

    If the inner producted is defined for ##f(x)## and ##g(x)##
    $$ \langle f|g \rangle = \int_{-\infty}^{\infty} f^{*}(x)g(x) \hspace {0.03 in} dx $$

    Then what is wrong with
    $$ \langle \psi |\psi \rangle = \int_{-\infty}^{\infty} \psi^{*}(x) \psi(x) \hspace {0.03 in} dx $$

    A side note, it seems like the author is using Dirac notation before formally introducing it as the section in chapter 3...
     
    Last edited: Mar 15, 2015
  14. Mar 15, 2015 #13

    vela

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    Nothing, but that's not what you wrote before.

    It may help to connect the new notation to what you've learned before. Suppose you have a vector ##\vec{a}##. In the new notation, you'd write ##\lvert a \rangle##. To keep things simple and familiar, let's say it lives in three-dimensional space. Now you decide how you're going to orient the x, y, and z axes. In linear algebra speak, you're choosing a basis. Once you've done this, it makes sense to talk about the x-component of ##\vec{a}## and so on, and you can now write
    $$\vec{a} = a_1 \hat{e}_1 + a_2 \hat{e}_2 + a_3 \hat{e}_3 = \sum_{i=1}^3 a_i \hat{e}_i$$ where the ##\hat{e}_i##'s correspond to the unit vectors (basis vectors) pointing along the various axes. In the new notation, you'd write
    $$\lvert a \rangle = a_1 \lvert 1 \rangle + a_2 \lvert 2 \rangle + a_3 \lvert 3 \rangle = \sum_{i=1}^3 a_i \lvert i \rangle.$$ To find ##a_i##, you'd calculate ##a_i = \hat{e}_i \cdot \vec{a}##. In Dirac notation, you have ##a_i = \langle i \vert a \rangle##. So finally we can write
    \begin{align*}
    \vec{a} &= \sum_{i=1}^3 \hat{e}_i (\hat{e}_i \cdot \vec{a}) \\
    \lvert a \rangle &= \sum_{i=1}^3 \lvert i \rangle\langle i \vert a\rangle
    \end{align*}

    Now let's go back to quantum mechanics… You have the state vector ##\lvert \psi \rangle##. The set of vectors {##\lvert x \rangle##} form a basis; they correspond to {##\lvert i \rangle##} in the previous example where ##x## acts as a label as ##i## did earlier. ##x## is continuous, so we replace the summation by an integral
    $$\lvert \psi \rangle = \int \lvert x \rangle \langle x \vert \psi \rangle\,dx.$$ Just as each ##a_i = \langle i \vert a \rangle## is a number, ##\langle x \vert \psi \rangle## is a number. It's equal to the wave function ##\psi## evaluated at ##x##. That is, ##\psi(x) = \langle x \vert \psi \rangle##. You need to distinguish between the state vector ##\lvert \psi \rangle##, the wave function ##\psi##, and ##\psi(x)##, the value of the wave function maps ##x## to.

    Now consider the dot product of two vectors ##\vec{a}## and ##\vec{b}##. Once we choose a basis, we can write
    \begin{align*}
    \vec{a}\cdot\vec{b} &= \sum_{i=1}^3 a_i b_i = \sum_{i=1}^3 (\vec{a}\cdot\hat{e}_i)(\hat{e}_i \cdot \vec{b}) \\
    \langle a \vert b \rangle &= \sum_{i=1}^3 a_i b_i = \sum_{i=1}^3 \langle a \vert i \rangle \langle i \vert b \rangle
    \end{align*} In the continuous (and complex-valued) case, we have
    $$\langle f \vert g \rangle = \int f^*(x)g(x)\,dx = \int \langle f \vert x \rangle \langle x \vert g \rangle.$$ In what you wrote back in post 6, you didn't put ##\psi(x)## in the integral; you used ##\lvert \psi \rangle##. That's why you got a non-sensical expression. It was analogous to saying
    $$\vec{a}\cdot\vec{a} = \sum_{i=1}^3 \vec{a}\cdot\vec{a} = \sum_{i=1}^3 (a_1^2+a_2^2+a_3^2) = 3(a_1^2+a_2^2+a_3^2).$$ In the very first step, instead of correctly using ##a_i## and regular scalar multiplication, I inserted ##\vec{a}## and used vector multiplication, which leads to an erroneous result.
     
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