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State transformation - Position to momentum space (Dirac)

  1. Jun 16, 2015 #1
    Hey everyone!
    It's my first semester with quantum mechanics and I'm uncertain if my solution of this problem is correct, would be nice if someone could check and let me know :smile:

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

    I have to calculate the representation of the state:
    [itex]|\alpha \rangle \equiv exp[-i \textbf b \cdot \textbf x]|\phi\rangle [/itex], in momentum space.

    Where [itex]\textbf b[/itex] is a constant vector.

    2. Relevant equations
    - Dirac notation
    - Position operator in momentum space: [itex]\textbf x=i\hbar\nabla [/itex](1)
    - [itex]\langle \textbf p | \textbf p' \rangle = \delta(\textbf p-\textbf p') [/itex] (2)

    3. The attempt at a solution

    First I multiply by [itex]\langle \textbf p |[/itex]
    [itex]\langle \textbf p |\alpha \rangle = \langle \textbf p | exp[-i \textbf b \cdot \textbf x]|\phi\rangle [/itex]

    Then I add a one to the equation:
    [itex] =\int \langle \textbf p |exp[-i \textbf b \cdot \textbf x]|\textbf p'\rangle \langle \textbf p'|\phi\rangle d^3p'[/itex]

    With (1):
    [itex] =\int \langle \textbf p |exp[-i \textbf b \cdot \textbf i\hbar\nabla]|\textbf p'\rangle \langle \textbf p'|\phi\rangle d^3p'[/itex]

    I'm unsure about this step. Can I move the bra [itex]\langle \textbf p |[/itex] past the operator [itex]exp[-i \textbf b \cdot \textbf i\hbar\nabla] [/itex] like this:
    [itex] \int exp[-i \textbf b \cdot \textbf i\hbar\nabla] \delta(\textbf p-\textbf p') \langle \textbf p'|\phi\rangle d^3p'[/itex] ?

    If I can then I would get:
    [itex]= exp[-i \textbf b \cdot \textbf i\hbar\nabla] \langle \textbf p|\phi\rangle [/itex]

    [itex]\Rightarrow \alpha(\textbf p) = exp[-i \textbf b \cdot \textbf i\hbar\nabla]\phi(\textbf p) [/itex]

    I think I'm doing something wrong somewhere since I didn't do a fouriertransform anywhere.
    Is this even the right approach?

    Any hints/tips or corrections are very appreciated.

    Kind regards
    Alex
     
  2. jcsd
  3. Jun 16, 2015 #2

    blue_leaf77

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    It might be clearer to you if you do this problem by first calculating the given state in position representation ##\langle \mathbf{x}|\alpha \rangle##. Then make use of the fact that the momentum representation of ##|\alpha \rangle## is just the Fourier transform of ##\alpha (\mathbf{x}) = \langle \mathbf{x}|\alpha \rangle##,
    $$\langle \mathbf{p} | \alpha \rangle = \int \langle \mathbf{p} |\mathbf{x'}\rangle \langle \mathbf{x'} |\alpha \rangle d^3\mathbf{x'} = \int e^{-i\mathbf{p}\cdot \mathbf{x'}/\hbar} \alpha (\mathbf{x'}) d^3\mathbf{x'}$$

    Upon doing this you may see the property of the operator ##e^{-i\mathbf{b}\cdot \mathbf{x}}##, especially when it acts on ##|\mathbf{p} \rangle##.
     
    Last edited: Jun 16, 2015
  4. Jun 17, 2015 #3
    Thanks for the fast reply.
    When calculating the position representation of the state do I have to use the series representation of the exponential function? What does the solution look like? Is it an exponential function or some wave function?
     
  5. Jun 17, 2015 #4

    blue_leaf77

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    The position representation means the state ket be projected onto the eigenstate of position operator ##\hat{\mathbf{x}}##, which means
    $$ \langle \mathbf{x} | \psi \rangle = \langle \mathbf{x} | e^{-i\mathbf{b}\cdot \mathbf{x}} | \phi \rangle $$
    Now since the exponential operator in the middle is only a function of the operator ##\hat{\mathbf{x}}##, you can simply make use of the identity ##f(\hat{\mathbf{x}}) | \mathbf{x'} \rangle = f(\mathbf{x'}) | \mathbf{x'} \rangle ## where the ##f(\mathbf{x'}) ## on the right is just a number, that is one replaces any appearing ##\hat{\mathbf{x}}## in ##f(\hat{\mathbf{x}})## with ##\mathbf{x'}##.

    EDIT: there was a mistake in the first equation of this comment, now corrected.
     
    Last edited: Jun 17, 2015
  6. Jun 17, 2015 #5
    Thanks for the help.
    I think I got it now, let me know what you think :smile:
    [itex]\langle x' | exp[-i\textbf b \hat x] |\phi \rangle = exp[i\textbf b x] \langle x' |\phi \rangle [/itex] and then I would do the fourier transform by adding a one:
    (Our prof always adds a one :rolleyes:)
    [itex]exp[i\textbf b x] \langle x' |\phi \rangle = \int exp[i\textbf b x] \langle x' | p' \rangle \langle p' |\phi \rangle d^3p' = \int exp[i\textbf b x] \frac{exp[ix'p']}{\sqrt{2\pi}^3} \langle p' |\phi \rangle d^3p' [/itex] If it is correct how do I continue from here? :)
    Is this identity you've given valid because [itex] x[/itex] is an eigenvalue to [itex] \hat x[/itex]?
     
  7. Jun 17, 2015 #6

    blue_leaf77

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    You forgot the minus sign in the exponent on the right.
    yes

    What you want to calculate is the state in momentum representation and now what you have is the same state represented in position space, right? This is the point where the equation in comment #2 comes into play.
     
  8. Jun 17, 2015 #7
    I got the minus because I thought I had to do the complex conjugate of the operator when using it to the left.
    Using the fourier transform I get this:
    [itex] \int exp(-i(\textbf b x + \textbf p \textbf x' / \hbar)) \langle x' | \phi \rangle d^3 x' = \langle \textbf p | \phi \rangle[/itex]
    Is this [itex]| \alpha \rangle[/itex] in momentum space?
    By the way, thanks for guiding me :smile:
     
  9. Jun 17, 2015 #8

    blue_leaf77

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    Yes you did, but the complex conjugation applies on both sides of the equation.
    Except for the ##x## which should be primed ##x'## and the ## | \phi \rangle ## on the RHS which should be ## | \alpha \rangle ##, it's correct. But that expression can actually be cast further in a more compact form by noting that ##exp(-i(\textbf b x' + \textbf p \textbf x' / \hbar))## is actually a wavefunction of momentum operator. Consider ##\langle p | x \rangle = e^{-ipx/\hbar} ##, what will ##exp(-i(\textbf b x' + \textbf p \textbf x' / \hbar))## look like when changed into braket notation?
    EDIT: Sorry I forget to write the ##\hbar## in the wavefunction of momentum operator above, I'm getting used to atomic unit.
     
    Last edited: Jun 17, 2015
  10. Jun 17, 2015 #9
    Something like this? [itex]\langle \textbf {b} + \textbf p/\hbar | x' \rangle [/itex]
    Is [itex] \textbf p /\hbar [/itex] the wave vector?
     
  11. Jun 17, 2015 #10

    blue_leaf77

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    Actually ## \langle \textbf {b}/\hbar + \textbf p | x' \rangle##. After you make this replacement in the integrand and making use of the completeness relation in the ##|x'\rangle ## space which I assume you have been familiar with from you prof's lectures, you can get the braket notation of the momentum representation of ##|\alpha \rangle##. Just for advice, it may be interesting to slightly play with the final form of ##\langle p | \alpha \rangle## in order to investigate how the operator ##e^{-i\mathbf{b} \cdot \mathbf{\hat{x}} }## acts on ##| \mathbf{p} \rangle##.
     
    Last edited: Jun 17, 2015
  12. Jun 17, 2015 #11
    [itex] \int \langle \textbf b /\hbar + \textbf p| x' \rangle \langle x' | \phi \rangle d^3 x' = \langle \textbf p | \phi \rangle \Rightarrow \langle \textbf b /\hbar + \textbf p| \phi \rangle = \langle \textbf p | \phi \rangle[/itex] Is it some sort of momentum translation?
     
  13. Jun 17, 2015 #12

    blue_leaf77

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    Yes exactly. However be careful you are being sloppy in writing the proper maths here and there, the last relation should be ## \langle \mathbf{b} /\hbar + \mathbf{p}| \phi \rangle = \langle \mathbf{p} | \alpha\rangle = \langle \mathbf{p} | e^{-i\mathbf{b} \cdot \mathbf{\hat{x}}} |\phi\rangle ## otherwise it makes no sense to say that the operator in question is a momentum translation operator.
    Needless to say, you can also convince yourself that the operator of the form ##e^{i\mathbf{\hat{p}} \cdot \mathbf{a} }## also does the same behavior but in position space, due to which it's called the position translation operator.
     
    Last edited: Jun 17, 2015
  14. Jun 17, 2015 #13
    Sorry, I don't quite understand what you mean there. Could you maybe elaborate?
     
  15. Jun 17, 2015 #14

    blue_leaf77

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    Ok, first I actually have to correct your first equation in comment #11 as well:
    ## \int \langle \textbf b /\hbar + \textbf p| x' \rangle \langle x' | \phi \rangle d^3 x' = \langle \mathbf{b}/\hbar+ \mathbf{p} | \phi \rangle ##
    Then coming to the second equation next to the right of the arrow sign in comment #11, no matter how you look at it, equation like ##\langle \textbf b /\hbar + \textbf p| \phi \rangle = \langle \textbf p | \phi \rangle## cannot be justified. The LHS is just plainly different from the RHS, isn't it? Remember you got ##\langle \textbf b /\hbar + \textbf p| \phi \rangle ## from the first equation which was obtained after modifying ##\langle \mathbf{p} | \alpha \rangle##, this means they are actually equal, that is ##\langle \mathbf{p} | \alpha \rangle = \langle \textbf b /\hbar + \textbf p| \phi \rangle##. If you are still confused, just carefully trace back your derivation to see which comes from which.
     
    Last edited: Jun 17, 2015
  16. Jun 18, 2015 #15
    Hey, I think I've understood it now. Thanks for staying with me through this problem :smile:
    Did the same transformation for another similar operator into position space:
    [itex]|\beta \rangle \equiv e^{-i \textbf a \hat p} |\phi \rangle [/itex]

    [itex]\langle p' | \beta \rangle = \langle p' |e^{-i \textbf a \hat p}|\phi \rangle = e^{-i\textbf a p}\langle p' |\phi \rangle [/itex]
    Then I do the fourier transform: (Other sign in front of the i, right?)
    [itex]\langle x | \beta \rangle = \int e^{i(p'x/\hbar-\textbf a p)}\phi(p')d^3p' = \langle x-\textbf a / \hbar | \phi \rangle[/itex]
    Then [itex] \langle x | \beta \rangle = \langle x-\textbf a / \hbar | \phi \rangle [/itex], which is a position translation in position space.
     
  17. Jun 18, 2015 #16

    blue_leaf77

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    That's right.
     
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