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Operators and eigenfunctions

  1. Jul 19, 2015 #1
    Hello, so I have a couple of related questions.

    1) If you have a wavefuction Ψ, and act on it with some operator, does it have to give you the same wavefunction back (ie. does the wavefunction have to be an eigenfunction of the operator)?

    Could you have a wavefunction like e-iħtSin(x)? Since this is an eigenfunction of the energy operator but not the momentum operator.

    I ask this because if Ψ is an eigenfunction of the momentum operator then it would have a definite momentum given by the eigevalue. I believe this is correct anyway. So if Ψ is not an eigenfunction of the operator you would not have a definite value for measuring that observable, then would you return some uncertainty in the observed value?

    2) If you have Ψ= ekx, then is Ψ not an eigenfunction of both the position and momentum operators? I'm struggling with this one since that would mean it has both a definite position and momentum, but since these operators are non-commutative there should be some uncertainty and so surely you can't have a definite answer for both?
     
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  3. Jul 19, 2015 #2

    andrewkirk

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    Strictly speaking, e-iħtSin(x) is not a wavefunction but a representation of the wavefunction in a particular basis. Based on what you wrote it is the energy basis in this case. The wavefunction itself is just an element of a Hilbert space, and we use a representation of it in a particular basis in order to understand and manipulate it better. Representations in different bases - eg energy vs momentum vs position - are always different. Even if their functional form is the same - as in your 2nd question - they will refer to different variables - eg p rather than x. One wavefunction has as many different representations as there are bases.

    When you apply an operator to a wavefunction you get a new wavefunction, and that will usually change the representation in any given basis as well.

    I find that if, whenever things get confusing, I remind myself that functions like those written above are only representations of wavefunctions, rather than the wavefunction itself, it often clarifies things.

    The terminology 'wave function' unfortunately tends to promote confusion on this point rather than clarify. It would be simpler if we always just referred to kets. But 'wave function' has been used for so long that we are unlikely to be able to get rid of it now. Plus it sounds cooler and more Dirk Gently-ish than ket (which just reminds me of kettles).

    Given that, a simple way to avoid confusion that helps a bit is to always write 'wavefunction' rather than 'wave function', as a reminder that it is not really a function. I've tried to do that above, but I may have missed some.
     
    Last edited: Jul 19, 2015
  4. Jul 20, 2015 #3
    1) that depends on if the wave function corresponds to an eigenfunction of that operator. if it is an eigenfunction, then you will get the function back times the eigenvalue. if the state is a superposition then the operator will just map it to something, and not much can be said about the action of the operator on such a state. i think you are confusing the action of an operator with one of the postulates of QM, namely that measurement can only yield eigenvalues of the corresponding operator. the action of an operator ALWAYS takes one vector into another vector, eigenvectors just get scaled. MEASUREMENT will reduce the state to a eigenstate, acting on a state with an operator IS NOT the same thing as making a measurement. Uncertainty in a measurement is introduced when you have 2 operators that don't commute. position and momentum do not commute, hence the uncertainty between them. absolute states of momentum and position do not exist in nature, we use delta functions to approximate such a thing.

    2)Generally it depends on the problem you are talking about, however momentum and position NEVER commute in any problem you will come across. Remember that an eigenfunction of any continuous variable is a delta function in its respective representation. ekx is clearly not a delta function, and when you transform it into a momentum representation it is still not a delta function either so this is not an eigenfunction of either position or momentum ie it is not an eigenfunction of position OR momentum
     
  5. Jul 20, 2015 #4
    Ok, so I'm still having a bit of a problem here. If Ψ= ei(kx-ωt) and you act on it with the position operator you get xΨ, so is it not an eigenfunction of that operator. Likewise if you act on it with the momentum operator -iħd/dx you return ħkΨ which through de Broglie is momentum p. So since you got the same state back is this not also an eigenfunction of the momentum operator.

    There's probably something simple that I just can't grasp.
     
  6. Jul 20, 2015 #5

    blue_leaf77

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    At this point it's important to understand the distinction between a state and the representation of a state in certain basis. The former is a vector in some vector space, while the latter is a scalar as this is calculated as the inner product between two vectors, the state being represented and the basis vector in which the state is represented. In quantum mechanics, it's a convention to write a state in the braket notation, so ##|a\rangle## is a state. But if you form an inner product between this vector state and a basis vector, let's say ##|u\rangle##, then ##\langle u | a \rangle## is called the representation of ##|a\rangle## in the basis ##|u\rangle##.
    Now attacking the actual problem, let's denote ##|x\rangle## an eigenfunction of position operator ##\hat{x}## with eigenvalue ##x## (note the difference in meaning between ##x## with hat and that without). We then have ##\hat{x} |x\rangle = x |x\rangle##. Now let's represent this state in the basis of momentum eigenstate ##|p\rangle##, we simply need to multiply the previous equation with ##\langle p|## from left to obtain ##\langle p| \hat{x} |x\rangle = x \langle p|x\rangle = x e^{ikx}## where we have used the fact that ##\langle p|x\rangle = e^{ikx}##. The point here is that it's not complete to just say that ##e^{ikx}## be the eigenfunction of position operator ##\hat{x}##, instead it's the eigenfunction of position operator represented in momentum basis. For another example, instead of representing ##\hat{x} |x\rangle = x |x\rangle## in momentum basis, you can try representing this in position basis (that is, by multiplying it with ##\langle x'|## from the left). The answer will of course look different from a complex exponential function we have before.
     
    Last edited: Jul 20, 2015
  7. Jul 20, 2015 #6
    Just to build on what blue said, x^|x>=x|x> Where x is a number corresponding to the ket. An arbitrary wavefunction is a sum, or a integral in the continuous case, over the basis kets. When x acts on an arbitrary wavefunction xΨ is to be integrated over when you perform any calculation, thus x is essentially a variable in this case and not a number like the case of the position kets. Viewed this way, it is clear xΨ is not just a scalar times the original function, ie not an eigenfunction of x operator.
     
  8. Jul 25, 2015 #7
    Where can I find a prove of the last relation: ##\langle p|x\rangle = e^{ikx}##?

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  9. Jul 25, 2015 #8

    blue_leaf77

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    From the relation
    $$\langle x |\hat{p}|\psi \rangle = -i\hbar \frac{\partial}{\partial x} \langle x | \psi \rangle $$
    If you make substitution ##|\psi \rangle = |p \rangle## where ##|p\rangle## is the eigenstate of momentum operator ##\hat{p}## and solve the differential equation, you will see immediately that it leads to ##\langle x|p \rangle## being a complex exponential function, then take the complex conjugate to get ##\langle p|x \rangle##.
     
  10. Jul 25, 2015 #9
    But $$\langle x |\hat{p}|\psi \rangle$$ shouldn't be equal to$$ \langle x |-i\hbar \frac{\partial}{\partial x}|\psi \rangle ?$$

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    Last edited: Jul 25, 2015
  11. Jul 25, 2015 #10

    blue_leaf77

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    No it's not. ##|\psi\rangle## is just a ket vector, not a function. It doesn't make sense to apply derivative to a ket (or bra) vector, instead ##\langle x|\psi \rangle## is a (scalar) function since it's formed from an inner product. If you want to know more about the derivation I recommend Sakurai's Modern Quantum Mechanics in chapter 1.
     
  12. Jul 25, 2015 #11
    the momentum operator isn't a derivative until the last step. the matrix elements of the momentum operator in position space are derivatives of delta functions, they are only a derivative of the wave function at the last step.
     
  13. Jul 25, 2015 #12
    Ok, so, since I don't have that book, how can I prove that $$\langle x |\hat{p}|\psi \rangle = -i\hbar \frac{\partial}{\partial x} \langle x | \psi \rangle ?$$

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  14. Jul 25, 2015 #13

    andrewkirk

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    In Shankar's 'Principles of Quantum Mechanics' his postulates of QM define the position ##\hat{x}## and momentum ##\hat{p}## operators by:
    $$\langle x|\hat{x}|x'\rangle = x\delta(x-x')$$
    $$\langle x|\hat{p}|x'\rangle = -i\hbar x\frac{\partial}{\partial x}\delta(x-x')$$
    The relationship follows fairly readily from that:

    $$\langle x |\hat{p}|\psi \rangle
    = \sum_{x'}\langle x |\hat{p}|x'\rangle\langle x'|\psi \rangle
    = \sum_{x'}-i\hbar x\frac{\partial}{\partial x}\delta(x-x')\langle x'|\psi \rangle
    = -i\hbar x\frac{\partial}{\partial x}\sum_{x'}\delta(x-x')\langle x'|\psi \rangle
    \\= -i\hbar x\frac{\partial}{\partial x}\sum_{x'}\langle x|x'\rangle
    \langle x'|\psi \rangle
    = -i\hbar x\frac{\partial}{\partial x}\langle x|\psi \rangle
    $$
     
    Last edited: Jul 25, 2015
  15. Jul 26, 2015 #14

    vanhees71

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    Everything follows from the commutation relations of operators, which themselves follow from the symmetry group underlying the space-time structure (in this case Galilei symmetry).

    Here, we only need the Heisenberg algebra, which is defined by the canonical commutator relation
    $$[\hat{x},\hat{p}]=\mathrm{i} \hbar \mathbb{1}.$$
    Now from classical mechanics we know that ##\hat{p}## is the generator for spatial translations. That leads us to investigate the operator
    $$\hat{X}(\xi)=\exp(\mathrm{i} \hat{p} \xi/\hbar) \hat{x} \exp(-\mathrm{i} \hat{p} \xi/\hbar).$$
    Taking the derivative wrt. ##\xi## leads to
    $$\mathrm{d}_x \hat{X}(\xi)=\frac{\mathrm{i}}{\hbar} \exp(\mathrm{i} p \xi/\hbar) [\hat{p},\hat{x}] \exp(-\mathrm{i} p \xi/\hbar) =\mathbb{1}.$$
    Since ##\hat{X}(0)=\hat{x}## we have
    $$\hat{X}(\xi)=\hat{x}+\xi \mathbb{1}.$$
    Now let ##|x \rangle## be a generalized eigenvector of ##\hat{x}## then you have
    $$\hat{x} \exp(-\mathrm{i} \hat{p} \xi/\hbar)|x \rangle=\exp(-\mathrm{i} \hat{p} \xi/\hbar) \hat{X}(\xi) |x \rangle=(x+\xi) \exp(-\mathrm{i} \hat{p} \xi)|x \rangle.$$
    That means that ##\exp(-\mathrm{i} \hat{p} \xi/\hbar)|x \rangle## is an eigenvector of ##\hat{x}## with eigenvalue ##x+\xi##.

    Thus we can define a position eigenbasis by
    \begin{equation}
    \label{1}
    |x \rangle=\exp(-\mathrm{i} \hat{p} x/\hbar)|x=0 \rangle.
    \end{equation}
    In doing so we have with a momentum eigenvector ##|p \rangle##
    $$\langle p|x \rangle=\langle p|\exp(-\mathrm{i} \hat{p} x/\hbar)|x=0 \rangle = \langle \exp(+\mathrm{i} \hat{p} x/\hbar) p|x=0 \rangle=\langle \exp(+\mathrm{i} p x/\hbar) p|x=0 \rangle=\exp(-\mathrm{i} \hat{p} x/\hbar) \langle p|x=0 \rangle=A^*(p) \exp(-\mathrm{i} p x/\hbar).$$
    This implies
    $$\langle x|p \rangle=A(p) \exp(+\mathrm{i} p x/\hbar).$$
    The normalization constant ##A(p)## is usually chosen such that
    $$\langle p|p' \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|p' \rangle=\int_{\mathbb{R}} \mathrm{d}x A^*(p) A(p') \exp[\mathrm{i}(p'-p) x] = |A(p)|^2 2 \pi \delta[(p-p')/\hbar]=|A(p)|^2 \hbar \delta(p-p') \stackrel{!}{=} \delta(p-p').$$
    From this we have
    $$A(p)=\frac{1}{\sqrt{2 \pi \hbar}}.$$
    With (\ref{1}) it's also easy to calculate the momentum operator in position representation. For arbitrary state ket ##|\psi \rangle## the corresponding position-space wave function is
    $$\psi(x)=\langle x|\psi \rangle=\langle \exp(-\mathrm{i} \hat{p} x/\hbar) 0|\psi \rangle=\langle 0|\exp(+\mathrm{i} \hat{p} x/\hbar) \psi \rangle.$$
    From this you get
    $$\partial_x \psi(x)=\langle 0|\exp(+\mathrm{i} \hat{p} x/\hbar) \mathrm{i} \hat p/\hbar \psi \rangle=\frac{\mathrm{i}}{\hbar} \langle x| \hat{p} \psi \rangle$$
    or rewritten
    $$\langle x|\hat{p} \psi \rangle=:\hat{p} \psi(x)=-\frac{\mathrm{i}}{\hbar} \partial_x \psi(x).$$
    Note that the symbol ##\hat{p}## on the very left-hand side of the equation has a different meaning than in the middle expression. In the former case it's the abstract momentum operator in abstract Hilbert space, i.e., representation independent, while in the latter case it's the momentum operator in the position representation, acting on (a dense subset of) the Hilbert space of square-integrable functions, ##\mathrm{L}^2(\mathbb{R})## .
     
    Last edited: Jul 26, 2015
  16. Jul 26, 2015 #15

    blue_leaf77

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    You must mean position translation.
     
  17. Jul 26, 2015 #16

    vanhees71

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    Sure, I'll correct the typo.
     
  18. Jul 26, 2015 #17
    Thanks to both of you!

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