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Hamiltonian and momentum operator acting on a momentum eigenstate

  1. Feb 17, 2016 #1
    suppose that the momentum operator [itex]\hat p[/itex] is acting on a momentum eigenstate [itex] | p \rangle [/itex] such that we have the eigenvalue equation [itex]\hat p | p \rangle = p| p \rangle[/itex]
    Now lets project [itex]\langle x | [/itex] on the equation above and use the completeness relation [itex]\int | x\rangle \langle x | dx =\hat I[/itex]
    we get that [itex]\psi_{p}(x)=Ae^{ipx/\hbar}[/itex] where [itex]\psi_{p}(x)=\langle x | p \rangle[/itex]

    Now let the operator [itex]\hat H[/itex] act on the eigenstate [itex]| p\rangle [/itex],such that[itex]\hat H | p \rangle=E | p \rangle[/itex]
    but the two operators [itex]\hat p and \hat H [/itex] have the same eigenstate [itex]| p\rangle[/itex],since[itex]\hat H | p\rangle=\hat p^{2}/2m | p \rangle = p/2m\hat p |p\rangle =p^{2}/2m| p \rangle [/itex]
    Now lets solve the eigenvalue equation [itex]\hat H | p \rangle=p^{2}/2m | p \rangle[/itex] in the [itex]X [/itex] basis, what we will certainly get is [itex]\psi_{p}(x)=\alpha e^{ipx/\hbar}+ \beta e^{-ipx/\hbar}[/itex] which is different from [itex]\psi_p(x)[/itex] above.
    This is what confuses me, why when use different operators( that have same eigenstates like the energy and momentum operator)in the eigenvalue problem, mathematics gives different [itex]\psi_p(x)[/itex]?
    As [itex] | p\rangle[/itex] is the same why we get different [itex]\langle x |p \rangle [/itex]s?
     
  2. jcsd
  3. Feb 17, 2016 #2

    blue_leaf77

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    The spectrum of a free space Hamiltonian is degenerate in ##p## - all eigenstates with equal magnitude of the momenta share the same energy.
     
  4. Feb 17, 2016 #3
    what confuses me is that the [itex]| p \rangle[/itex] for both eigenvalue equations[itex]( \hat H | p \rangle=p^{2}/2m | p \rangle and \hat p =p | p \rangle)[/itex] is the same, so why we get different [itex]\langle x | p \rangle [/itex]s?
    Frankly,I couldn't relate your "answer" to my "question" maybe because the picture is not clear to me yet. :cry:
     
  5. Feb 17, 2016 #4

    stevendaryl

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    I'm not sure I understand the question, but if you let [itex]|p\rangle[/itex] be the state where [itex]\hat{p}|p\rangle = p |p\rangle[/itex], then:

    [itex]\hat{H} |p\rangle = \frac{p^2}{2m} |p \rangle[/itex], and also [itex]\hat{H} |-p\rangle = \frac{p^2}{2m} |-p\rangle[/itex]. So knowing [itex]\hat{H} |\psi\rangle = E |\psi\rangle[/itex] doesn't uniquely determine [itex]|\psi\rangle[/itex]. It could be any state of the form [itex]\alpha |p_E\rangle + \beta |-p_E\rangle[/itex], where [itex]p_E = \sqrt{2m E}[/itex]
     
  6. Feb 17, 2016 #5

    blue_leaf77

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    When you solve ##\langle x|\hat H | p \rangle=p^{2}/2m \langle x| p \rangle##, there are three solutions you can get: ##Ae^{ipx/\hbar}##, ##Be^{-ipx/\hbar}##, or a linear combination of the previous two, which is the one you have computed. The first two solutions are also the eigenfunctions of the momentum operator, the last one however is not.
     
  7. Feb 18, 2016 #6
    @blue_leaf77
    @stevendaryl

    suppose we have the eigenvalue equation [itex]\hat p |p \rangle=p |p\rangle[/itex] then lets operate [itex]\hat p /2m [/itex] on [itex] p | p \rangle[/itex] so we will get [itex]\hat p^{2}/2m | p \rangle=p^{2}/2m | p \rangle[/itex] so we are still now dealing with the same [itex]| p \rangle[/itex]! Name what I did here "step 1".

    suppose that the solution of the second eigenvalue equation,which is after acting on the next operator [itex]\hat p /2m[/itex]is [itex]\psi_{p}(x)=Ae^{ipx/\hbar}+Be^{-ipx/\hbar}[/itex],the solution to the first eigenvalue equation is certainly not the same it will be[itex]\psi_{p}(x)=Ae^{ipx/\hbar} or \psi_{p}(x)=Be^{-ipx/\hbar}[/itex](therfore we get different eigenfunctions while |p> is the same !!)

    I know that |p>s are not unique,as you said in your replies,but in "step 1" the operations didn't affect | p>. So why is this happening?I'm still not convinced yet.
    What is going wrong?where is my mistake here?
     
  8. Feb 18, 2016 #7

    blue_leaf77

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    The point where you go wrong in this case is that you assume ##\langle x|p \rangle## in
    $$
    -\hbar^2 \frac{d^2}{dx^2} \langle x|p \rangle = p^2 \langle x|p \rangle
    $$
    and
    $$
    -i\hbar \frac{d}{dx} \langle x|p \rangle = p \langle x|p \rangle
    $$
    to be the same in all cases. The fact that the ##\langle x|p \rangle##'s satisfy different equations means that they are generally different functions. It's just a math problem, to give another illustration, consider the equation
    $$
    -\frac{1}{\alpha} \frac{d}{dx}e^{-\alpha x} = e^{-\alpha x}
    $$
    and then you apply ##d/dx## to both sides to get
    $$
    \frac{1}{\alpha^2} \frac{d^2}{dx^2}e^{-\alpha x} = e^{-\alpha x}
    $$
    This particular function ##e^{-\alpha x}## certainly satisfies the last equation, but is it the only solution of this equation? Definitely no, the last equation is a homogenous linear second order differential equation, so it must have one more solution, which is ##e^{\alpha x}##.
    Be careful with the math:
    1. It should be ##\hat{p} = p|p\rangle##, not ##\hat{p}|p\rangle = p|p\rangle##.
    2. When you apply ##\hat p /2m## to ##p|p\rangle##, you will get ##\frac{p^3}{2m}|p\rangle##.
     
  9. Feb 18, 2016 #8

    stevendaryl

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    I don't understand why you think there is a mistake. All the following statements are true:
    1. [itex]\hat{p} e^\frac{ipx}{\hbar} = p e^\frac{ipx}{\hbar}[/itex]
    2. [itex]\frac{\hat{p}^2}{2m} e^\frac{ipx}{\hbar} = \frac{p^2}{2m} e^\frac{ipx}{\hbar}[/itex]
    3. [itex]\hat{p} e^\frac{-ipx}{\hbar} = -p e^\frac{-ipx}{\hbar}[/itex]
    4. [itex]\frac{\hat{p}^2}{2m} e^\frac{-ipx}{\hbar} = \frac{p^2}{2m} e^\frac{-ipx}{\hbar}[/itex]
     
  10. Feb 18, 2016 #9
    @blue_leaf77
    @stevendaryl

    I'm convinced with all the math and I know that math gives the different eigenfunctions you have written.
    My problem is that I see that the math is not consistent here, because lets forget about the differential equations and look at |p> by itself,by looking at "step 1" above we see that |p> didn't change. project <x| on |p> as |p> is same in both cases then <x|p> is the same.
    but I guess I know where I'm wrong, I think because in dealing with infinite-dimensional spaces focusing in |p> is somehow meaningless in solving the eigenvalue equation we should focus on <x|p>, to get the <x|p> there is no way to get it unless by solving the differential equation,to get the differential equation we should insert the identity relation ∫|x><x|dx=I,so the identity relation is the reason behind all of this.
    So I conclude that the answer of my question is to understand the identity relation and its derivation,and this is beyond my level.
    Is this right?
     
  11. Feb 18, 2016 #10

    stevendaryl

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    I really don't understand why you think there is anything inconsistent going on. And I don't think that it being infinite-dimensonal has anything to do with it. What is it that you think is inconsistent?
     
  12. Feb 18, 2016 #11

    blue_leaf77

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    It has nothing to do with the dimensionality, what you are encountering is a ubiquitous problem in QM: degeneracy. Another example of such problem is found in the angular momentum operators - the operator ##L^2## is degenerate in the eigenstate of the operator ##L_z##.
    I think the problem is all about the line of reasoning. First you have the equation
    $$
    -i\hbar \frac{d}{dx} \langle x|p \rangle = p \langle x|p \rangle,
    $$
    this equation is satisfied only by ##\langle x|p \rangle## corresponding to positive ##p##. Now, you apply another derivative to this equation to get
    $$
    -\hbar^2 \frac{d^2}{dx^2} \langle x|p \rangle = p^2 \langle x|p \rangle
    $$
    this last equation is also satisfied by ##\langle x|p \rangle## corresponding to positive ##p##. The point I would like to stress here is that, the function ##\langle x|p \rangle## appearing in the last equation is 100% known to be that corresponding to the positive momentum. If you ask why, the answer is because the last equation was obtained by differentiating the first equation, which is satisfied by ##\langle x|p \rangle## with positive momentum. Knowing this, don't solve the last equation again because you know how ##\langle x|p \rangle## looks like. You were being inconsistent (instead of the math) when doing:
    because ##|p\rangle## in your equation [itex]\hat H | p \rangle=p^{2}/2m | p \rangle[/itex] is the ##|p\rangle## which satisfies ##\hat p|p\rangle = p|p\rangle##. By writing [itex]\psi_{p}(x)=\alpha e^{ipx/\hbar}+ \beta e^{-ipx/\hbar}[/itex], you are going against your own mathematical chronology in which you started with the equation ##\hat p | p \rangle = p| p \rangle##.
     
  13. Feb 18, 2016 #12
    I think your last two lines convinced me.If we started with the equation [itex]\hat p | p \rangle=p | p\rangle[/itex] to end up with [itex]\hat H | p \rangle=p^{2}/2m |p\rangle[/itex] we must have the same [itex]\psi_{p}(x)[/itex] because we are talking about the same p.
    suppose we are dealing in the case of finite-dimensional case, [itex]\hat A|1 \rangle=a| 1\rangle[/itex](1) now Let [itex]\hat A[/itex] act another time in this equation what we will get is [itex]\hat A^{2}| 1 \rangle=a^{2}| 1\rangle[/itex](2).In (1) and (2) [itex]| 1\rangle[/itex] didn't change,it is the same.
    lets project a vector named [itex]| L\rangle[/itex] on [itex]|1\rangle[/itex],<L|1> in (1) and<L|1> in (2) are the same.
    In infinite dimensional space:
    here I guess |p> didn't change too,so if we want to project <x| in the first |p>(before acting [itex]\hat p[/itex] again) and <x| in the second |p>(after acting [itex]\hat p[/itex] again) we will get the same result <x|p> unless we insert the identity relation∫|x><x|dx=I because this will lead to the differential equations that will give us the <x|p>s in each case,and what I saw is inconsistent is that <x|p> in first case will be different from <x|p> of second case not like what intuition told me above.Thats why I talked about dimensionality in the previous reply, I hope you get my point!
    But I think blue leaf77 convinced me by his last two lines.
     
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