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Prove that 0=1 in quantum mechanics

  1. Aug 6, 2010 #1
    Of course every prove of this type have some mistake.

    If nobody will see I will post solution!

    [tex]\hat{A}[/tex], [tex]\hat{B}[/tex] - linear, hermitian operator which commutator is

    [tex][\hat{A},\hat{B}]=i\hbar\hat{I}[/tex]

    [tex]\hat{I}[/tex] - unit operator
    Eigen problems of operators are:

    [tex]\hat{A}|\psi \rangle=a|\psi \rangle[/tex]

    [tex]\hat{B}|\psi \rangle=b|\psi \rangle[/tex]


    [tex]|\psi \rangle[/tex] - normalized eigen vector

    [tex]\langle \psi|\psi\rangle=1[/tex]

    Look now

    [tex][\hat{A},\hat{B}]=i\hbar\hat{I}[/tex]

    Take a sandwich of right and left side with vector [tex]|\psi\rangle[/tex]

    [tex]\langle \psi|[\hat{A},\hat{B}]|\psi\rangle=i\hbar\langle \psi|\hat{I}|\psi\rangle[/tex]

    [tex]\frac{1}{i\hbar}\langle \psi|(\hat{A}\hat{B}-\hat{B}\hat{A})|\psi \rangle=1[/tex]

    [tex]\frac{a}{i\hbar}(\langle \psi|\hat{B}|\psi\rangle-\langle \psi|\hat{B}|\psi\rangle)=1[/tex]

    [tex]0=1[/tex]

    Where is mistake? :D
     
  2. jcsd
  3. Aug 6, 2010 #2
    If

    [tex]A|a>=a|a>[/tex]

    consider the state ket

    [tex]|\psi>=e^{-ibB}|a>[/tex]

    where b is a real number. Then

    [tex]A|\psi>=Ae^{-ibB}|a>=\left(e^{-ibB}A+[A,e^{-ibB}]\right)|a> = \left(e^{-ibB}A-ib[A,B]e^{-ibB}\right)|a> = (a+b)e^{-ibB}|a>[/tex]

    In other words

    [tex]e^{-ibB}|a>=|a+b>[/tex]

    and this is true for every [tex]a,b\in\mathbb{R}[/tex]. Thus A and B have continuous spectrum, and you can't normalize [tex]<a|a>=1[/tex]. Rather

    [tex]<a|a'>=\delta(a-a')[/tex]

    So you get something like

    [tex]-ia<a|B|a'>=\delta(a-a')[/tex]

    and this is not 1 = 0 when you put a = a'.
     
    Last edited: Aug 6, 2010
  4. Aug 6, 2010 #3
    No, it's far simpler than that. If [itex]A \lvert\psi\rangle = a \lvert\psi\rangle[/itex], and [itex]B \lvert\psi\rangle = b \lvert\psi\rangle[/itex], then
    [tex][A,B] \lvert\psi\rangle = AB \lvert\psi\rangle - BA \lvert\psi\rangle = ab \lvert\psi\rangle - ba \lvert\psi\rangle = 0 \ne i\hbar I \lvert\psi\rangle.[/tex]
    Hence [itex][A,B] \ne i\hbar I[/itex].
     
  5. Aug 6, 2010 #4
    You simply proved that, since A and B do not commute, they cannot be simultaneously diagonalized. Petar Mali didn't use the fact that psi is also an eigenvector of B.
     
  6. Aug 6, 2010 #5
    If [tex]
    [\hat{A},\hat{B}]=i\hbar\hat{I}
    [/tex]

    then [tex]
    \langle \psi|\psi\rangle\ne1
    [/tex]

    because operators [tex]\hat{A}, \hat{B}[/tex] have continual spectrum!
     
  7. Aug 6, 2010 #6
    Yeah we solved the arcane! :biggrin:
     
  8. Aug 6, 2010 #7

    George Jones

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  9. Aug 6, 2010 #8

    George Jones

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  10. Aug 6, 2010 #9
    [tex]<A><B>\ne<AB>[/tex]
     
  11. Aug 6, 2010 #10
    Do you know how to prove this claim?
     
  12. Aug 6, 2010 #11
    If

    [tex]
    A = \left(\begin{array}{cccc}
    A_{11} & 0 & 0 & \cdots \\
    0 & A_{22} & 0 & \cdots \\
    0 & 0 & A_{33} & \cdots \\
    \vdots & \vdots & \vdots & \ddots \\
    \end{array}\right)
    [/tex]

    and

    [tex]
    B = \left(\begin{array}{cccc}
    B_{11} & B_{12} & B_{13} & \cdots \\
    B_{21} & B_{22} & B_{23} & \cdots \\
    B_{31} & B_{32} & B_{33} & \cdots \\
    \vdots & \vdots & \vdots & \ddots \\
    \end{array}\right)
    [/tex]

    then

    [tex]
    AB-BA = \left(\begin{array}{cccc}
    0 & B_{12}(A_{11} - A_{22}) & B_{13}(A_{11}-A_{33}) & \cdots \\
    B_{21}(A_{22}-A_{11}) & 0 & B_{23}(A_{22}-A_{33}) & \cdots \\
    B_{31}(A_{33}-A_{11}) & B_{32}(A_{33}-A_{22}) & 0 & \cdots \\
    \vdots & \vdots & \vdots & \ddots \\
    \end{array}\right)
    [/tex]

    so it is impossible to find [tex]A,B\in\mathbb{C}^{\mathbb{N}\times\mathbb{N}}[/tex] such that [itex]A[/itex] is diagonalizable and [itex]AB-BA=1[/itex].

    On the other hand [itex]M_x(-\partial_x) - (-\partial_x)M_x = 1[/itex], where [itex]\partial_x[/itex] is the differentation operator and [itex]M_x[/itex] is the multiplication operator [itex](M_x\psi)(x)=x\psi(x)[/itex]. The claim about continuous spectrum and non-existence of (normalizable) eigenvectors seems to be correct.
     
  13. Aug 6, 2010 #12
    The problem is not the continuous spectrum, although what you (and jostpuur) wrote is certainly right. The (bigger) problem is that two operators which don't commute don't have the same eigenvectors. The line above is impossible unless A and B commute. This whole exercise could be done with a discrete spectrum with a few minor changes. Or am I missing something?
     
  14. Aug 6, 2010 #13
    Petar Mali didn't mean that the [itex]|\psi\rangle[/itex] would be the same on both of those lines.

    Take a closer look at the steps leading to the contradiction:

    It is only assumed that [itex]A|\psi\rangle=a|\psi\rangle[/itex], and nothing about [itex]B[/itex].
     
  15. Aug 6, 2010 #14

    George Jones

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    Here is a proof of something close to this.

    Set [itex]\hbar = 1[/itex] and assume

    [tex]AB - BA = iI.[/tex]

    Multiplying the commutation relation by [itex]B[/itex] and reaaranging gives

    [tex]
    \begin{equation*}
    \begin{split}
    AB - BA &= iI \\
    AB^2 - BAB &= iB \\
    AB^2 - B \left( BA + iI \right) &= iB \\
    AB^2 - B^2 A &= 2iB.
    \end{split}
    \end{equation*}
    [/tex]

    By induction,

    [tex]AB^n - B^n A = niB^{n-1}[/tex]

    for every positive integer [itex]n[/itex]. Consequently,

    [tex]
    \begin{equation*}
    \begin{split}
    n\left\| B \right\|^{n-1} &=\left\| AB^n - B^n A \right\| \\
    &\leq 2\left\| A \right\| \left\| B \right\|^n\\
    n &\leq 2\left\| A \right\| \left\| B \right\| .
    \end{split}
    \end{equation*}
    [/tex]

    Because this is true for every [itex]n[/itex], at least one of [itex]A[/itex] and [itex]B[/itex] must be unbounded. Say it is [itex]A[/itex]. Then, by the Hellinger-Toeplitz theorem, if [itex]A[/itex] is self-adjoint, the domain of physical observable [itex]A[/itex] cannot be all of Hilbert space!
     
  16. Aug 6, 2010 #15
    I think I pretty much answered my own question already. My calculation proves that if [itex]A[/itex] can be diagonalized so that it has eigenvectors in the Hilbert space, then [itex][A,B]\neq 1[/itex].

    At this moment I'm not 100% sure that the non-existence of eigenvectors in the Hilbert space would be the same thing as the spectrum being continuous, though. Rigorous definitions should be recalled from somewhere...

    George Jones, notice that you are proving a different thing. You proved unboundedness, but for example

    [tex]
    A = \left(\begin{array}{cccc}
    1 & 0 & 0 & \cdots \\
    0 & 2 & 0 & \cdots \\
    0 & 0 & 3 & \cdots \\
    \vdots & \vdots & \vdots & \ddots \\
    \end{array}\right)
    [/tex]

    is diagonalized so that it has eigenvectors in the Hilbert space, despite the fact that it is unbounded.
     
  17. Aug 6, 2010 #16
    If you ignore the inconsistent and unnecessary assertion that [tex]|\psi\rangle[/tex] is an eigenvector of B, then there's no reason that [tex]|\psi\rangle[/tex] can't be normalizable. I think what the "proof" tells you, and what you'll find if you plug in some real operators with this property, is that the expression

    [tex]\langle \psi|\hat{B}|\psi\rangle-\langle \psi|\hat{B}|\psi\rangle[/tex]

    is an indeterminate expression like [tex]\infty - \infty[/tex] which can't be simplified down to 0.
     
  18. Aug 6, 2010 #17
    I bet you can't find two hermitian operators with discrete spectrum, in any Hilbert space you like, with the commutation relation above...
     
  19. Aug 6, 2010 #18
    I repeat the original paradox with a "concrete" representation.

    Operators are [itex]M_x[/itex] and [itex]-i\hbar\partial_x[/itex]. Then [itex][M_x,-i\hbar\partial_x]=i\hbar\;\textrm{id}[/itex].

    We'll pick some eigenvector of [itex]M_x[/itex]. [itex]|\psi\rangle[/itex]! So that [itex]M_x|\psi\rangle = x_0|\psi\rangle[/itex] and [itex]\langle\psi|\psi\rangle=1[/itex]. In position representation this means

    [tex]
    \psi(x) = \frac{1}{\sqrt{\delta(0)}}\delta(x-x_0)
    [/tex]

    :wink:

    Let's sandwich the commutator between brackets then!

    [tex]
    \langle\psi| [M_x,-i\hbar\partial_x] |\psi\rangle = \int\limits_{-\infty}^{\infty} \psi(x)^*
    \big( M_x (-i\hbar\partial_x) - (-i\hbar\partial_x)M_x\big)\psi(x) dx
    [/tex]
    [tex]
    = \frac{-i\hbar}{\delta(0)}\int\limits_{-\infty}^{\infty} \delta(x-x_0)\big(x \partial_x\delta(x-x_0)
    -\partial_x(x\delta(x-x_0))\big) dx = \cdots
    [/tex]

    Now there is two different ways to continue. We can substitute

    [tex]
    \partial_x(x\delta(x-x_0)) = x\partial_x\delta(x-x_0) + \delta(x-x_0)
    [/tex]

    and then the calculation comes to end like this

    [tex]
    \cdots = \frac{-i\hbar}{\delta(0)}\int\limits_{-\infty}^{\infty}\delta(x-x_0)\big(-\delta(x-x_0)\big)dx = i\hbar
    [/tex]

    But we can also substitute

    [tex]
    x\delta(x-x_0) = x_0\delta(x-x_0)
    [/tex]

    Then the calculation comes to end like this

    [tex]
    \cdots = \frac{-i\hbar}{\delta(0)} x_0\int\limits_{-\infty}^{\infty} \delta(x-x_0)\big(\partial_x\delta(x-x_0)
    -\partial_x\delta(x-x_0)\big)dx = 0
    [/tex]

    So [itex]i\hbar = 0[/itex]. Did it look rigorous? :biggrin:
     
  20. Aug 6, 2010 #19
    Ah; you're right; for some reason I thought that e.g. position eigenstates were normalizable in the desired way but of course they aren't.
     
  21. Aug 6, 2010 #20
    Good point. This by the way is nice proof that the commutation relation [tex][A,B][/tex] [tex]= \alpha I[/tex] is impossible in the discrete case where operators can be diagonalized.
     
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