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Consider infinite potential well i.e. Hilbert space $L^2 \bigl([0,1]\bigr)$. Next we consider subset $$D_\theta = \left\{ \psi \in L^2 \bigl([0,1]\bigr) | \; \psi \; \text{is absolutely continuos and } \psi (0) = e^{i \theta} \psi (1) \right\} $$ on which we define operator $p_\theta = i \frac{\partial}{\partial x}$. Denote by $\psi_{n, \theta} = e^{i (2\pi n - \theta) x}, \; n \in \mathbb{Z}$ eigenfunctions of $p_\theta$ to the eigenvalues $\lambda_{n, \theta} = 2\pi - \theta$. Now move on to commutator $[x,p_\theta]$. Typically it would be equal to $-i$, but one can write: $$\langle \psi_n | [x, p] \psi_n \rangle = \langle \psi_n | (xp - p x) \psi_n \rangle = \lambda_n \langle \psi_n | (x-x) \psi_n \rangle = 0 \neq -i \langle \psi_n | \psi_n \rangle = -i$$

My question is: how one should cope with uncertainty principle in infinite potential well?

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How did you do the third step? If you have $p$ acting leftward, you should realize that $\langle \psi_{n}|$ is the complex conjugate of $|\psi_{n}\rangle$ –  Jerry Schirmer May 14 '12 at 12:48
    
$\langle \psi_n | (xp-px) \psi_n \rangle = -\langle p^* \psi_n | x \psi_n \rangle + \langle \psi_n | x | p \psi_n \rangle = - \lambda \langle \psi_n | x | \psi_n \rangle + \lambda \langle \psi_n | x | \psi_n \rangle$ –  qoqosz May 14 '12 at 14:22
    
The eigenvalue of $p^{*}$ is $-\lambda$ –  Jerry Schirmer May 14 '12 at 14:36
    
But $p$ is selfadjoint - check slu.cz/math/cz/knihovna/ucebni-texty/… page 32 example 1. That's my source. –  qoqosz May 14 '12 at 14:40
    
related: physics.stackexchange.com/questions/10230/… –  user2963 May 14 '12 at 15:23
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1 Answer 1

up vote 5 down vote accepted

The problem in the evaluation is that the function $x \psi_n$ lies outside the domain $D_{\theta}$ since it does not satisfy the boundary condition.

The operator $p{_\theta}$ is not self adjoint on this class of functions. Therefore, the step of its evaluation on the ket is not correct.

The correct way to perform the computation is by integration by parts and in this case, the boundary term at $x = 1$ gives the correct answer.

Please see the following article by: F. Gieres: "Mathematical surprises and Dirac's formalism in quantum mechanics", where examples of similar errors are given. Please see especially, example 5 in page 6 and its solution in page 39, which is very similar to the problem at hand.

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I'm basing on the following article slu.cz/math/cz/knihovna/ucebni-texty/… -- page 32, example 1. Author there states that the momentum operator is selfadjoint. The problem is that $x$ operator in not compatible with boundary condition. Still I'd like to see how to overcome uncertainty problem in well. –  qoqosz May 14 '12 at 14:19
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The Hilbert space $D_{\theta}$ is not stable under the action of the position operator. The solution is allow for operators to act on a much larger space than this Hilbert space, please see page 18 in the reference given in the main text. Now, the operators are just multiplication and derivative,the uncertainty relation is verifiable by elementary calculus. What is wrong to to write is the following: $(p_{\theta}\Psi_1, \Psi_2) = (\Psi_1, p_{\theta}\Psi_2) $ in the case that $\Psi_1 = \psi_n$ and $\Psi_2 = x \psi_n$. You can verify that equality is wrong again by using elementary calculus. –  David Bar Moshe May 14 '12 at 14:57
    
Ok, thank you. Things are getting much clearer for me. If I were to check uncertainty principle for infinite well, while computing $\sigma_x$, $\sigma_p$ and $\left| \langle [p,x] \rangle \right|$ I should just use standard integrals for this expression and it's fine? –  qoqosz May 14 '12 at 15:18
    
Yes, in this case, it is true. Gieres reference emphasizes that care must be taken in handling the Dirac bra and ket formalism in the case of infinite dimensional Hilbert spaces. The operator domains must be taken into account in deciding self adjontness. –  David Bar Moshe May 14 '12 at 15:32
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