I am considering a particle of mass m in a symmetric infinite square well of width a in the fundamental state.

$$V(x)= \begin{cases} 0 & \mbox{$|x|<\frac{a}{2}$} \\ \infty & \mbox{otherwise} \end{cases}$$

I want to know what values are obtained from the measurement of energy $E$, position $x$ and impulse $p$ and the corresponding probabilities.

So I did:

$$\psi_{n=1}(x)=\sqrt{\frac{2}{a}}\cos\left(\frac{\pi}{a}x\right)$$ $$E_{1}=\frac{\hbar^2\pi^2}{2ma^2} \,\,\,\,\,\,\,\,\,\,\ P(E_1)=100\%$$ I can not calculate the eigenvalues of the operator position that I imagine is a continuous set of values in the interval $\left[ -\frac{a}{2},\frac{a}{2} \right]$.

Those that I have considered up until now are the eigenfunctions of the Hamiltonian and not of the position operator so I do not think it makes sense:

$$\hat{x}| 1\rangle=\sqrt{\frac{2}{a}}\int_{-\frac{a}{2}}^{\frac{a}{2}}x\cos\left(\frac{\pi}{a}x\right)dx=0$$ However I do not know how to do it or even for the momentum. I also have a suggestion that to calculate the probability of the momentum it is sufficient to calculate the wave function in the space of the impulses, but I honestly can not understand it

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    $\begingroup$ The equation in the title is wrong. The right-hand side is the expectation value, $\langle 1 | \hat{x} | 1 \rangle$, not $\hat{x} |1 \rangle$. $\endgroup$ – knzhou Jun 17 '18 at 17:28
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    $\begingroup$ Is not $\langle 1| \hat{x} | 1\rangle =\frac{2}{a}\int_{-\frac{a}{2}}^{\frac{a}{2}}x\cos^2\left(\frac{\pi}{a}x\right)dx$ $\endgroup$ – Stefano Barone Jun 17 '18 at 17:33
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    $\begingroup$ Oh yeah, you're right. The point is, what you have is not $\hat{x} |1 \rangle$ at all. $\endgroup$ – knzhou Jun 17 '18 at 17:35
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    $\begingroup$ An operator acting on a state gets you a state, not a number. If you wanted the expectation value, you would have to integrate x cos^2(…) and the answer would be zero by symmetry. $\endgroup$ – Bert Barrois Jun 17 '18 at 18:03
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    $\begingroup$ You claim to be trying to evaluate the eigenvalues of x but you are using the eigenstates of H, this is not even valid. The eigenvalues of the operator x are x, the eigen functions are Dirac delta functions. $\endgroup$ – ggcg Jun 17 '18 at 20:59

@knzhou already indicated the perfect appositeness of your title.

You apply the definitions of your text, as the chthonian pundit suggests, $$\hat{x}| 1\rangle= \hat{x}\int dx ~|x\rangle\langle x| 1\rangle= \sqrt{\frac{2}{a}}\int_{-\frac{a}{2}}^{\frac{a}{2}}dx~~\left( x\cos\left(\frac{\pi}{a} x\right)\right ) ~~~|x\rangle . $$ The position operator just multiplies the wavefunction by x for every position x, but, of course, the state $|1\rangle$ is an x -integral of eigenstates of this operator with x -dependent coefficients, the very heart of Dirac's Bra-Ket formalism.

So, then, $$\langle 1|\hat{x}| 1\rangle= \sqrt{\frac{2}{a}}\int_{-\frac{a}{2}}^{\frac{a}{2}}dx~~\left( x\cos\left(\frac{\pi}{a} x\right)\right ) ~\langle 1|x\rangle = \frac{2}{a} \int_{-\frac{a}{2}}^{\frac{a}{2}}dx~~\left( x\cos^2\left(\frac{\pi}{a} x\right)\right ) =0 , $$ $\langle 1|\hat{x}^2| 1\rangle= a^2(1/12- 1/2\pi^2) $, etc.

You clearly naively calculated the eigenvalue of $\hat {p}^2$, since you have the eigenvalue of the energy; however, a subtlety prevents $|1\rangle$ from being an eigenstate of of $\hat p$, as the symmetric wave packet $\langle p|1\rangle$ is not infinitely sharp. In any case, you may avoid all this; confirm directly that $\langle 1| \hat{p}|1\rangle =0$, which might not be surprising; and, of course, $\langle 1| \hat{p}^2|1\rangle =\hbar^2\pi^2/a^2$. This is just 10% off saturating the uncertainty principle bound!

  • Small footnote to be avoided until the fourth reading. The moot self-adjointness of this $\hat p$ is fully discussed in here.

    • Terms of use agreement, so in the smallest print possible, to only read with mental lawsuits in mind. In his book, Dirac defines $|x\rangle$ via the ''standard ket'' which, up to a normalization, is but the translationally invariant momentum eigenstate $|\varpi\rangle=\lim_{p\to 0} |p\rangle$ in the momentum representation, i.e., $\hat{p}|\varpi\rangle=0$. Consequently, the corresponding wavefunction is a constant, $\langle x|\varpi\rangle \sqrt{2\pi \hbar}=1$. The definition is then $~~~|x\rangle= \delta(\hat{x}-x) |\varpi\rangle \sqrt{2\pi \hbar}$.
  • $\begingroup$ Chthonian! Now that's something I've never been called before! $\endgroup$ – knzhou Jun 17 '18 at 20:06
  • $\begingroup$ @knzhou misdirected adjective landing... I was reading the last comment.... will redress. $\endgroup$ – Cosmas Zachos Jun 17 '18 at 20:08
  • $\begingroup$ "You clearly calculated the eigenvalue of p^, since you have the eigenvalue of the energy, effectively its square, no? " No. The momentum is ill defined in the infinite square well. The momentum operator is not self adjoint. $\endgroup$ – thermomagnetic condensed boson Jun 17 '18 at 20:09
  • $\begingroup$ @ofthe... fair enough, but this is not the subtlety the OP is reaching for. The operatively word is "naively"... $\endgroup$ – Cosmas Zachos Jun 17 '18 at 20:12
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    $\begingroup$ Ach, Stefano, the integral is over both the function, and the labels of the eigenstates |x> --they are part of the integrand too. Think of the sum over all dummy xs so a sum over |x>s, each with its peculiar coefficient , the function in the big parenthesis. This is a formal integral you can't do to get a number. You get a sum of states. $\endgroup$ – Cosmas Zachos Jun 17 '18 at 20:22

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