1
$\begingroup$

I have a question about some formulars our professor wrote on the black board.

Let $\hat{Q}_{1},...,\hat{Q}_{N}$ be operators, which are a CSCO. We know now that there exists a set of eigenvectors $\{|q\rangle\}$, which are simultanously eigenstates of all commuting operators:

$\hat{Q}_{n}|q\rangle = q_{n}|q\rangle$

Now we learnt the following relations:

(1) $\langle q^{\prime}|q\rangle = \prod_{n=1}^{N}\delta(q^{\prime}_{n} - q_{n})$

(2) $\mathbb{1} = \int\,\big (\prod_{n=1}^{N}\mathrm{d}q_{n}\big )|q\rangle\langle q|$

I don`t unterstand this relations. First of all, why does he use the delta-distribution instead of the Kronecker-delta? Secondly, why does the completness relation (2) looks like that? Shouldn't it be

$\mathbb{1} = \int\,\mathrm{d}q |q\rangle\langle q|$?

Thanks!

$\endgroup$
1
$\begingroup$

It might help to apply these relations to a familiar example such as the cartesian components of the 3D position operator, which forms a complete set of commuting observables.

In the 1st relation, I think he is using the Dirac delta as opposed to the Kronecker because he is assuming the eigenvalues are continuous as opposed to discrete. If we are in 3D space, then the 1st relation would read:

$$ \langle q'|q\rangle=\delta(x'-x)\delta(y'-y)\delta(z'-z),$$

which is sensible since position is a continuous variable.

In the 2nd relation, the integral is simply going over the entire dimensionality of space. Again for position, we have

$$1 = \int dx\,dy\,dz\,|q\rangle\langle q|,$$

which is also sensible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.