0
$\begingroup$

Suppose I have a wave function $\psi $ we express it in a continous states as

$$\psi= \int_{-\infty}^{\infty} dxC (x)\rvert x\rangle = \int_{-\infty}^{\infty} dx\rvert x\rangle \langle x \rvert \psi (x)$$

This can be expanded in Riemann sum as

$$   \psi= \lim_{\Delta x\to 0} \sum_{i=-\infty}^{\infty} \Delta xC (x_i)\rvert x_i\rangle.$$

This is not symmetric with the expression for $\psi $ in other discrete Hilbert spaces.

Which is

$$\psi= \sum_{i=-\infty}^{\infty} C (\phi_i)\rvert \phi_i\rangle,$$ where this $i$ takes discrete values.

Symmetry is lost due to the appearance of the factor $\Delta x$.

Can any one please help me to solve this paradox?

$\endgroup$
2
$\begingroup$

There is no paradox. If you have the continuous relation as $$| \psi \rangle = \int_{-\infty}^\infty dx \ \psi(x) |x \rangle = \lim_{\Delta x \rightarrow 0} \sum_i \Delta x \psi(x_i) |x_i \rangle,$$ to draw an analogy with the discrete basis expansion $\sum_i C_i |e_i \rangle$, you need to realize that the analogy goes as $C_i \leftrightarrow \Delta x \psi(x_i)$, not $C_i \leftrightarrow \psi(x_i)$.

For a better intuition, think of it this way, the continuous basis has infinitely many more basis vectors than a discrete basis, meaning that the component of any state $|\psi \rangle$ along any of these basis vectors $|x\rangle$ has to be infinitesimal for the expansion to make sense. So $\psi(x) =\langle x | \psi \rangle$ is more of a component density (for lack of a better term).

This is similar to any other part of physics where densities are involved. For example, the total mass of a system of discrete point masses is: $$M = \sum_i m_i,$$ whereas for a continuous mass distribution it's $$M = \int_\mathcal D d^3\mathbf x \rho(\mathbf x) = \lim_{\Delta V \rightarrow 0} \sum _{i} \Delta V \rho(\mathbf{x}_i).$$ Here you can make the analogy $m_i \leftrightarrow \Delta V \rho(\mathbf x_i)$, i.e. the mass of an infinitesimal volume element in the continuous distribution is $\Delta V \rho(\mathbf x_i)$ (and not $\rho$ itself).

Similarly, the component of $|\psi \rangle$ along the basis vector $|x \rangle$ is $\Delta x \psi(x)$, with $\psi(x)$ playing the role of a "density".

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

There is no paradox. You're comparing two different formulas: one is an approximation that applies to a continuous basis of states $|x\rangle$, the other is an exact formula that applies to a discrete Hilbert space. Just because two formulas look different doesn't mean one of them is wrong.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you derive this formulae for continuous basis of states @Hans Moleman $\endgroup$ – ROBIN RAJ Dec 18 '19 at 14:07

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.