Quantum Negative Value For <p^2>

[laser1]

Homework Statement

A particle is described by the equation Psi, which is $5x^2$ if $-L\leq x \leq L$. The normalisation constant is $A=\sqrt {1/10L^5}$. Calculate $<p^2>$.

Relevant Equations

N/A

When I do this I keep getting a negative answer. Why?

My workings: $<O> = \int \psi* O \psi dx$ in general. And $\hat{p}=\frac{\hbar}{i} \frac{d}{dx}$ so $\hat{p^2} = -\hbar^2 \frac{d^2}{dx^2}$... And by plugging in $\Psi$, I get $<p^2>=-\frac{10\hbar^2}{3L^2}$.

Any thoughts on why $<p^2>$ is negative, which isn't possible?

 

[PeroK]

Someone had a similar problem recently with a discontinuous derivative. In fact it was you!

 

[laser1]

Someone had a similar problem recently with a discontinuous derivative. In fact it was you!

ah ya it's 0 if it's outside isn't it so the 2nd derivative isn't continuous :(. Cheers

 

[pines-demon]

Someone had a similar problem recently with a discontinuous derivative. In fact it was you!

I was going to repost my solution but I guess OP already got the my response in the past. I really like PF time-travelling posting feature.

 

[kuruman]

Homework Statement: A particle is described by the equation Psi, which is $5x^2$ if $-L\leq x \leq L$.

I don't think the above assertion is justified. The function, normalized or not, should be a solution of the time-independent Schrodinger equation. However, $$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\left(5x^2\right)=-\frac{\hbar^2}{2m}(10)\neq E \left(5x^2\right).$$

 

[vela]

The function, normalized or not, should be a solution of the time-independent Schrodinger equation.

That's only true for the energy eigenstates. Even a superposition of energy eigenstates with different energies won't satisfy the time-independent Schrödinger equation.

Superpositions, however, will satisfy the time-dependent Schrödinger equation, and you can presumably come up with a superposition that converges to the given wave function at $t=0$.

 

[kuruman]

That's only true for the energy eigenstates. Even a superposition of energy eigenstates with different energies won't satisfy the time-independent Schrödinger equation.

Superpositions, however, will satisfy the time-dependent Schrödinger equation, and you can presumably come up with a superposition that converges to the given wave function at $t=0$.

Fair enough. The statement of the problem says that the particle "is described" by the given unnormalized wavefunction. How can I make sense of what it describes? How do I interpret $$A^2\int_{- L}^{L}\psi^*(x) \psi(x)~dx=1$$ to mean anything other than that the particle can be found in the region $-L\leq x \leq L$?

If that's the case, then this wavefunction must be a superposition of particle-in-the-box eigenstates which all vanish at $~\pm L$ something that $\psi(x)=5x^2$ doesn't do. If that is not the case and the wavefunction extends beyond $x=\pm L$, then how do I interpret the above normalization condition? Are we trying to have our cake and eat it too? Am I missing something?

 

[vela]

It's the magic of the Fourier series. To keep the math simpler, I used a particle in a box from 0 to $L$ and the normalized wave function
$$\psi(x) = \sqrt{80/L} (\tfrac xL-\tfrac12)^2$$ for $0\le x\le L$ and 0 elsewhere. In terms of the energy eigenstates $\phi_m$, the wave function is
$$\psi(x) = \sum_{n~{\rm odd}} \frac{\sqrt{40}\,(n^2 \pi^2 - 8)}{n^3 \pi^3} \phi_n.$$ Here's what it looks like if you sum the first 100 terms:

 

[kuruman]

It's the magic of the Fourier series. To keep the math simpler, I used a particle in a box from 0 to $L$ and the normalized wave function
$$\psi(x) = \sqrt{80/L} (\tfrac xL-\tfrac12)^2$$ for $0\le x\le L$ and 0 elsewhere. In terms of the energy eigenstates $\phi_m$, the wave function is
$$\psi(x) = \sum_{n~{\rm odd}} \frac{\sqrt{40}\,(n^2 \pi^2 - 8)}{n^3 \pi^3} \phi_n.$$ Here's what it looks like if you sum the first 100 terms:
View attachment 359849

That's what I missed. My intuition deceived me and undermined my trust in the magic of the Fourier series to do its thing at the boundaries. In other words, I guessed instead of calculating. Thanks for the plot.