Particle in a box: Finding <T> of an electron given a wave function

Mayhem
Homework Statement:
An electron in a carbon nanotube of length ##L## is described by the wavefunction ##\psi(x) = (2/L)^{1/2}\sin{(\pi x/L)}##. Compute the expectation value of the kinetic energy of the electron.
Relevant Equations:
##\left \langle \Omega \right \rangle = \int \psi^* \hat{\Omega}\psi ~d\tau##
If ##\hat{T} = -\frac{\hbar}{2m}\frac{\mathrm{d^2} }{\mathrm{d} x^2}##, then the expectation value of the kinetic energy should be given as:
\begin{align*} \left \langle T \right \rangle &= \int_{0}^{L} \sqrt{\frac{2}{L}} \sin{\left(\frac{\pi x}{L}\right)} \hat{T}\sqrt{\frac{2}{L}}\sin{\left(\frac{\pi x}{L}\right)} dx \\ &= \frac{-\hbar^2}{mL} \int_{0}^{L} \sin{\left(\frac{\pi x}{L}\right)} \frac{\mathrm{d^2} }{\mathrm{d} x^2} \sin{\left(\frac{\pi x}{L}\right)} dx \\ &= \frac{\pi^2 \hbar^2}{mL^3} \int_{0}^{L} \sin^2{\left(\frac{\pi x}{L}\right)} dx \\ &=\frac{h^2}{8mL^2} \end{align*}
Are my calculations correct?

To solve the integral ##\int \sin^2(f(x))) dx## I use the half angle identity for ##\sin(x)##.

bob012345, vanhees71 and Delta2

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Looks good, although generally ##\hbar## is used in QM.

vanhees71 and Mayhem
Mayhem
Looks good, although generally ##\hbar## is used in QM.
Wouldn't the ##\hbar## reduce to ##h## in the last step, though?

vanhees71
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Wouldn't the ##\hbar## reduce to ##h## in the last step, though?
Yes, but it's not conventional to switch between ##\hbar## and ##h## depending on whether you can cancel ##2\pi##. Most QM texts will stick with ##\hbar## by convention.

vanhees71 and Mayhem
Mayhem
Yes, but it's not conventional to switch between ##\hbar## and ##h## depending on whether you can cancel ##2\pi##. Most QM texts will stick with ##\hbar## by convention.
Confusing, but who am I to tell quantum physicists that they are wrong.

Twigg, vanhees71 and Delta2
Mayhem
Just reviewed my book: it actually uses ##h## when ##\hbar## reduces as shown. I suppose I'm right in this particular case.

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To solve the integral ##\int \sin^2(f(x))) dx## I use the half angle identity for ##\sin(x)##.
An important point of this problem is to realize that, in this case, the expectation value of the kinetic energy is equal to the energy of the ground state because the particle is in an energy eigenstate and the potential energy is zero. There is no need to integrate.

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Just reviewed my book: it actually uses ##h## when ##\hbar## reduces as shown. I suppose I'm right in this particular case.
Modern texts and tracts use ##\hbar## exclusively I am told.
The reason is probably having to be different.
Same applies when all of a sudden ## m_0 ## in relativity is taboo. So we said goodbye to ## E = mc^2 ## for moving particles. It was good enough for Feynman but not for us.

Mayhem
Modern texts and tracts use ##\hbar## exclusively I am told.
The reason is probably having to be different.
Same applies when all of a sudden ## m_0 ## in relativity is taboo. So we said goodbye to ## E = mc^2 ## for moving particles. It was good enough for Feynman but not for us.
IMO, it is good to keep in mind that ##\hbar## is used interchangeably with ##h##, but I honestly err on the side of unambiguous notation.

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To elaborate, perhaps a bit unnecessarily, on post 7:
Given ## \psi(x) = (2/L)^{1/2} sin(\pi x/L) ## in a box of length L,
The energy eigenstate n (collapsed wave function) of a particle in that box is
## \psi(x)_{E_n} = (2/L)^{1/2}~sin (n\pi x /L) , 0~<=x~<=L ##
and ## \psi(x) = \Sigma_n~ A_n~ \psi_{E_n} (x) ##
with ##A_n = \int_0^L ~\psi^*_{E_n} (x) ~ \psi(x) ~dx ##
= 1, n=1 and = 0 for any other n
so there is only one allowed energy state with ## E = \hbar^2 \pi^2/2mL^2 ##
(from the eigenstate Schroedinger equation).

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