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

In summary, the ground state of a particle in a box of length L is energy eigenstate with the particle in the lowest energy state.
  • #1
Mayhem
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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)##.
 
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  • #2
Looks good, although generally ##\hbar## is used in QM.
 
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  • #3
PeroK said:
Looks good, although generally ##\hbar## is used in QM.
Wouldn't the ##\hbar## reduce to ##h## in the last step, though?
 
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  • #4
Mayhem said:
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.
 
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  • #5
PeroK said:
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.
 
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  • #6
Just reviewed my book: it actually uses ##h## when ##\hbar## reduces as shown. I suppose I'm right in this particular case.
 
  • #7
Mayhem said:
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.
 
  • #8
Mayhem said:
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. :frown:
 
  • #9
rude man said:
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. :frown:
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.
 
  • #10
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).
 
Last edited:

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

1. What is a particle in a box?

A particle in a box refers to a theoretical model used to understand the behavior of a particle confined within a finite region. In this model, the particle is assumed to be completely confined within the boundaries of the box and cannot escape.

2. How is the wave function of an electron in a box determined?

The wave function of an electron in a box is determined by solving the Schrödinger equation, which is a mathematical equation that describes the behavior of quantum particles. The solution to this equation gives the wave function, which describes the probability of finding the electron at a certain position within the box.

3. What does represent in the equation for finding the average energy of an electron?

In this context, represents the kinetic energy of the electron. This is the energy that the electron possesses due to its motion within the box.

4. How is the average energy of an electron in a box calculated?

The average energy of an electron in a box is calculated by finding the expectation value of the Hamiltonian operator, which represents the total energy of the system. This is done by integrating the product of the wave function and the Hamiltonian operator over the entire region of the box.

5. What is the significance of finding the average energy of an electron in a box?

Finding the average energy of an electron in a box allows us to understand the behavior of quantum particles in confined spaces. It also helps us to make predictions about the behavior of other particles in similar systems, and is an important concept in the field of quantum mechanics.

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