Quantum Mechanics-I

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Chapter#4: One Dimensional Problems

Team Quanta presents all possible short questions of BS Physics book Quantum Mechanics-I’Chapter#4: One Dimensional Problems.

Q.4.1 Describe various properties of one dimensional motion.

Answer: Properties of one dimensional motion are listed below:

  • The spectrum of bound states is discrete and non-degenerate.
  • The bound state Eigen functions in an even potential have definition parity.

Q.4.2 Explain degenerate spectrum.

Answer: If spectrum of Hamiltonian corresponding to a symmetric potential is degenerate, its Eigen states are expressed only in terms of even and odd states. It means that Eigen states do not have definite parity.

Q.4.3 what is meant by potential step?

Answer: A simple problem consisting of a particle that is free anywhere, but beyond a particular point say x = 0, the potential repulsive becomes repulsive or attractive; a potential of this type is called a potential step.

    \[\left{\begin{matrix} 0 & for \ \ x<0 \\ V_0 & for \ \ x>0 \end{matrix}\right.\]

Q.4.4 Evaluate zero point energy of harmonic oscillator using uncertainty principle.

Answer: Energy of harmonic oscillator is given by,

    \[E=\frac{1}{2}mV^2+\frac{1}{2}kx^2=\frac{m^2V^2}{2m}+\frac{1}{2}m\omega^2x^2=\frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2\]

Since motion of particle is confined to a region

    \[\frac{-L}{2}\le x\le\frac{L}{2},\]

, so may take ∆x ~ L By uncertainty principle, minimum values of momentum of particles is,

    \[p\approx \frac{\hbar}{2 \Delta x} ~ \frac{\hbar}{2L}\]

So,

    \[E~\frac{1}{2m}\left ( \frac{\hbar}{2L} \right ) +\frac{1}{2}m\omega^2L^2\]

To find minimum energy, put

    \[\frac{dE}{dL}=0\Rightarrow- \frac{\hbar}{4mL^3} +m\omega^2L=0\ \ \ \Rightarrow L=\sqrt{\frac{\hbar}{2m\omega}}\]

Hence,

    \[E_{min}=\frac{1}{2m}\left(\frac{\hbar}{2\sqrt{\frac{\hbar}{2m\omega}}}\right)+\frac{1}{2}m\omega^2\left(\sqrt{\frac{\hbar}{2m\omega}}\right)^2\]

    \[\Rightarrow\ E_{min}=\frac{\hbar\omega}{2}\]

This is called zero point energy of harmonic oscillator.

Q.4.5 Show that zero point energy of harmonic oscillator is consistent with uncertainty principle.

Answer: Energy of harmonic oscillator is given by,

    \[E=\frac{1}{2}mV^2+\frac{1}{2}kx^2=\frac{m^2V^2}{2m}+\frac{1}{2}kx^2=\frac{p^2}{2m}+\frac{1}{2}kx^2\]

According to classical mechanics, least value of energy is zero i.e.

    \[\frac{p^2}{2m}+\frac{1}{2}kx^2=0\ \ \ \ \ \ \Rightarrow p^2=0,\ \ \ \ x^2=0\ \ \ \ \ \Rightarrow<\ p^2\geq\ =0,\ \ \ \ \ <\ x^2\geq\ =0\]

Also for harmonic oscillator x=0, p=0.

∆x = 0 & ∆p = 0

Thus position and momentum of particle can be determined simultaneously, which is contradiction to uncertainty principle. Hence minimum energy of harmonic oscillator cannot be zero, which is in accordance with uncertainty principle.

Q.4.7 An electron is moving freely inside one dimensional infinite potential box with walls at  x=0 and x=L If electron is initially in ground state(n =1) of box and we suddenly quadrupole the size of box, evaluate probability of finding the electron in ground state of new box.

Answer: Wave function of particle, initially is

    \[\mathrm{\Psi}_1\left(x\right)=\sqrt{\frac{2}{L}}\sin{\left(\frac{\pi x}{L}\right)}\]

In new box, x = 0 & x = 4L, wave function is,

    \[\mathrm{\Psi}_1^\prime\left(x\right)=\sqrt{\frac{1}{2L}}sin\left(\frac{\pi x}{4L}\right)\]

Probability of finding the electron in ground state of new box is,

    \[P=\left|\int_{0}^{L}{\mathrm{\Psi}<em>1^{\prime^\ast}\left(x\right)\mathrm{\Psi}_1\left(x\right)dx}\right|^2=\left|\int</em>{0}^{L}\sqrt{\frac{1}{2L}}\ sin\left(\frac{\pi x}{4L}\right)\sqrt{\frac{2}{L}}\ sin\left(\frac{\pi x}{L}\right)dx\ \right|^2=0.058=5.8%\]

Q.4.9 Assuming the potential seen by a neutron in a nucleus to be schematically represented by a one dimensional, infinite rigid walls potential of length 10fm, estimate minimum kinetic energy of neutron.

Answer: Energy of a particle in one dimensional box is,

    \[E_n=\frac{n^2h^2}{8mL^2}\]

Assuming neutron to be a non-relativistic particle,

    \[E_{min}=\frac{h^2}{8m_nL^2}=\frac{h^2c^2}{8\left(m_nc^2\right)L^2}=2.04Mev\]

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