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Monday, 12 December 2016
Problem set 39
A state of a system with spherically symmetric potential has zero uncertainty in simultaneous measurement of operator L_x
and L_y. Which of the following statement is true?
The state must be l=0 state
Such a state can never exist as operators L_x and L_y do not commute
The state has l=1 with m=0
The state cannot be an eigenstate of L^2 operator
\begin{align*}
L_z\psi_{nlm}&=m\hbar\psi_{nlm} \text{ and } L^2\psi_{nlm}\\
&=l(l+1)\hbar^2\psi_{nlm}
\end{align*}\left < L_z \right > =m\hbar, \left < L_z^2 \right > =m^2\hbar^2\text{ and } \left < L^2 \right > =l(l+1)\hbar^2
The state \psi_{nlm} is symmetric in x and y, so
\begin{align*}
\left < L_x^2 \right >&=\left < L_y^2 \right > \\
&=\frac{\left(\left < L^2 \right >-\left < L_z^2 \right >\right)}{2}\\
&=\frac{\left(l(l+1)-m^2\right)\hbar^2}{2}
\end{align*}
It can be verified that \left < L_x \right >=\left < L_y \right >=0. Using, \Delta L_i=\sqrt{\left < L_i^2 \right >-\left < L_i \right >^2}, we have
\Delta L_x=\Delta L_y=\sqrt{\frac{\left(l(l+1)-m^2\right)\hbar^2}{2}}\Delta L_x\Delta L_y=\frac{\left(l(l+1)-m^2\right)\hbar^2}{2}
From this relation one can see that uncertainty relation will be zero when l=0 and henc, m=0.
Hence, answer is (A)
The wave function for identical fermions is antisymmetric under particle exchange. Which of the following is a consequence of
this property?
Heisenberg's uncertainty principle
Bohr correspondence principle
Bose-Einstein condensation
Pauli exclusion principle
(D) Pauli exclusion principle
The spin part of two electron wave function is described as a triplet state. The space part of the wave function is given by
(\psi_1 and \psi_2 are two different functions):
\psi_1(r_1)\psi_2(r_2)
\psi_1(r_1)\psi_2(r_2)-\psi_2(r_1)\psi_1(r_2)
\psi_1(r_1)\psi_2(r_2)+\psi_2(r_1)\psi_1(r_2)
\psi_1(r_1)\psi_1(r_2)+\psi_2(r_1)\psi_1(r_2)
Triplet state means S=s_1+s_2=1 and M_s=-1,0,+1.
For M_s=-1 we must have m_{s1}=-\frac{1}{2} and m_{s2}=-\frac{1}{2}
For M_s=0 we must have either m_{s1}=\frac{1}{2} and m_{s2}=-\frac{1}{2} or m_{s1}=-\frac{1}{2} and m_{s2}=\frac{1}{2}
For M_s=+1 we must have m_{s1}=+\frac{1}{2} and m_{s2}=+\frac{1}{2}
Hence, we have three possible spin eigenfunctions
\phi\left(+\frac{1}{2},+\frac{1}{2}\right)\frac{1}{\sqrt{2}}\left[\phi\left(+\frac{1}{2},-\frac{1}{2}\right)+\phi\left(-\frac{1}{2},+\frac{1}{2}\right)\right]\phi\left(-\frac{1}{2},-\frac{1}{2}\right)
All these functions are symmetric. However, total wavefunction must be antisymmetric. Hence, space part must be antisymmetric.
Hence, answer is (B) i.e. \psi_1(r_1)\psi_2(r_2)-\psi_2(r_1)\psi_1(r_2)
A transition in which one photon is radiated by the electron in a hydrogen atom when the electron's wave function changes from \psi_1 to \psi_2 is forbidden if \psi_1 and \psi_2
A. FALSE: It's not related to the selection rules.
B. TRUE: If both initial and final states have spherically symmetrical wave functions, then they have the same angular momentum. l = 0 \rightarrow l = 0 is forbidden as \Delta l=0 in this case.
C. FALSE: In any transition, eigenstates should always be mutually orthogonal.
D. FALSE: Eigenstates zero at the center \Rightarrow l> 0 could change, for example from 3 to 2. This is allowed.
Hence, answer is (B)
The puzzle of magic numbers for nuclei was resolve by :
introducing hard-core potential
introducing Yukawa potential for shell model
introducing tensor character to nuclear force
introducing spin-orbit part in the nuclear potential
(D) introducing spin-orbit part in the nuclear potential
Good questions
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