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All Solutions of Type: Quantum Mechanics
 0 Click here to jump to the problem! GR8677 #18 Quantum Mechanics$\Rightarrow$}Schrodinger Equation The problem gives the wave function, wherein the hidden mysteries of the problem are contained. The potential is referenced as $V(x)$, which means that it's time-independent. Thus, the Time-Independent Schrodinger Equation can be used: $-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi+V(x)\psi=E\psi$. The second derivative of the given $\psi$ is $\frac{d^2}{dx^2}\psi=(-b^2+(b^2x)^2)\psi$. Plug that into the TISE, and one gets $-\frac{\hbar^2}{2m}(-b^2+(b^2x)^2)=E-V(x)$ \par Now, plugging in the potential condition $V(0)=0$, one gets, $\frac{\hbar^2}{2m}((b^2x)^2)=E-V(0)=E$. This implies that the term on the left that disappeared from that substitution is the $V(x)$ term. Therefore, one deduces that $V(x)=\frac{\hbar^2}{2m}(b^2x)^2$. \par The right answer would be (B). Click here to jump to the problem!
 1 Click here to jump to the problem! GR8677 #19 Quantum Mechanics$\Rightarrow$}Bohr Theory Recall the Rydberg energy. QED Click here to jump to the problem!
 2 Click here to jump to the problem! GR8677 #27 Quantum Mechanics$\Rightarrow$}Spin Spin explains a lot of things. (A) Remember orbitals? Whether a shell is full or not determines the properties of each column of the periodic table. A full shell has all electron spins paired together, while a partially filled or empty shell doesn't have that. So, spin is definitely in the Periodic Table. (B) The specific heat of metals differs if one calculates it using the Fermi-Dirac or Bose-Einstein distributions; the first is used for fermions and the second for bosons. So, spin plays a role here. (C) The Zeeman effect has to do with splitting caused by spin. (D) The deflection of a moving electron is due to the magnetic field contribution to the Lorentz Force. This is a classical non-spin related phenomenon, on first analysis. This is the best choice. (E) Fine structure has to do with splitting caused by spin. Click here to jump to the problem!
 3 Click here to jump to the problem! GR8677 #28 Quantum Mechanics$\Rightarrow$}Normalization Recall that $\int |\psi(x)|^2 dx=1$ is the condition for a normalized function. $\begin{eqnarray} |\psi|^2&=&\int^{2\pi}_0 |A|^2 d\phi\\ &=&2\pi|A|^2\\ &=&1\\ \Rightarrow A&=&\frac{1}{\sqrt{2\pi}}, \end{eqnarray}$ where the condition $|e^{im\phi}|^2=e^{im\phi}e^{-im\phi}=1$ is used. This is choice (D). Click here to jump to the problem!
 4 Click here to jump to the problem! GR8677 #31 Quantum Mechanics$\Rightarrow$}Photoelectric Effect Perhaps one recalls the hot phrase" stopping potential. The $|eV|$ used in the photoelectric equation is essentially the stopping potential (energy). This memory-recall immediately narrows the choices down to just (A) and (C). Now, to find out if the potential is positive or negative. FYI: The Photoelectric effect equation is basically just conservation of energy. One has $eV+W=hf$, where $eV$ is the kinetic energy of the electron as it accelerates through the medium between the cathode and collector, $W$ is the energy to free the electron from the metal, and $hf$ is the energy given by the light source. Essentially, the energy from the photon first frees the electron from a sort of (valence) electron sea it lives in while in the metal, and then the excess energy propels it from cathode to collector in order to keep the current running. The minimum kinetic energy required to get the current going is the stopping potential (energy) $eV$. (According to the fight analogy below, the photon is the one who starts the fight, socking the electron off and away across the currents...) The electron charge is negative and the potential $V$ must be a negative quantity in order to make $eV$ positive overall. (On the lighter side... The Photoelectric effect is also related to the Compton Effect. The effects can be seen as an arena fight. Think WWWF, but with electrons and photons. In PEE, the photon knocks the electron out, while in CE, the electron retaliates! It knocks the photon off course.) Click here to jump to the problem!
 5 Click here to jump to the problem! GR8677 #32 Quantum Mechanics$\Rightarrow$}Photoelectric Effect (A) The photoelectric effect was derived before formal quantum mechanics and that angular momentum mess came along. Moreover, electron orbits don't really apply to the valance electron sea. (B) This is true, but it doesn't help derive the photoelectric effect. (C) Nah, there's also something weird called electron-electron annihilation. Basically, two electrons crash into each other and a photon is created. (Perhaps some other particles, too.) Also, think of your regular desktop lamp---light is emitted, but the electrons are probably not jumping between orbits. (D) Right. Einstein won the Nobel Prize about a hundred years ago via his proposal that photons have quantized energy $E=h\nu$. (E) As a pure-ideal theory, the photoelectric effect depends on a single photon exciting a single electron. It favors the particle view of light. Choice (E) is out. Click here to jump to the problem!
 6 Click here to jump to the problem! GR8677 #33 Quantum Mechanics$\Rightarrow$}Photoelectric Effect The quantity $W$ is also known as the work function, and it's the minimum amount of work that has to be done to free the electron from the electron sea. If one forgets whether it's the electron or photon that's being moved, then here's a useful mnemonic. Remember that the photoelectric effect has a photon knocking out an electron and the Compton effect has an electron knocking a photon off course. (For the full fight-club analogy, See Prob 32) Click here to jump to the problem!
 7 Click here to jump to the problem! GR8677 #45 Quantum Mechanics$\Rightarrow$}Compton Effect This problem can be solved via the Compton Effect equation $\Delta\lambda=\lambda_c(1-\cos\theta)$, where $\theta$ is the angle of the scattered photon and $\lambda_c=h/(m_ec)$ is the so-called Compton Wavelength. In this case, since the particles are protons, one has $\lambda_c=h/(m_pc)$. Since the photon scatters off at 90 degrees, the equation simplifies to $\Delta\lambda=\lambda_c$. This is the increase in wavelength, as in choice (D). To a certain degree, one can hand-wave this problem via the following method: Recall the de Broglie relation, $p=\frac{h}{\lambda}$, where $\lambda$ is the wavelength, $p$ is the momentum and $h$ is Planck's constant. The momentum of the proton is $m_p c$, thus $\lambda=\frac{h}{m_p c}$. Click here to jump to the problem!
 8 Click here to jump to the problem! GR8677 #47 Quantum Mechanics$\Rightarrow$}Franck-Hertz Experiment Like the typical experiment in QM, the Frank-Hertz experiment has to do with shooting a bunch of particles through some oven. Its significance is in early observations of energy levels. Also, its key conclusion is that electrons are scattered elastically. But, if one doesn't know the above, one can always use MOE---the Method of Elimination. (A) Elastic collision requires conservation of kinetic energy. A bit restraining, but keep the choice. (B) Never scattered elastically? Never is too strong a word to be favored by ETS. (C) Seems reasonable-ish. Keep it. (D) This is, again, too restraining. It makes sense that depending on the impact angle and momentum, different amount of energy would be lost. (E) Discrete energy lost is mentioned in (C), but again, this choice is much too restraining, stating that there is no energy range..."---too strong of a phrase to be favored by ETS. Click here to jump to the problem!
 9 Click here to jump to the problem! GR8677 #49 Quantum Mechanics$\Rightarrow$}Schrodinger Equation Recall the Time-independent Schrodinger Equation, $H\psi=E\psi\Rightarrow -\frac{\hbar^2}{2m}\psi '' + V(x) \psi = E\psi, $ where $H=-\frac{\hbar^2}{2m}\psi '' + V(x) \psi = E\psi$. The classical Hamiltonian is $H=\frac{p^2}{2m} + V(x)$, where one sees the only difference is $p \rightarrow \frac{\hbar}{i}\frac{d}{dx}$. From the classical Hami, one can directly reach the Hami operator in the TISE via substituting a differential operator for momentum, as in choice (B). Click here to jump to the problem!
 10 Click here to jump to the problem! GR8677 #56 Quantum Mechanics$\Rightarrow$}Expectation Value Recall the following result of the Born Assumption, $\langle Q \rangle = \psi^* Q \psi = \langle \psi | Q | \psi \rangle, $ where the above yields the average value of Q---i.e., the expectation value. Click here to jump to the problem!
 11 Click here to jump to the problem! GR8677 #57 Quantum Mechanics$\Rightarrow$}Operators One doesn't really need QM to solve this. Just plug and chug each of the five functions into the following equation, $-i\hbar \frac{\partial}{\partial x} f = \hbar k f$ Applying the operator $-i\hbar \frac{\partial}{\partial x}$ to each function included in the choice, one gets: (A) $i \hbar \sin(k x)$ ... which isn't an eigenfunction (B) $-i \hbar \cos(k x)$ ... which isn't an eigenfunction (C) $-\hbar k f$ ... eigenvalue is off by a sign (D) $\hbar k$ ... this is the wanted eigenvalue! (E) $-i\hbar k$ ... off by a sign and imaginary term. Moreover, operators representing observables in QM have real eigenvalues. Click here to jump to the problem!
 12 Click here to jump to the problem! GR8677 #89 Quantum Mechanics$\Rightarrow$}Identical Particles According to Griffiths, the proof to this rule comes from QFT, $\psi = \frac{1}{\sqrt{2}}\left( \psi_\alpha(x_1) \psi_\beta(x_2) \pm \psi_\beta(x_1) \psi_\alpha(x_2) \right),$ where it's $+$ for bosons and $-$ for fermions. The given identical particle wave function contains a plus sign, so the particles must be bosons. Bosons have integer spin (while fermions have half-integer spin). Electrons, protons, and neutrons are all fermions. A positron is just a positive electron, so that is presumably, also, a fermion. Thus, the remaining choice would be deuteron---which is a boson. Click here to jump to the problem!
 13 Click here to jump to the problem! GR8677 #90 Quantum Mechanics$\Rightarrow$}Particle in a Box The particle is in an infinite well (or box, if you will) of length 2a. (It's stuck forever bouncing around between the two walls.) The number of nodes in the wave function determines the energy level. In this case, there is one load, thus this is $E_2$. The lowest state would be $E_1$ $E_n=k n^2$ eV for particle in a box. Given that $E_2 = 2 = 4 k$, one determines $k=1/2$ eV. Thus, $E_1 = k = 1/2$ eV. The answer is thus (C). Click here to jump to the problem!
 14 Click here to jump to the problem! GR8677 #92 Quantum Mechanics$\Rightarrow$}Perturbation Theory One can derive the selection rules by applying the electric dipole approximation in time-dependent perturbation theory. The results are the following: $\Delta m = 0, \pm 1$, $\delta l = \pm 1$. Choices (B) and (C) are (exactly this, thus) immediately out. There is no selection rule for spin, and thus choice (D) is it. (The correction is due to user snim1.)Click here to jump to the problem!
 15 Click here to jump to the problem! GR8677 #96 Quantum Mechanics$\Rightarrow$}Perturbation Technically, one can solve this problem without knowing anything about perturbation theory. Just remember the useful fact that the eigenfunctions of the unperturbed infinite deep well form a complete set (or complete basis). This means that any function can be represented by the old unperturbed infinite deep well. Thus, the solution to the perturbed eigenfunction should look like $\psi'_0 = \sum_n a_{0n} \psi_n$. Also, since the perturbed potential is also symmetrical with respect to the origin (as the original unperturbed potential was, too), one knows that all the odd terms should go to 0. This leaves choice (B). Click here to jump to the problem!
 16 Click here to jump to the problem! GR8677 #17 Quantum Mechanics$\Rightarrow$}Probability The careless error here would be to just directly square the grids. When one remembers the significance of the meaning of the probability $P=\int |\langle \psi | \psi \rangle|^2 dV$, one finds that one must square the wave function, and not the grids. The total probability is, $\int_0^6|\psi|^2 =1+1+4+9+1+0=16$ The un-normalized probability from $x=2$ to $x=4$ is, $\int_2^4|\psi|^2 =4+9=13$ The normalized probability is thus $13/16$, as in choice (E). Click here to jump to the problem!
 17 Click here to jump to the problem! GR8677 #18 Quantum Mechanics$\Rightarrow$}Scattering This is a conceptual scattering question. No calculations needed. (A) A particle incident from the left would have an oscillating wave function until it meets the barrier... not the other way around. (B) For $E, It is true that the barrier would decrease the amplitude of the wave function, however, when it emerges, the tunneled part of it should have an even smaller amplitude. (This graph would be good for $E>V_0$, however.) (C) This is the only graph that shows the wanted characteristics: oscillating wave before incidence, decay while in barrier, and tunneled-decrease amplitude when exit. (D) A typical particle is probably least likely to be found inside the barrier, so this is the least likely choice. (E) This wave function shows no change, when the potential barrier demands a change! Click here to jump to the problem!
 18 Click here to jump to the problem! GR8677 #33 Quantum Mechanics$\Rightarrow$}Probability Recall that $P(apple)=\int |\langle \psi_{apple} | \psi_{fruit} \rangle |^2 dx$ Given the wave function in terms of the spherical harmonic eigenfunctions, one has it totally easy. One has $\langle \psi | = (3Y^1_5 + 2Y^{-1}_5)/\sqrt{38}$. Ketting the bra, one has, $P=(9 + 4)/38$, where one recalls the orthonormality of the spherical harmonic eigenfunctions. This is choice (C). Click here to jump to the problem!
 19 Click here to jump to the problem! GR8677 #35 Quantum Mechanics$\Rightarrow$}Identical Particles Because of the antisymmetric interchange of identical particles, one gets $\psi=0$ if two fermions are in the same state. This is basically the foundation behind the familiar Pauli exclusion principle. Click here to jump to the problem!
 20 Click here to jump to the problem! GR8677 #40 Quantum Mechanics$\Rightarrow$}Bohr Theory It's amazing how far one can get with the Bohr formula. To start with, one should calculate the ground-state energy of the singly ionized Helium (i.e., the ionization energy). $E_1 = Z^2 E_{H1} = 4 \times 13.6 eV$, since Helium has 2 protons. (The general formula is $E_n = Z^2/n^2 E_1$.) The Bohr formula gives $E=E_1\left(1/n_f^2-1/n_i^2\right)=E_1(1/n_f^2-1/16)$, since $n_i^2=4^2=16$. $E=hc/\lambda\approx 1.24E-6/4.7E-7$ gives $E\approx 2.5$ eV. The only unknown expression above is $n_f$. Plugging everything in and solving for that, $n_f^{-2}\approx 8^{-1} \Rightarrow n_f\approx 3$. This yields choice (A). One can check via $E_f=E_1/n_f^2 4\times 13.6/9 \approx 6$, which verifies (A). Click here to jump to the problem!
 21 Click here to jump to the problem! GR8677 #42 Quantum Mechanics$\Rightarrow$}Photoelectric Effect Recall the photoelectric equation relating the incident electromagnetic wave to the kinetic energy and the work function $hc/\lambda = K+\phi \ approx 12E-7/500E-9 = K+2.28 \Rightarrow K = 12E-7/5E-7-2.28\approx 0.2 eV$, as in choice (B). Click here to jump to the problem!
 22 Click here to jump to the problem! GR8677 #51 Quantum Mechanics$\Rightarrow$}Infinite Well The even wave functions always have nodes in the middle, and thus the probability density for states of even n vanish. (One can deduce this by fitting curves inside a box. The first state has no nodes in the middle, but a node at each end of the well. The second state has one node in the middle. The third state has two nodes, neither of which are in the middle. The fourth state has three nodes, one of which is in the middle.) Click here to jump to the problem!
 23 Click here to jump to the problem! GR8677 #52 Quantum Mechanics$\Rightarrow$}Spherical Harmonics $Y^m_l$ is the convention used for a spherical harmonic of eigenvalue m, l. The only spherical harmonic proportional to $\sin\theta$ is $Y^{\pm 1}_1$. Recalling that the eigen-equation, $L_z \psi = m \hbar \psi$, one deduces that since $m=\pm 1$, the eigenvalues are $\pm \hbar$. (Open call for a better solution: Is there a method that does not require memorizing the first few spherical harmonics?) Reproduced for the reader's pleasure and convenience, (proportional values of the first few SH's: $\begin{eqnarray} Y^0_0 & = & const \\ Y^0_1 & = & \cos\theta \\ Y^{\pm 1}_1 &=& \sin\theta e^{\pm i\phi} \end{eqnarray}$ Click here to jump to the problem!
 24 Click here to jump to the problem! GR8677 #59 Quantum Mechanics$\Rightarrow$}Laser Not much understanding of lasers is required to solve this one; the basic idea of the problem tests the relation between photon energy and energy from the laser. Recalling the equation $E=hf=hc/\lambda\approx 12E-7/600E-9=2 eV$, equates that to the energy (in eV) calculated from $Pt = 10E3 \times 1E-15/1.602E-19$. The answer comes out to choice (B). Click here to jump to the problem!
 25 Click here to jump to the problem! GR8677 #76 Quantum Mechanics$\Rightarrow$}Uncertainty One can make a good stab at this problem by applying the uncertainty principle. I. If the average momentum of the packet is 0, then one violates the uncertainty principle. See IV. II. Maybe. III. Maybe. IV. True, recall the Gaussian uncertainty principle $\Delta x \Delta k = \hbar/2$. Since I is false, choices (A), (C), and (E) are out. Choices (B) and (D) remain. Take the conservative approach and choose (B). Click here to jump to the problem!
 26 Click here to jump to the problem! GR8677 #77 Quantum Mechanics$\Rightarrow$}Operators This problem can be solved (without much knowledge of quantum mechanics) by noting the following general arithmetic trick: $ab=\frac{1}{2}\left((a+b)^2 - a^2 - b^2\right)$. The problem gives the Hamiltonian $H=-JS_1 \cdot S_2$, which has the same form as the arithmetic trick above. Thus, $H=-J\frac{1}{2}\left((S_1+S_2)^2 - S_1^2 - S_2^2\right)$. Recalling some basic linear algebra, one can make use of the eigenvalue equations supplied with the problem defining the eigenvalues of the wanted operators, $S_i^2 \psi_i = S_i(S_i+1) \psi_i$. Thus, $\langle H \rangle = \langle \psi_i | H | \psi_i \rangle = -J\frac{1}{2}\left((S_1+S_2)(S_1+S_2+1) - S_1(S_1+1) - S_2(S_2+1) \right)$, where one has applied the eigenvalue equation above and generalized it for the case $(A+B^2)\psi = (A+B)(A+B+1) \psi$. From a bit of math manipulation, one has arrived at choice (D). Click here to jump to the problem!
 27 Click here to jump to the problem! GR8677 #97 Quantum Mechanics$\Rightarrow$}Probability Density Current If one has forgotten the expression for the probability density current, then one needs not despair! If one remembers the vaguest definition of a probability current," one can solve this problem without the usage of the forgotten formula: Recall that the probability current density J is the difference between incoming P(in) and outgoing P(out) probability densities. This is just the difference between the probability densities of a rightwards moving plane wave and a leftwards moving plane wave, since the probability density is related to the wave function by $P=|\psi\|^2$ $J\propto P(in)-P(out)=|f(e^{-i k x})|^2 - |f(e^{i k x})|^2$ The given wave function can be written in terms of plane waves, $\begin{eqnarray} \psi &=& e^{i\omega t} \left( \frac{\alpha}{2}(e^{ikx}+e^{-ikx}) + \frac{\beta}{2i}(e^{ikx}-e^{-ikx}) \right)\\ &=&\left( (\alpha+\beta/i)\frac{e^{ikx}}{2} + (\alpha-\beta/i)\frac{e^{-ikx}}{2} \right) \end{eqnarray}$ $f(e^{ikx})$ gives the coefficient for the leftwards (negative-direction) wave, while $f(e^{-ikx})$ gives the coefficient for the rightwards traveling wave. The probability density for each wave is given by, $\begin{eqnarray} |f(e^{ikx})|^2 &=& \frac{1}{4}\left(|\alpha|^2+|\beta|^2 + \frac{\alpha\beta^{*}}{-i} +\frac{\alpha^{*}\beta}{i}\right)\\ |f(e^{-ikx})|^2 &=& \frac{1}{4}\left(|\alpha|^2+|\beta|^2 + \frac{\alpha\beta^{*}}{i} -\frac{\alpha^{*}\beta}{i}\right) \end{eqnarray}$ Plugging these quantities into the formula for the probability current density above (J), one gets, $J \propto \frac{1}{2i} (\alpha\beta^{*}-\alpha^{*}\beta),$ which is choice (E). Alternatively: The formal expression for the probability current density can be effortlessly derived from recalling the definition of probability and Schrodinger's Equation---both of which every physics (or engineering) major should know by heart. Probability is defined (in the Born Interpretation) as $P=\int |\Psi(x,t)|^2 dx$. One should recall that $|A|^2=A^*A=AA^*$ in general (to wit: the absolute value squared of a complex expression is itself times its complex conjugate). The time-dependent Schrodinger's Equation is $\hbar i \frac{\partial \Psi}{\partial t}=H\Psi=-\frac{\hbar^2}{2m}\Psi^{''}+V\Psi$, where $H=-\hbar^2/2m \frac{\partial}{\partial x}+V$ has the form of the familiar time-independent Hamiltonian. From this, one finds that $\frac{\partial \psi}{\partial t}=\frac{-i}{\hbar}\left( -\frac{\hbar^2}{2m}\Psi^{''}+V\Psi \right)$. Generalizing the idea of a current from classical physics to the idea of a probability current, one takes the time derivative of the probability to get $\frac{d}{dt}P =\frac{d}{dt} \int |\psi|^2 dx =\int \frac{d}{dt} |\psi|^2 dx=\int \left(\psi \frac{\partial \psi^*}{\partial t}+\psi^* \frac{\partial \psi}{\partial t}\right)dx$, where the product-rule for baby-math derivatives has been used and the derivative has been taken inside the integral because the integral and derivative are with respect to different variables. Plugging in the expression for $\dot{\psi}$ from the Schrodinger's Equation, one gets $dP/dt= \int \left(\Psi \frac{i}{\hbar}\left( -\frac{\hbar^2}{2m}\Psi^{*''}+V\Psi^* \right)+\Psi^* \frac{-i}{\hbar}\left( -\frac{\hbar^2}{2m}\Psi^{''}+V\Psi \right)\right)dx$, where the terms involving V's cancel out, and thus, $dP/dt = \int \frac{i}{\hbar}\frac{\hbar^2}{2m}\left( -\Psi\Psi^{*''}+ \Psi^*\Psi^{''}\right)dx = \frac{i\hbar}{2m}\int\left( -\Psi\Psi^{*''}+ \Psi^*\Psi^{''}\right)dx$ Rewriting $\int \left( -\Psi\Psi^{*''}+ \Psi^*\Psi^{''}\right)dx = \int \frac{\partial}{\partial x} \left( -\Psi\Psi^{*'}+ \Psi^*\Psi^{'}\right)dx$, one can eliminate the integral in the probability current by applying the fundamental theorem of calculus (to wit: $\int_a^b \frac{\partial \psi}{\partial x}dx = \psi(b)-\psi(a)$), $dP/dt=\frac{i\hbar}{2m}\left( -\Psi\Psi^{*'}+ \Psi^*\Psi^{'}\right)$. But, since the probability current is usually define as $dP/dt = J(a)-J(b)$, one has $dP/dt = \frac{i\hbar}{2m}\left(\Psi\Psi^{*'}- \Psi^*\Psi^{'}\right)$. (Aside:) One can print-out a cool poster or decent T-shirt iron-on to remember the Schrodinger's Equation (among other miscellanai) at a site the current author made several years ago, \begin{quote} http://anequationisforever.com/ds.php \end{quote} One can remember the general form of the probability current by recalling that it has to do with the difference of $\Psi$ times its conjugate. Right, so onwards with the problem: The problem gives the wave function, so one needs just chunk out the math to arrive at the final answer, $\Psi^{'} = e^{i\omega t} k \left( -\alpha \sin(kx) +\beta \cos(kx) \right)$ $\Psi^{*'}=e^{-i\omega t}k \left( -\alpha \sin(kx) +\beta \cos(kx) \right)$ Thus, $\Psi^{*}\Psi^{'}=k \left( \alpha^{*} \cos(kx) +\beta^{*} \sin(kx) \right) \left( -\alpha \sin(kx) +\beta \cos(kx) \right)$ and, $\Psi \Psi^{*'}=k \left( \alpha \cos(kx) +\beta \sin(kx) \right) \left( -\alpha^{*} \sin(kx) +\beta^{*} \cos(kx) \right)$, where one notes that the imaginary terms go to unity from the complex conjugate. Plugging this into the probability current, one arrives at the expression for choice (E). Click here to jump to the problem!
 28 Click here to jump to the problem! GR8677 #98 Quantum Mechanics$\Rightarrow$}Symmetry One recalls the simple harmonic oscillator wave functions to be symmetric about the vertical-axis (even) for the 0th energy level, symmetric about the origin (odd) for the first energy level, and so on. If there is a wall in the middle of the well, then all the 0th energy level wave function would disappear, as would all even wave functions. Recall the formula for SHO $E=\hbar \omega \left(n+\frac{1}{2}\right)$. The first few odd states (the ones that remain) are $E_1 = 3/2 \hbar \omega, E_3=7/2 \hbar \omega$, etc. This is choice (D). Click here to jump to the problem!
 29 Click here to jump to the problem! GR8677 #99 Quantum Mechanics$\Rightarrow$}Two-State Systems Recall the mnemonic for remembering what a LASER is---Light Amplified Stimulated Emission Radiation. A laser consists of two states, with a metastable-state in between the top and bottom state. Initially, one has all the atoms in the ground-state. But, photons come in to excite the atoms (through absorption), and the atoms jump into the top state; this is called a population inversion, as the ground-state atoms are now mostly in the top inverted" state. More photons come in to excite these already excited atoms, but instead of absorption, emission occurs, and the atoms jump to a lower meta-stable state while emitting photons (in addition to the incident photons). The atoms stay in this metastable state due to selection rules, where a transition back to the ground-state is forbidden. One doesn't need to know all that to solve this problem. Instead, merely the idea of a laser requiring two main states and a metastable state in between would suffice. Since the question gives the bottom state as $n=1$ and top state as $n=3$, one deduces that the metastable state must be $n=2$, as in choice (B). Click here to jump to the problem!
 30 Click here to jump to the problem! GR8677 #100 Quantum Mechanics$\Rightarrow$}Raising Operator $a^\dag = a^* = \sqrt{\frac{m\omega_0}{2\hbar}}\left(x+-ip/(m\omega_0)\right) \neq a$, and thus choice III must be true. This eliminates all but choice (C) and (E). Since one knows that the raising operator acts to raise the energy level, $[a,H]\neq 0$ implies that they don't commute. This leaves just choice (C). (Choice II is false because, from above, the condition for a Hermitian operator is violated $a^\dag \neq a$.) Click here to jump to the problem!
 31 Click here to jump to the problem! GR8677 #27 Quantum Mechanics$\Rightarrow$}Hermitian Operator The eigenvalues of a Hermitian operator are always real. This is one of those quantumisms one memorizes after even a semi-decent course in QM. One can quickly prove it by the other quantumism-proof one memorized for that same course: Suppose one has $A|\psi\rangle = a |\psi\rangle$. Then, $A^\dag |\psi\rangle = a^* |\psi\rangle$. $A | \psi \rangle - A^\dag | \psi \rangle = (a-a^*)|\psi\rangle$. But, the left side is just 0 from, as $A^\dag = A$ from the definition of Hermitian operator. Then, the right side must be 0, too, since $\langle \psi |\psi \rangle \neq 0$ in general. Click here to jump to the problem!
 32 Click here to jump to the problem! GR8677 #28 Quantum Mechanics$\Rightarrow$}Orthogonality States or orthogonal when their dot-product (bra-ket) is 0. Since the eigenvectors $|i\rangle$ are given as orthonormal, $\langle i | i \rangle=1$. $\langle \psi_i | \psi_2 \rangle =5-15+2x=0$ Solve the equation to find that $x=-10$, as in choice (E). Click here to jump to the problem!
 33 Click here to jump to the problem! GR8677 #29 Quantum Mechanics$\Rightarrow$}Expectation Value $\langle O \rangle = \langle \psi | O \psi \rangle = -1/6 + 1/2 +2/3 = 1$. (Recall that $\langle \psi | A | \psi \rangle = a \langle \psi | \psi \rangle =a$.) Click here to jump to the problem!
 34 Click here to jump to the problem! GR8677 #43 Quantum Mechanics$\Rightarrow$}Commutators Write it out and recall the commutator identities $[B,AC]=A[B,C]+[B,A]C$ and $[A,B]=-[B,A]$. Thus $-[L_z,L_xL_y]=-L_x[L_z,L_y]-[L_Z,L_x]L_y=i\hbar (L_x^2-L_y^2)$, using the identities supplied by ETS. Click here to jump to the problem!
 35 Click here to jump to the problem! GR8677 #44 Quantum Mechanics$\Rightarrow$}Measurements When one measures something, one can get something that is the eigenvalue of the operator. Energy eigenvalues are proportional to $n^2$. The only choice that is a squared quantity is choice (D), for $n=3$. Click here to jump to the problem!
 36 Click here to jump to the problem! GR8677 #45 Quantum Mechanics$\Rightarrow$}Simple Harmonic Oscillator The expectation value is the average value, and it is calculated by $\langle \psi | H \psi \rangle$. Using orthonormality, one has $E=(1/14)(3/2)+(4/14)(5/2)+(9/14)(7/2)\hbar \omega = 43/14 \hbar \omega$ Click here to jump to the problem!
 37 Click here to jump to the problem! GR8677 #46 Quantum Mechanics$\Rightarrow$}de Broglie Wavelength Since $E=p^2/(2m)$ and the de Broglie wavelength gives $\lambda=h/p$, one can find the initial wavelength $\lambda=h/sqrt{2mE}$. This yields an expression for $h=\lambda \sqrt{2mE}$ When the particle enters the well, its energy becomes $E^{'}=E-V$, and thus $\lambda^{'}=h/sqrt{2m(E-V)}$. Plugging h in, one has $\lambda^{'}=\lambda \sqrt{2mE}/sqrt{2m(E-V)}$, which is choice (E). (This above solution is due to the lout Marko.) Also, the current author's original approach leads to the right result, but it hand-waves the massiveness of the particle: The initial kinetic energy of the free particle is its total energy $E=h v/\lambda \Rightarrow \lambda =h v/E$. But, since one's dealing with de Broglie, one has $p=h/\lambda=mv$. Thus, $\lambda^2 = h^2/(mE)$. In the potential, its energy becomes $E-V$. Replace E above with that to get $\lambda^{'2}=h^2/(m(E-V))$. But, from the same relation above, one has $h^2 = mE \lambda^2$. Thus, $\lambda^{'2} = \lambda^2 E/(E-V) = \lambda^2 (1-V/E)^-1$. This is choice (E). Click here to jump to the problem!
 38 Click here to jump to the problem! GR8677 #81 Quantum Mechanics$\Rightarrow$}Angular Momentum Recall the angular momentum eigen-equations, $L^2 \psi = \hbar^2 l(l+1) \psi$ and $L_z \psi = m\hbar \psi$. The problem wants $L^2 \psi = 6\hbar^2 \psi$ and $L_z \psi = -\hbar \psi$. Matching coefficients with the above equations, one finds that $l(l+1) = 6$ and $m=-1$. Solving, one finds that $l=2,-3$. Any one of the spherical harmonics with $Y^{-1}_{2}$ or $Y^{-1}_{-3}$ would work. So, since the second spherical harmonic isn't listed, take choice (B). Click here to jump to the problem!
 39 Click here to jump to the problem! GR8677 #82 Quantum Mechanics$\Rightarrow$}Addition of Angular Momentum For 2 electrons, one recalls that the there are three spin triplet states and one spin singlet state (hence their names). One can apply the lowering operator multiple times to the up-up state to arrive at, respectively, the up-down state, the down-up state, and then the down-down state (the spin-singlet state). Applying the lowering operator to choice I, one gets choice III. Applying the lowering operator to that, one gets down-down. These are the 3 spin-triplet states. By orthogonality, the down-down state has a negative sign. So, only choices I and III are in the triplet configuration. (David J. Griffiths vanity alert---the QM problems thus far are all straight out of his textbook, An Introduction to Quantum Mechanics.) Click here to jump to the problem!
 40 Click here to jump to the problem! GR8677 #83 Quantum Mechanics$\Rightarrow$}Basis One can chunk out the 2-by-2 matrices by choosing a basis for z. Choose the easiest, $| \uparrow \rangle =(1,0)$ and $| \downarrow \rangle = (0,1)$. One can tell by inspection that choice (A) can't be it, nor can choices (D) and (E) (since there aren't any imaginary numbers involved for x). Recall that $S_i = \hbar/2 \sigma_i$. Plug in the deductions above into choice (B) to get $\psi_b =(| \uparrow \rangle + | \downarrow \rangle )/\sqrt{2} = (1,1)/\sqrt{2}$. Multiply that with the Pauli matrix to get $S_x \psi_b = \hbar/2 \psi_b$. Plug in the deductions above into choice (C) to get $\psi_c = (| \uparrow \rangle - | \downarrow \rangle )/\sqrt{2} = (1,-1)/\sqrt{2}$. Multiply that with the Pauli matrix to get $S_x \psi_c = -\hbar/2 \psi_b$. Click here to jump to the problem!
 41 Click here to jump to the problem! GR8677 #84 Quantum Mechanics$\Rightarrow$}Selection Rules The selection rules are $\Delta l = \pm 1$ and $\Delta m =0,\pm 1$. The transitions have the following quantum numbers: A: $\Delta l = 0$ and $\Delta j = 0$ B: $\Delta l = -1$ and $\Delta j = -1$ C: $\Delta l = -1$ and $\Delta j = 0$ By the selection rules above, transition A is forbidden in $\Delta l$. Transitions B and C both work out, since they do not violate the $\Delta l$ rule. Take choice (D). Click here to jump to the problem!
 42 Click here to jump to the problem! GR8677 #93 Quantum Mechanics$\Rightarrow$}Most probable value Recall the definition of probability, $P=\int |\psi|^2dV=\int|\psi|^2 4\pi r^2 dr$. (The radial probability distribution $P_r$ is related to the probability $P$ by $P=\int P_r dr$.) The most probable value is given by the maximum of the probability distribution. Taking the derivative with respect to r, one has this condition for a maximum (the second derivative shows that it's concave up) $dP_r/dr = 4\pi r^2 d|\psi|^2/dr + 8\pi r |\psi|^2=0$. The problem gives $\psi = \frac{1}{\sqrt{\pi a^3_0}}e^{-r/a_0}$. Thus, $|\psi|^2=\frac{1}{\pi a^3_0}e^{-2r/a_0}$ and $d|\psi|^2/dr = \frac{-2}{\pi a^3_0 a_0}e^{-2r/a_0}$. Plug this into the expression for $dP/dr$, solve for $r$ to find that the most probable distance is just the Bohr radius, as in choice (C). Click here to jump to the problem!
 43 Click here to jump to the problem! GR8677 #94 Quantum Mechanics$\Rightarrow$}Perturbation Theory The energy for first-order perturbation theory of $H=H_0+\Delta H$ ($H_0$ is the known Hamiltonian and $\Delta H$ is the perturbed Hamiltonian) is given by $E_1=\langle \psi_0 | \Delta H \psi \rangle_0$, where the wave-functions are the unperturbed ones. Thus, the problem amounts to calculating $E_1=\langle n | V(a+a^\dag)^2 | n \rangle$. This is just raising and lowering operator mechanics. $(a+a^\dag)^2=a^2+a^{\dag^2}+aa^\dag + a^\dag a$. But, after bra-ketting, one finds that the expectation value of $a^2$ and $a^{\dag^2}$ are 0, since $n$ and $n+2$, $n-2$ are orthogonal. Thus, the problem becomes, $E_1=\langle n | (a a^\dag + a^\dag a) | n \rangle$. Applying the given eigen-equations, one finds that $E_1=\langle n | (a a^\dag + a^\dag a) | n \rangle = (2n+1)V$. For $n=2$, one finds $E_1=5V$, as in choice (E). (Note that: $a^\dag a |n\rangle = a^\dag \sqrt{n} | n-1 \rangle = n | n \rangle$ and $a a^\dag |n\rangle = a \sqrt{n+1} | n+1 \rangle = (n+1)|n\rangle$ and $a^2 = a \sqrt{n}|n-1\rangle = \sqrt{n}\sqrt{n-1}|n-2\rangle$.) Click here to jump to the problem!
 44 Click here to jump to the problem! GR8677 #1 Quantum Mechanics$\Rightarrow$}Momentum Operator $\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}\psi \Rightarrow p \psi = \hbar k \psi $ Click here to jump to the problem!
 45 Click here to jump to the problem! GR8677 #3 Quantum Mechanics$\Rightarrow$}Bohr Theory Recall the Bohr Equation, $E_n = Z^2/n^2E_1$, which applies to both the Hydrogen atom and hydrogen-like atoms. One can find the characteristic X-rays from that equation (since energy is related to wavelength and frequency of the X-ray by $E=\lambda f$). The ratio of energies is thus $E(Z=6)/E(Z=12)=6^2/12^2=1/4$, as in choice (A). Click here to jump to the problem!
 46 Click here to jump to the problem! GR8677 #27 Quantum Mechanics$\Rightarrow$}Uncertainty This problem looks much more complicated than it actually is. Since $k$ and $x$ are fourier variables, their localization would vary inversely, as in choice (B). Click here to jump to the problem!
 47 Click here to jump to the problem! GR8677 #28 Quantum Mechanics$\Rightarrow$}Probability One doesn't actually need to know much (if anything) about spherical harmonics to solve this problem. One needs only the relation $P=\sum_i |\langle Y^3_i |\psi (\theta,\phi)\rangle|^2$. Since the problem asks for states where $m=3$, and it gives the form of spherical harmonics employed as $Y^m_l$, one can eliminate the third term after the dot-product. So, the given wave function $\psi(\theta,\psi)=\frac{1}{\sqrt{30}}\left( 5 Y^3_4 + Y^3_6 - 2Y^0_6\right)$ gets dot-product'ed like $|\langle Y^3_i |\psi (\theta,\phi)\rangle|^2\left(\frac{1}{\sqrt{30}}\left( 5 Y^3_4 + Y^3_6\right)\right)\left(\frac{1}{\sqrt{30}}\left( 5 Y^3_4 + Y^3_6 - 2Y^0_6\right)\right)=\frac{25+1}{30}=\frac{13}{15}$, as in choice (E). Click here to jump to the problem!
 48 Click here to jump to the problem! GR8677 #29 Quantum Mechanics$\Rightarrow$}Bound State Tunneling should show exponential decay for a finite-potential well, and thus choice (E) is eliminated. Choice (C) is eliminated because the wave function is not continuous. One eliminates choice (D) because the bound-state wave functions of a finite well isn't linear. The wave function for a bound state should look similar to that of an infinite potential well, except because of tunneling, the well appears larger---thus the energy levels should be lower and the wave functions should look more spread out. Choice (B) shows a more-spread-out version of a wave function from the infinite potential well. Click here to jump to the problem!
 49 Click here to jump to the problem! GR8677 #30 Quantum Mechanics$\Rightarrow$}Bohr Theory The ground state binding energy of positronium is half of that of Hydrogen. This is so because the energy is proportional to the reduced mass, and that of the positronium has a reduced mass of half that of Hydrogen. Thus, from the Bohr formula, one has $E = Z^2 E_1/n^2$, where $E_1=E_0/2$ and $E_0$ is the ground state energy of Hydrogen. (The ground state energy of positronium is half that of Hydrogen because its reduced mass is half that of Hydrogen's.) Since $Z=1$, then for $n=2$, the energy is $E = E_0/8$, as in choice (E). Click here to jump to the problem!
 50 Click here to jump to the problem! GR8677 #50 Quantum Mechanics$\Rightarrow$}Simultaneous Eigenstates QM in verse... Two operators, both alike in state functions, In fair bases, where we lay our scene, From ancient grudge break new mutiny... Two operators unlike in eigenvalues Yet star-crossed lovers commute. So anyway, the problem gives $A | a \rangle = \alpha | a \rangle$ and $B | a \rangle = \beta | a \rangle$. That is, both A and B share the same eigenstate $| a \rangle$. Consider $\langle a^{'} | [A,B]| a^{''} \rangle =\langle a^{'}| AB-BA| a^{''} \rangle = \langle a^{'} |(A\beta^{'}-B\alpha^{''}) | a^{''} \rangle = \langle a^{'} |(\beta^{'}A-\alpha^{''}B) | a^{''} \rangle = \langle a^{'} |(\alpha^{''}\beta^{''}-\beta^{''}\alpha^{''}) | a^{''} \rangle = (\alpha^{''}\beta^{''}-\beta^{''}\alpha^{''}) \langle a^{'} | a^{''} \rangle$. But, the scalar term is 0. This implies that in order for both operators to have the same eigenstate, $[A,B]=0$. (Also, one knows that $\langle a^{'} | a^{''} \rangle = \delta_{',''}$ by definition of orthonormal eigenfunctions.) Click here to jump to the problem!
 51 Click here to jump to the problem! GR8677 #51 Quantum Mechanics$\Rightarrow$}Momentum The momentum operator in position space is given by $p = \hbar/ i \frac{\partial}{\partial x}$. Thus, given the wave function, one can calculate the expectation value as $\langle \psi_n | p | \psi_n \rangle = \langle \psi_n | \hbar/ i \frac{\partial | \psi_n \rangle}{\partial x}\propto \int_0^a \cos(n\pi x/a)\sin(n\pi x/a)dx =0$, since sine's and cosine's are orthogonal over a whole period. The answer is thus (A). Click here to jump to the problem!
 52 Click here to jump to the problem! GR8677 #52 Quantum Mechanics$\Rightarrow$}Orthonormality $\langle \psi_m | \psi_n \rangle = \delta_{nm}$ This is the definition of orthonormality, i.e., something that is both orthogonal (self dot others = 0) and normal (self dot self = 1). Click here to jump to the problem!
 53 Click here to jump to the problem! GR8677 #53 Quantum Mechanics$\Rightarrow$}Energy If one forgets the energy of an infinite well, one can quickly derive it from the time-independent Schrodinger's Equation $-\frac{\hbar}{2m}\psi^{''}+V\psi = E\psi$. However, since $V=0$ inside, one has $-\frac{\hbar}{2m}\psi^{''} = E\psi$. Plug in the ground-state wave function $\psi = A \sin(kx)$, where $k=n\pi/a$. Chunk out the second derivative to get $E=\frac{k^2 \hbar^2}{2m}$. Plug in k to get $E=\frac{\hbar^2 n^2 \pi^2}{2ma^2}$. Note that $k=n\pi/a$ can be deduced from boundary conditions, i.e., the wave function vanishes at both ends ($\psi(0)=0$ and $\psi(a)=0$). The second boundary condition forces the n's to be integers. Since one can't have a trivial wave function, $|n| \geq 1$, and thus $n^2 \geq 1$. One finds that $E \geq \frac{\pi^2 \hbar^2}{2m a^2}$, since $n=1,2,3...$, as in choice (E). Click here to jump to the problem!
 54 Click here to jump to the problem! GR8677 #56 Quantum Mechanics$\Rightarrow$}Simple Harmonic Oscillator The energy of a simple harmonic oscillator is given by $E_n=(n+\frac{1}{2})h\nu$. Thus, the ground state energy is simply $E_0=h\nu/2$, as in choice (C). Click here to jump to the problem!
 55 Click here to jump to the problem! GR8677 #77 Quantum Mechanics$\Rightarrow$}Gyromagnetic Ratio The intrinsic magnetic moment is defined in terms of the gyromagnetic ratio and spin as $\vec{\mu}_s = \gamma \vec{S}$, where $\gamma = \frac{eg}{2m}$ (g is the Lande g-factor). Thus, one sees that the magnetic moment is inversely related to mass. The ratio of the magnetic moment of a nucleus to that of an electron is $\mu_n/\mu_e = m_e/m_n << 1$, as in choice (E). (One can cancel out the S since ETS is nice enough to have the nucleus have the same spin as the electron.). Click here to jump to the problem!
 56 Click here to jump to the problem! GR8677 #80 Quantum Mechanics$\Rightarrow$}Planck Energy The key equation is $E=hc/\lambda$. Since $hc=1.24E-6eVm$ and $E=25E3eV=2.5E4eV$, one can immediately plug the quantities in to solve for $\lambda=hc/E = 1.24E-6/2.5E4=0.5E-10$, which is just choice (B). No knowledge of X-rays required, other than the elementary knowledge that it's a electromagnetic wave and allows one to write the Planck energy as $E=hf = h c/\lambda$. Click here to jump to the problem!
 57 Click here to jump to the problem! GR8677 #89 Quantum Mechanics$\Rightarrow$}Symmetry There is now a node in the middle of the well. By symmetry, the ground state will disappear $n=0$, as well all the even n states. Thus, the remaining states are the odd states, as in choice (E). Click here to jump to the problem!
 58 Click here to jump to the problem! GR8677 #90 Quantum Mechanics$\Rightarrow$}Rotational Energy Level Rotational energy is related to angular momentum by $E_{rot}=L^2/(2I)$. Quantum mechanics quantizes the angular momentum $L^2=\hbar^2 l(l+1)$. Thus, $E_{rot}=\hbar^2 l(l+1)/(2I)$. The approximate spacing between two levels is given by $\Delta E = E(l=1) - E(l=0)\approx\hbar^2/I$. The moment of inertia of $H_2$ is just that of two point-masses rotating about a center-point, thus $I=2mr^2$, taking $r=0.5E-10m$ and $m=E-27kg$ (mass of proton). $I\approx E-47kgm^2$. Now, $\hbar^2 \approx (6E-34)^2/(6)^2=E-68$. Plug everything in to get the right answer. $\hbar^2/I \approx E-68/E-47= E-21 J$. Converting J to eV, one has $E-21 J/E-19 = E-3$ eV, as in choice (B). Click here to jump to the problem!
 59 Click here to jump to the problem! GR8677 #98 Quantum Mechanics$\Rightarrow$}Characteristic Equation The characteristic equation of the matrix solves for the eigenvalues. It is $-\lambda(\lambda^2)+1=0$. Not all solutions are real, since $\lambda = e^{2i\pi/3} = \cos(2\pi/3)+i\sin(2\pi/3)$, where Euler's relation is used. Click here to jump to the problem!
 60 Click here to jump to the problem! GR8677 #99 Quantum Mechanics$\Rightarrow$}Perturbation Theory The perturbed Hamiltonian is given by $\Delta H = eEz = eEr\cos\theta$, where the last substitution is made for spherical coordinates. The first-order energy-shift is given by $\langle \psi_0 | \Delta H | \psi_0 \rangle$, where $\psi_0\propto e^{-kr}$. $dV=r^2drd\Omega=r^2\sin^2\theta d\phi d\theta dr$. Thus, $E_0 = \int_0^\infty \int_0^{\pi} \int_0^{2\pi} e^{-kr}\cos\theta\sin^2\theta d\theta d\phi =0$, since the integral of $\cos\theta\sin^2\theta$ over $0$ to $\pi$ is 0. Search this site for keyword perturbation for more on this. Click here to jump to the problem!

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