I have only recently started to learn quantum mechanics and I have only gone so far as to solve the hydrogen atom using Schrödinger's equation and the corresponding Hamiltonian, but all this has me wondering if it is possible to find a wave function for a planet-sun or moon-planet system. I know that this would be amazingly(and needlessly) more complicated than the classical solution. So is this possible, have people done it, and can we learn more from this approach?
1503
Many introductory quantum mechanics textbooks include simple exercises on computing the de Broglie wavelength of macroscopic objects, often contrasting the results with that of a proton, etc.  For instance, this example, taken from a textbook:     Calculate the de Broglie wavelength for      (a) a proton of kinetic energy 70 MeV      (b) a 100 g bullet moving at 900 m/s   The pedagogical motivation behind these questions is obvious: when the de Broglie wavelength is small compared to the size of an object, the wave behaviour of the object is undetectable.  But what is the validity of the actual approach of applying the formula naively? Of course, a 100g bullet is not a fundamental object of 100g but a lattice of some $10^{23}$ atoms bound by the electromagnetic force. But is the naive answer close to the actual one (i.e. within an order of magnitude or two)? How does one even calculate the de Broglie wavelength for a many body system accurately?
24229
I'm currently taking Quantum Mechanics in school and I came across a table displaying the following in the given order:   $E_n (\text{energy levels}), g_n (\text{eigen functions})$   $n \ l, [m]$  $E_1, 1$   $1s, [0]$  $E_2, 4$   $2s, [0]$ $2p, [-1,0,1]$  $E_3, 9$   $3s, [0]$ $3p, [-1,0,1]$ $3d, [-2,-1,0,1,2]$    Now, as I understand $n$ defines the size of the orbital, $l$ the shape and $m$ the orientation.  My question is, how does the atoms shell model fit into this if one can have two electrons per orbital? Are $1s$ the same as the orbital $2s$, but just renamed for the sake of higher energy levels? Or do two similarly looking orbits exist for different energy levels? For example two "layers" of spherical electron clouds for an atom with $4$ electrons.  I just can't see how $2$, $8$, $18$ becomes $2$, $8$, $8$ as we were taught in Physics at high school? Am I missing something?
23700
I am currently trying to follow Leonard Susskind's "" lecture series on quantum mechanics. (I know a bit of linear algebra and calculus, so far it seems definitely enough to follow this course, though I have no university physics education.)  In general, I find these lectures focus a bit too much on the math and not really on the physical motivation behind it, but so be it (if there are other courses aimed at those with reasonable math skills that focus more on physical meaning, let me know!). However, that's only indirectly related to what my question is about.  In , Susskind sets out to derive an expression for what he earlier labelled the time-development operator $U$:  $$|\psi(t)\rangle = U(t)|\psi(0)\rangle$$  He starts out as such:  $$U(\epsilon) = I + \epsilon H$$  which makes sense because the change in time will have to be small, i.e. on the order of a small $\epsilon$. However, he then goes ahead and changes this into:  $$U(\epsilon) = I - i\epsilon H$$  which of course is still fine, because we still don't know what $H$ is supposed to be. Now my problem lies with the fact that Susskind then proceeds to derive an expression for $H$ and, subsequently, the Schrödinger equation in which it figures, from the above equation. The $i$ never gets lost and ends up in that equation.  Could we not just as easily have left the $i$ out, or put a 6 or whatever there? Why put $i$ there? I finished the entire lecture hoping Susskind would get back to this, but he never does, unfortunately. (Which is, I guess, another symptom of this course, with which I'm otherwise quite happy, occasionally lacking in physical motivation.)   For those of you who know this lecture, or similar styles of teaching: am I missing something here? Alternatively, a general answer as to why there is an $i$ in the Hamiltonian and the Schrödinger equation?
57443
In an hydrogen-like atoms the orbitals are solutions to the Schrodinger equation suitable for the problem.  They describe the regions where an electron can be found.  So, why don't they have spherical symmetric?  As in this picture:    Why, in a certain distance from the nucleus, are there points where an electron can be found, and points where it can't?  EDIT: Is there any intuation about the symmetry breaking?
107284
suppose we have a potential that's independent of time $V(x,t) = V(x)$ so in Schrödinger equation we get: $$i\hbar \frac{\partial \Psi (x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi (x,t)}{\partial x^2}+V(x)\Psi(x,t)$$ since the LHS involves a variation of $\Psi$ with $x$ and the LHS involves a variation of $\Psi$ with $t$ can we say that a basis for the solutions will be with the form $\Psi(x,t)=\psi(x)T(t)$?
24421
I have done the Problem 2.1 in Griffiths' quantum mechanics,  and it seems not making sense to me.  What if the wave function isn't symmetric at all?  Then obviously the proof doesn't work. The solution confuses me.      If $V(x)=V(-x)$ then the "position" wave function can always be taken to be either even or odd.
22936
So, I was reading Atom: Journey Across The Subatomic Cosmos by Isaac Asimov in order to better understand quantum mechanics when I came across this sentence:      The electron couldn't spiral into the proton because it couldn't take up an orbit with a length less than a single wave.    I was wondering why that is? Why can't an electron not have non-whole number wavelengths in atomic energy levels?
46537
I'm trying to derive $\left\langle \frac{1}{r} \right\rangle = \frac{1}{n^2 a_0}$ (where $a_0$ is the Bohr radius) for the $| nlm\rangle$ state of hydrogen.  I've separated the radial and angular parts of the hydrogen wavefunction and split up the integration to yield  \begin{align}\left\langle \frac{1}{r} \right\rangle &amp;= \int R^*_{nl}(r) {Y^{m}_{l}}^*(\theta, \phi) \frac{1}{r} R_{nl}(r) Y_{l}^{m}(\theta, \phi) r^2 \sin \theta ~\mathrm dr ~\mathrm d\theta~\mathrm d\phi\\ &amp;= \int {Y^{m}_{l}}^*(\theta, \phi) Y_{l}^{m}(\theta, \phi) \sin \theta ~\mathrm d\theta ~\mathrm d\phi \int \limits_{0}^{\infty} r R^*_{nl}(r) R_{nl}(r) ~\mathrm dr\\ &amp;= \int \limits_{0}^{\infty} r R^*_{nl}(r) R_{nl}(r) ~\mathrm dr\end{align}  But from here, I'm not sure how to continue. Since $R_{nl}(r)$ is defined as $$R_{nl}(r) = \left( \frac{2 Z}{n a_0} r \right)^l \sum \limits_{k = 0}^{n - l - 1} a_k \left( \frac{2 Z}{n a_0} r \right)^k e^{-Zr/n a_0}$$ where $$a_{k + 1} = \frac{k + l + 1 - n}{(k + 1)(k + 2l + 2)}a_k$$  How do I deal with the sum inside the integral (especially a sum that requires recursion to compute coefficients)?  Have looked this  so far.
56049
I've gone through an intermediate classical mechanics course, and in solving the two-body problem, we reduce it to a one-body between a larger stationary mass, and a smaller reduced mass.  Most solutions I've seen of the hydrogen atom simply neglect the movement of the nucleus. Some solutions replace the electron mass with the reduced mass, for a more accurate answer.  Since the reduced mass is a result derived in classical mechanics, is it valid to apply it in quantum mechanics? Is it possible to derive the reduced mass within a quantum mechanical framework?
114252
In quantum mechanics any two level system can be mapped onto effective spin variables. If the system is defined by two energy levels, $|E_1\rangle$ and $|E_2\rangle$, the Hamiltonian is  $$ H = \left(\begin{array}{cc} E_1 &amp; 0 \\ 0 &amp; E_2 \end{array}\right) \, .$$  This can be recast as  $$ H = \frac{E_1+E_2}{2} 1\!\!1 + \frac{E_1-E_2}{2} \sigma_3\, .$$  $1\!\!1$ is the identity operator and $\sigma_3$ is the third pauli matrix. This is very nice because we can now apply our knowledge and intuition of spin dynamics to any two level system.  My question is the following: for a real spin I can measure expectation values of $\sigma_{1,2}$ by physically rotating my system and align its $x$ or $y$ axis with my measuring apparatus. How do I do this (in the lab!) with a pseudo-spin system? Imagine I am looking at a cold atom and I want to know if the system is in an symmetric or anti-symmetric superposition of $|E_1\rangle$ and $|E_2\rangle$, how do I do it?
122475
In  , the general solution for the wavefunction, with $k = R \sqrt{2 m E / \hbar^2}$, $R$ being the ring radius, $c_{+, -}$ being constants, $E$ the energy, and $m$ the particle mass, is  $$\psi(\theta) = c_{+} e^{i k \theta} +  c_{-} e^{-i k \theta}$$  Periodic boundary conditions give the eigenvalues $k = n$, with $n=0, \pm 1, \pm 2, \ldots$.  At first glance the problem looks "underdetermined" - it seems that there are two unknowns ($c_{+}$ and $c_{-}$) but only one condition to find them, normalization of the wavefunction $\langle \psi \lvert \psi\rangle = 1$. But in my naive understanding, the problem is actually not underdetermined because one may set $c_{-} = 0$, as is seemingly done on  in order to normalize the wavefunction.  Why may one set $c_{-} = 0$? Or, why do  and textbooks (below) seemingly set $c_{-} = 0$? And how can we treat the general case with $c_{-} \ne 0$?  Even if we do not trust , most textbooks on introductory quantum look at the ring problem and do the same thing. For example:   "Introduction to Quantum Mechanics: in Chemistry, Materials Science, and Biology" By Sy M. Blinder "Atkins' Physical Chemistry" By Peter Atkins, Julio de Paula "Quantum Mechanics for Chemists" David O. Hayward "Case studies in atomic physics 4" edited by E. McDaniel "Quantum Chemistry" By John P. Lowe
81415
In a quantum mechanics script I'm reading, the Schrödinger equation is "derived" (more precisely, motivated) by the De Broglie hypothesis. It starts at  $$ \lambda = \frac{2\pi h}{p} $$ $$ \omega = \frac{E}{h} $$  then takes the partial derivatives of the wave $\Psi(x,t) = \Psi(0,0)e^{\frac{2\pi ix}{\lambda}-it\omega}$  $$ \tag{1} \frac{\partial}{\partial x}\Psi(x,t) = \frac{ip}{h}\Psi(x,t) $$ $$ \frac{\partial}{\partial t}\Psi(x,t) = -i\frac{E}{h}\Psi(x,t) $$  With the non-relativistic free particle $E=\frac{1}{2m}p^2$ one gets  $$ \tag{2} ih\frac{\partial}{\partial t}\Psi(x,t) = E\Psi(x,t) = \frac{p^2}{2m}\Psi(x,t) $$  From there, they miraculously get the time-dependent Schrödinger equation. I cannot understand this step. If I insert formula (1) for $p^2$ in (2), I get something with $(\frac{\partial}{\partial x}\Psi)^2$, which is not the second partial derivative of $\Psi$ with respect to $x$.  Any hints?
83748
I'm working on some exercise regarding the spin coupling of two electrons. There we have the wavefunctions corresponding to the S values as  $$\begin{align} S = 1: &amp;\begin{array}{c}\uparrow\uparrow \\ \dfrac{1}{\sqrt 2}(\uparrow\downarrow+\downarrow \uparrow) \\ \downarrow\downarrow\end{array} \\[5mm] S = 0: &amp;\ \frac{1}{\sqrt{2}} (\uparrow\downarrow-\downarrow\uparrow) \end{align}$$  I understand that the three wavefunctions for $S=1$ are symmetric and the one for $S=0$ is antisymmetric. My question is, why is the combination with the minus sign is the one for $S=0$?  My thinking is that in a $\uparrow \downarrow$ or $\downarrow\uparrow$ combination the $S_z$ components would already add up to $0$ so that the minus sign would not change anything in the fact that $S=0$.   Or is it that $\downarrow\uparrow$ or $\uparrow\downarrow$ each represent a state with $S=1$ and $S_z = 0$ so that subtracting the one from the other gives $S=1-1=0$?
20052
The  is the basis to understanding quantum mechanics, but how can one derive it? I asked my instructor but he told me that it came from the experience of Schrödinger and his experiments. My question is, can one derive the Schrödinger equation mathematically?
23316
So when learning about the  of hydrogen and de Broglie waves, it was shown that treating the electron of hydrogen as a de Broglie wave results in the relationship $$L=n\hbar, \qquad n\in\mathbb{N}.$$ However, when learning about the azimuthal quantum number, it was stated that $$L=\sqrt{\ell(\ell+1)}\hbar.$$ So how come in the ground state ($n=1, \ell=0$), these two equations give different values for angular momentum? I feel like I'm missing something really important here. If it's the case that the Bohr model doesn't accurately describe the angular momentum of the electron in the ground state, why is the angular momentum zero?
78641
In deriving the quantization condition for a bound state in a potential with "no verticle walls" we start with the WKB connection formulas to find the wavefunction in the interior of the well ($x_1&lt;x&lt;x_2$), namely $$\psi(x) \cong \begin{cases}\displaystyle\frac{2C_1}{\sqrt{p(x)}}\sin\left[\frac{1}{\hbar}\int_{x_1}^x{dx\ p(x)}+\frac{\pi}{4}\right]&amp;\quad\text{for}\quad x_1&lt;x\\ \\\displaystyle \frac{2C_2}{\sqrt{p(x)}}\sin\left[\frac{1}{\hbar}\int_x^{x_2}{dx\ p(x)}+\frac{\pi}{4}\right]&amp;\quad\text{for}\quad x&lt;x_2 \end{cases}$$  Our task is to connect the two wave functions in the interior region together. Seeing that the coefficients are easily handled, it is then necessary that the arguments of the sine functions be equivalent. Denoting  $$\theta_1(x) = -\frac{1}{\hbar}\int_{x_1}^x{dx\ p(x)}-\frac{\pi}{4}$$  $$\theta_2(x) = \frac{1}{\hbar}\int_{x}^{x_2}{dx\ p(x)}+\frac{\pi}{4}$$ we have  $$\theta_1(x)+n\pi = \theta_2(x)\quad\text{for}\quad n = 0,1,2,3..$$  since  the relative minus can be absorbed into either coefficient. Thus $$-\frac{1}{\hbar}\int_{x_1}^x{dx\ p(x)}-\frac{\pi}{4}+n\pi= \frac{1}{\hbar}\int_{x}^{x_2}{dx\ p(x)}+\frac{\pi}{4}$$ \begin{equation}\int_{x_1}^{x_2}{dx\ p(x)} = \left(n-\frac{1}{2}\right)\pi\hbar\quad\text{for}\quad n=0,1,2,3..\end{equation} This is our quantization condition for the allowed energies.   However, following the derivation in Griffiths or Sakurai, $n=0$ is not included. Why?   My ideas: Well this would imply  $$\theta_1(x) = \theta_2(x)$$ \begin{equation}\int_{x_1}^{x_2}{dx\ p(x)} = -\frac{\pi}{2}\hbar\end{equation} \begin{equation}\int_{x_1}^{x_2}{dx\ \sqrt{2m(E-V(x))}} = -\frac{\pi}{2}\hbar\end{equation} I am trying to imagine why a negative value of the action integral might be prohibited. Am I correct in assuming that if we were to restrict all possible values of the energy to be real, it would that imply $$p(x)&gt;0\ $$ Can someone provide a proof as to why  $$p(x)&gt;0\rightarrow E\in\mathbb{R}. $$It would then follow the the integral of a strictly positive integrand must itself be positive, which means we cannot have $-\frac{\pi}{2}\hbar$.
17652
I've just finished a Classical Mechanics course, and looking back on it some things are not quite clear. In the first half we covered the Lagrangian formalism, which I thought was pretty cool. I specially appreciated the freedom you have when choosing coordinates, and the fact that you can basically ignore constraint forces. Of course, most simple situations you can solve using good old $F=ma$, but for more complicated stuff the whole formalism comes in pretty handy.  Then in the second half we switched to Hamiltonian mechanics, and that's where I began to lose sight of why we were doing things the way we were. I don't have any problem understanding the Hamiltonian, or Hamilton's equations, or the Hamilton-Jacobi equation, or what have you. My issue is that I don't understand why would someone bother developing all this to do the same things you did before but in a different way. In fact, in most cases you need to start with a Lagrangian and get the momenta from $p = \frac{\partial L}{\partial \dot{q}}$, and the Hamiltonian from $H = \sum \dot{q_i}p_i - L$. But if you already have the Lagrangian, why not just solve the Euler-Lagrange equations?  I guess maybe there are interesting uses of the Hamiltion formalism and we just didn't do a whole lot of examples (it was the harmonic oscillator the whole way, pretty much). I've also heard that it allows a somewhat smooth transition into quantum mechanics. We did work out a way to get Schrödinger's equation doing stuff with the action. But still something's not clicking.  My questions are the following: Why do people use the Hamiltonian formalism? Is it better for theoretical work? Are there problems that are more easily solved using Hamilton's mechanics instead of Lagrange's? What are some examples of that?