Solving the Schrodinger equation for hydrogen atom we arrive to the conclusion that quantum numbers $n$ and $\ell$ have the relation $$\ell=0,1,...,n-1.$$   Now,since we can not solve the Schrodinger equation for many electrons atoms we can approximate the potential $V(r_i)$ to be central than Hamiltonian becomes $$ \hat H= \sum_i \left( -\frac{\hbar^2}{2m} \nabla_i^2 +V(r_i) \right).$$  So for each electron we have the following equation  $$-\frac{\hbar^2}{2m} \nabla_i^2\psi_i +V(r_i)\psi_i=E\psi_i.$$  And solution is of the form   $$\psi_i=R_i^{n,\ell}(r)Y^m_{\ell}(\theta, \phi).$$  My question is why quantum numbers $n$ and $\ell$ have the relation $$\ell=0,1,...,n-1.$$  Is this an assumption or can we proof it?
88287
The solutions of the Schrödinger equation for hydrogen are the "electronic orbitals", shown in this picture:  ()  They have the following degeneracy structure:  ()  It is often said that atoms in other elements simply have more electrons "filling up" the orbitals in increasing order of energy, as required by the Pauli exclusion principle.  This is somewhat confirmed via xray spectroscopy:  ()  But why would non-hydrogen atoms have the same orbital degeneracy structure as hydrogen, if the actual Hamiltonian is different from one atom to another?
2248
I was playing around with a 3-D potential $V$ such that $V_{(r)} = 0$ for $r&lt;a$, and $V_{(r)} = V_0&gt;0$ otherwise. By using the Schrödinger Equation, I showed that: $$\frac{-\hbar}{2m}\frac{1}{r^2}\frac{d}{dr}\bigl(  r^2\frac{d}{dr}\bigr)\psi = E\psi$$  I then used the substitution $\psi_{(r)}=f_{(r)}/r$ and $k=\sqrt{2mE}/\hbar$ to get: $$\frac{1}{r}\frac{d^2f_{(r)}}{dr^2}=-\frac{k^2}{r}f_{(r)} \tag{I}$$  which describes the wavefunction $\psi_{(r)}=f_{(r)}/r$ inside the sphere. Hence, the differential equation has the domain $0\leq r&lt;a$, and I cannot multiply both sides by $r$. This is unfortunate, because there is a similar equation for the outside of the sphere: $$\frac{1}{r}\frac{d^2f_{(r)}}{dr^2}=\frac{k'^2}{r}f_{(r)}$$ As this is outside the sphere, I can multiply both sides by $r$ to get a familiar differential equation that can be solved easily: $$\frac{d^2f_{(r)}}{dr^2}=k'^2f_{(r)}$$  If I do the same thing to $(I)$, I obtain the equation for simple harmonic motion, but substituting the solution back into $(I)$ as a sanity check gives a division by zero when evalutating for $r=0$. After that, I tried a number of substitutions to make $(I)$ have a more recognisable form - to no avail. Then I had the idea of multiplying my trial solution by some other function of $r$ so that upon substitution into $(I)$, the evaluation of $r=0$ doesn't give an infinity... but I don't know quite how to do that...  Long story short..... my question is: what trick do I need to get a meaningful solution to $(I)$?
43963
In the Schrödinger equation, at any given time $t$ we should jointly add another sub equation, like $$||\psi_t(x)|| = 1$$ where $\psi_t(x) = \Psi(x,t)$, and then try to solve the two equations simultaneously. Why not? I know it does not yield, but I am always baffled, who is doing the normalization of the wave function? Observers, the system, the measuring process, God?
92904
I'm an undergraduate mathematics student trying to understand some quantum mechanics, but I'm having a hard time understanding what is the status of the Schrödinger equation.  In some places I've read that it's just a postulate. At least, that's how I interpret e.g. the following quote:     Where did we get that (equation) from? Nowhere. It is not possible to derive it from anything you know. It came out of the mind of Schrödinger.    -- Richard Feynman   (from the Wikipedia entry on the Schrödinger equation)  However, some places seem to derive the Schrödinger equation: just search for "derivation of Schrödinger equation" in google.  This motivates the question in the title: Is the Schrödinger equation derived or postulated? If it is derived, then just how is it derived, and from what principles? If it is postulated, then it surely came out of somewhere. Something like "in these special cases it can be derived, and then we postulate it works in general". Or maybe not?  Thanks in advance, and please bear with my physical ignorance.
121361
It is well established that:   'In quantum mechanics, the behavior of an electron in an atom is described by an orbital, which is a probability distribution rather than an orbit.   There are also many graphs describing this fact: : (hydrogen atomic orbital - one electron) In the figure, the shading indicates the relative probability to "find" the electron, having the energy corresponding to the given quantum numbers, at that point.  My question is: How do these symmetries shown in the above article occur?                 What about the 'preferable' axis of symmetries? Why these?
109718
If we do the wick rotation such that τ = it, then Schrödinger equation, say of a free particle, does have the same form of heat equation. However, it is clear that it admits the wave solution so it is sensible to call it a wave equation.   Whether we should treat it as a wave equation or a heat equation with imaginary time, or both? If it is a wave equation, how do we express it in the form of a wave equation? Is there any physical significance that Schrödinger equation has the same form of a heat equation with imaginary time? For example, what is diffusing?
93850
Given a single-particle system with Hamiltonian $H$, what constraints can be put on the wave function at a particular point in time $\psi(x)$? Of course $\psi(x)$ must obey boundary conditions given by $H$. However, in situations where $H$ does not yield strict boundary conditions (e.g. the harmonic oscillator) can $\psi(x)$ be any normalised function? My intuition says no for the following reason: QM textbooks say that any valid wave function can be represented as a linear combination of the eigenstates of $H$. In the case of the harmonic oscillator, the eigenstates of $H$ are a countably infinite set. However the set of all normalised functions is uncountably infinite. It seems to me that one cannot change between one complete set of basis functions to another and reduce the "dimensionality" somehow. This makes me think that the eigenstates of $H$ are not complete, that they imply some constraints on $\psi(x)$. This is not an area of mathematics that I understand well. Can somebody help me out?
65433
Why that exact number of electrons in one coulomb? who decided it? there is nothing wrong with the number, it just seems slightly messy. Why didn't the scientific community just settle on an easier number, such as $1\times10^{-19}$ for example?
21409
Given some potential $V$, we have the eigenvalue problem   $$ -\frac{\hbar^2}{2m}\Delta \psi + V\psi = E\psi $$   with the boundary condition   $$ \lim_{|x|\rightarrow \infty} \psi(x) = 0 $$   If we wish to seek solutions for   $$ E &lt; \lim_{|x|\rightarrow \infty}V(x), $$   I find textbooks like Sakurai and others assert that (without proof) set of all such eigenvalues $E$ that admits nontivial solutions is discrete, which gives us liberty to index them with natural numbers (principal quantum numbers). Can anyone help me with a little elaboration of this mathematical fact ? As a student of mathematics, I know that compact operators have discrete spectrums, so for the operator $ T : L^2 \rightarrow L^2 $given as  $$ T(\psi) = -\frac{\hbar^2}{2m}\Delta \psi + V\psi. $$   Are there methods to show that $T$ is compact? Or possibly other ways to establish discreteness of its eigenspectrum?
108406
Suppose $\mathcal{B}=\{|0\rangle, |1\rangle, |2\rangle, ... \}$ is the energy eigen-basis of a quantum simple harmonic oscillator. I want to create the state \begin{equation} |\Psi\rangle = c_{0}|0\rangle + c_{1}|1\rangle + c_{2}|2\rangle + ... \end{equation} Besides the normalization condition \begin{equation} \langle\Psi|\Psi\rangle = |c_{0}|^2+ |c_{1}|^2+ |c_{2}|^2+...=1, \end{equation} are there any other mathematical constrains on the complex amplitudes $c_0,c_1,c_2,...$?  Also, is there a protocol, perhaps from quantum optics, for creating an arbitrary $|\Psi\rangle$ as defined above?   For example, the complex amplitudes which define the state can be taken as \begin{equation} \{c_n\}=\{ \frac{1}{\sqrt{\zeta\left(3\right)}}\frac{1}{\left(1+n\right)^{3/2}} \}, \end{equation} where $\zeta\left(3\right)\approx 1.202$ is given by the . In this case, the normalization condition holds. Also, we get a nice finite value for the mean energy \begin{equation} \langle\Psi|\hat{n}|\Psi\rangle \approx0.019, \end{equation} where $\hat{n}=\hat{a}^{\dagger}\hat{a}$ is the number operator (i.e. the Hermitian operator corresponding to the eigen-basis $\mathcal{B}$). However,the variance in the energy \begin{equation} \text{Var}[\hat{n}]=\langle\Psi|\hat{n}^2|\Psi\rangle-\langle\Psi|\hat{n}|\Psi\rangle^2  \end{equation} is infinite. Can such a state exist in nature?
59403
Is there a good physical picture of why the energy levels in a hydrogen atom are independent of the angular momentum quantum number $\ell$ and $m$?
97545
I have found the hydrogen wave functions and would now like to calculate the function that describes the orbitals so that I can plot this function and see how they look.  I don't know how I can do that and it is crazy how I cannot find anything on the subject on the internet as if there was no relation between the wave functions and the orbitals.  Could you please tell me what I need to do to get a function $r(\theta,\phi)$ out of my wave functions $\psi(r,\theta,\phi)$.
112674
Why can't the Klein-Gordon equation with a Couloumb potential describe the hydrogen atom?  Why can the first order Dirac equation explain it? What are the failures?
32722
I'm having trouble finding a simple answer to this question (maybe because there isn't one), but I'm just confused about how the Schrödinger Equation would look for two electrons. I understand that it would exist in 6 dimensional configuration space, but how does the potential look? It's confusing that the potential would be different for each electron depending on where the other one is.
106623
I wonder what is the number of different possible values that the total angular momentum quantum number $j$ can take for fixed values of the angular quantum number $\ell$ and the spin quantum number $s$. If $j$ can take the following range of values, jumping only in integer steps, $$|\ell-s| \leq j \leq \ell+s$$ is there any formula that gives the number of different values it can take?
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 $S=1-1=0$?
91797
Consider the Hilbert space  $$\ell^2=\{(x_1,x_2,...,x_n),x_i\in\mathbb C\text{ for all }i\text{ and }\sum_{i=1}^\infty |x_i|^2&lt;\infty\}$$ with the inner product $$\langle(x_1,x_2,\dots,x_n)(y_1,y_2,\dots,y_n)\rangle = \sum_{i=1}^\infty x_i \overline y_i.$$ Ddefine $T:\ell^2 \to \ell^2$ by  $$T((x_1,x_2,...,x_n))=(x_1,\frac{x_2}2,\frac{x_3}3,...).$$ Then $T$ is a)neither self-adjoint or unitary b)both self-adjoint and unitary c)unitary but not adjoint d)self-adjoint but unitary  $$\langle Tx,y\rangle=\langle T((x_1,x_2,...,x_n)),(y_1,y_2,...,y_n)\rangle= \langle (x_1,x_2/2,x_3/3,...),(y_1,y_2,..,y_n)\rangle= \sum (x_i\overline y_i/i)= \sum x_i(\overline y_i/i)= \langle (x_1,x_2,...,x_n),(y_i,y_2/2,y_3/3,..y_n/n)\rangle=\langle x,Ty\rangle$$ so T is self-adjoint