I was thinking the other day, if you had the Schrodinger equation in 3-dimensions, and had a spherically symmetrical potential. Ie.: $$-\frac{ℏ^{2}}{2m}∇^{2}ψ+V(r)ψ=Eψ$$ Then you could simplify the Laplacian to: $$-\frac{ℏ^{2}}{2m}\frac{1}{r^{2}}\frac{∂}{∂r}\left(r^{2}\frac{∂ψ}{∂r}\right)+V(r)ψ(r)=Eψ(r)$$ The solution to this equation, doesn't correspond to the WKB solution, which if I recall correctly, is: $$\frac{A}{\sqrt{p\left(r\right)}}e^{\left(\frac{i}{ℏ}\int_{ }^{ }p\left(r\right)\ dr\right)}$$ Unless, you somehow remove the first-order term that turns up in the Laplacian - which I vaguely remember somewhere that you take the limit as $r$ approaches infinity. I don't know if this is what is suppose to be done, but if it is, then how is it viable? And if not, then could someone please explain how to solve this discrepancy. Edit: I want the states to also be spherically symmetrical. EDIT: I think I've found the solution, I made the substitution $$\frac{w\left(r\right)}{r}=ψ\left(r\right)$$ I then substituted this into the previous radial equation, and it gave me the regular TISE in 1 dimensions. This ultimately yields the WKB solution for ψ, which is : $$ψ\left(r\right)=\frac{1}{r}\frac{A}{\sqrt{p\left(r\right)}}e^{\left(\frac{i}{ℏ}\int_{ }^{ }p\left(r\right)\ dr\right)}$$
121390
I'm trying to learn how to apply the WKB approximation. Given the following problem:      An electron, say, in the nuclear potential   $$U(r)=\begin{cases}  &amp; -U_{0} \;\;\;\;\;\;\text{ if } r &lt; r_{0} \\   &amp; k/r \;\;\;\;\;\;\;\;\text{ if } r &gt; r_{0} \end{cases}$$   1. What is the radial Schrödinger equation for the $\ell=0$ state?      2. Assuming the energy of the barrier (i.e. $k/r_{0}$) to be high, how do you use the WKB approximation to estimate the bound state energies inside the well?     For the first question, I thought the radial part of the equation of motion was the following  $$\left \{ - {\hbar^2 \over 2m r^2} {d\over dr}\left(r^2{d\over dr}\right) +{\hbar^2 \ell(\ell+1)\over 2mr^2}+V(r) \right \} R(r)=ER(r)$$  Do I simply just let $\ell=0$ and obtain the following? Which potential do I use?  $$\left \{ - {\hbar^2 \over 2m r^2} {d\over dr}\left(r^2{d\over dr}\right) +V(r) \right \} R(r)=ER(r)$$  For the other question, do I use $\int \sqrt{2m(E-V(r))}=(n+1/2)\hbar π$, where $n=0,1,2,...$ ? If so, what are the turning points? And again, which of the two potentials do I use?
77033
What is $V_{\alpha\beta}$?  And what is a symmetric,  potential energy matrix?  And why is there a linearized equation like this?
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 $-\frac{\pi}{2}\hbar$.
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A form of the  is  $$     \left[-\frac{\hbar^2}{2m} \nabla^2 + V(\vec{r}, t)\right]\Psi = i\hbar \frac{\partial}{\partial t} \Psi $$  The bracketed term is of course the Hamiltonian, and the whole equation can hence be written as  $$     \hat{H}\Psi = i\hbar \frac{\partial}{\partial t} \Psi $$  My question is, is there a existing/named operator represented by $i\hbar \frac{\partial}{\partial t}$? That is, can we further rewrite the equation as  $$     \hat{H}\Psi = \hat{\Theta}\Psi $$  For some predefined operator $\hat{\Theta} = i\hbar \frac{\partial}{\partial t}$?
116108
The general formulation of quantum mechanics is done by describing quantum mechanical states by vectors $|\psi_t(x)\rangle$ in some Hilbert space $\mathcal{H}$ and describes their time evolution by the Schrödinger equation $$i\hbar\frac{\partial}{\partial t}|\psi_t\rangle = H|\psi_t\rangle$$ where $H$ is the Hamilton operator (for the free particle we have $H=-\frac{\hbar^2}{2m}\Delta$).  Now I have often seen used spaces like $\mathcal{H}=L^2(\mathbb{R}^3)$ (in the case of a single particle), but I was wondering whether this is correct or not.  In fact shouldn't we require to be able to derivate $\left|\psi_t\right&gt;$ twice in $x$ and thus choose something like $\mathcal{H} = H^2(\mathbb{R}^3)$?   If we treat directly $\psi(t,x) := \psi_t(x)$, shouldn't we require them to be in something like $H^1(\mathbb{R};H^2(\mathbb{R}^3))$? i.e., functions in $H^1(\mathbb{R})$ with values in $H^2(\mathbb{R}^3)$, e.g. the function $t\mapsto\psi_t$.
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I am studying Perturbation Theory from J.J. Sakurai's textbook Modern Quantum Mechanics. I am having trouble proving formulas on page 299 (5.2.5) and (5.2.6) from the previous ones [mainly (5.2.4)]. Can anyone help?  This is a problem of working with perturbation theory when the non-perturbed state has degeneracy. To elaborate I do not get why $(E-H_0-\lambda P_1VP_1)$ is not singular.  More importantly is the equation below correct and if so why:  $$(E-\lambda P_1VP_1)|l \rangle = E_D^{(0)}|l \rangle $$  That is the core of my problem. If the above equation - in the framework of Sakurai's statement of the problem - is correct I can work out all the rest on my own. However I do not get why this equation would hold true.
85360
I was reading Michio Kaku's Beyond Einstein. In it, I think, he explains that when physicsts treat a particle as a geometric point they end up with infinity when calculating the strength of the particle's field as you approach the particle.  First, did I get that part right?  Second, how does (in very simple terms) renormalization try to fix this in Quantum Mechanics? If you can include some type of word picture that would be great. Accepted answer will go to the clearest explanation.  Update  I got a little further in the book (lol) and Kaku talks about using symmetry to remove the divergences in the math. I'd appreciate an answer that incorporates this also. Thanks guys!
61693
Let me remind you about the following classical  examples in quantum mechanics.     Example 1. Bound states in 1-dim potential V(x).   Let $V(x)$ be a symmetric potential i.e.   $$V(x) = V(-x)$$   Let us introduce the parity operator $\hat\Pi$ in the following way:   $$\hat\Pi f(x) = f(-x).$$   It is obvious that    $$[\hat H,\hat\Pi] = 0.$$   Therefore, for any eigenfunction of $\hat H$ we have:   $$\hat H|\psi_E(x)\rangle = E|\psi_E(x)\rangle = E\hat\Pi|\psi_E(x)\rangle,$$   i.e. state $\hat\Pi|\psi_E(x)\rangle$ is eigenfunction with the same eigenvalue. Is $E$ a degenerate level? No, because of linear dependence of $|\psi\rangle$ and $\hat\Pi|\psi\rangle.$   Consider the second example.     Example 2. Bound states in 3-dim a potential $V(r)$. Where $V(r)$ possesses central symmetry, i.e. depends only on distance to center.   In that potential we can choose eigenfunction of     angular momentum $\hat L^2$ for basis    $$|l,m\rangle,$$   where $l$ is total angular momentum and $m$ - its projection on chosen axis (usually $z$). Because of isotropy eigenfunction with different $m$ but the same $l$ correspond to one energy level and linearly independent. Therefore, $E_l$ is a degenerate level.      My question is if there is some connection between symmetries and degeneracy of energy levels. Two cases are possible at the first sight:   Existence of symmetry $\Rightarrow$ Existence of degeneracy Existence of degeneracy $\Rightarrow$ Existence of symmetry     It seems like the first case is not always fulfilled as shown in the first example. I think case 1 may be fulfilled if there is continuous symmetry. I think the second case is always true.
119915
I have almost no background in physics and I had a question related to Schrodinger's Equation. I think, it is not really research level so feel free to close it, but I would request you to kindly suggest some existing literature which can help me develop a better understanding for the same.   While reading up about it from an introductory text on Quantum Physics, I wondered for a little while how does one derive this equation. Very soon, the author () added the detail that he just took it on faith when he was himself a student. He goes on to say Schrodinger equation rightly predicts the behavior of atomic transitions etc and people believe that it is a fundamental law of nature for quantum systems.  Then he raises the question himself which I wanted to know the answer to. Namely     What made Schrodinger write such an equation which became a fundamental equation, not to be derived from more fundamental equations?   He then adds that it will be an interesting topic for students of history of science which does not answer my question. Could you please try answering this question (or is this question really useless) ?  Thanks for your time
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I am aware of the debate on whether Schrödinger equation was derived or motivated. However, I have not seen this one that I describe below. Wonder if it could be relevant. If not historically but for educational purposes when introducing the equation.   Suppose that we have the time dependent Schrödinger equation for a free particle, $V=0$.   $$-\frac {\hbar i}{2m} \nabla^2 \Psi_\beta = \frac {\partial \Psi_{\beta}}{\partial t} $$  As the particle moves its heat is diffused throughout space. Now consider that we consider Heat equation or in general Diffusion equation:  $$\alpha\nabla^2 u= \frac {\partial u}{\partial t} $$  Where $u$ is temperature.   Also we have particle diffusion equation due to Fick's second law.  $$D \frac {\partial^2 \phi}{\partial x^2}= \frac {\partial \phi}{\partial t} $$  Where $\phi$ is concentration.  Furthermore, probability density function obeys Diffusion equation. So as the free particle moves, the heat, the temperature, or the density is diffused.   Now we can motivate Schrödinger equation in an intuitive way. Mathematically it is describing the same diffusion.  Am I right? Have you seen more like this motivation elsewhere?
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A typical example in textbooks about the application of Hellmann–Feynman theorem is calculating $\left\langle\frac{1}{r^2}\right\rangle$ in hydrogen-like atoms.  has a nice demonstration of this. At some point in the Wikipedia derivation is used that $$\frac{\partial n}{\partial \ell}~=~1. \tag{1}$$   But why is eq. (1) true? I know that  $$n=n_r+\ell+1,$$   but $n_r$ is just another variable with different physical meaning, so why is $n_r$ independent from $\ell$, whereas $n$ is not? The Wikipedia proof for Hellmann–Feynman theorem does not address the problem of independence of different parameters. What variables are kept fixed during the differentiation $(1)$ and why?  The Wikipedia page seems to have only a vague notion of $\frac{\partial\hat{H}}{\partial\lambda}$ and $\frac{\partial E}{\partial\lambda}$, unlike in, e.g., thermodynamics, where all partial derivatives are typically written like  $$\left( \frac{\partial U}{\partial V} \right)_S \qquad ,\qquad  \left( \frac{\partial U}{\partial V} \right)_p \qquad ,\qquad  \ldots $$ so that it is clear which variables are kept fixed during the differentiations.
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In Griffiths' Intro to QM [1] he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being   $$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right)$$  (cf. last formula on p. 101). He then says that these eigenfunctions are not square integrable because   $$\int_{-\infty}^{\infty}g_{\lambda}\left(x\right)^{*}g_{\lambda}\left(x\right)dx ~=~\left|B_{\lambda}\right|^{2}\int_{-\infty}^{\infty}\delta\left(x-\lambda\right)\delta\left(x-\lambda\right)dx ~=~\left|B_{\lambda}\right|^{2}\delta\left(\lambda-\lambda\right) ~\rightarrow~\infty$$  (cf. second formula on p. 102). My question is, how does he arrive at the final term, more specifically, where does the $\delta\left(\lambda-\lambda\right)$ bit come from?   My total knowledge of the Dirac delta function was gleaned earlier on in Griffiths and extends to just about understanding   $$\tag{2.95}\int_{-\infty}^{\infty}f\left(x\right)\delta\left(x-a\right)dx~=~f\left(a\right)$$  (cf. second formula on p. 53).  References:   D.J. Griffiths, Introduction to Quantum Mechanics, (1995) p. 101-102.
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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?
21670
I am trying to put the symplectic form of the 2-sphere defined by $\omega_u(v,w) := \langle u,v\times w\rangle,$ where $u \in \mathbb{S}^2$ and $v, w \in T_u\mathbb{S}^2$. I just can't do this in stereographic coordinates. Would you help me?
38405
I have the one dimensional free particle Schrödinger equation  $$i\hbar \frac{\partial}{\partial t} \Psi (x,t) =  -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi (x,t), \tag{1}$$  with general solution  $$\psi(x,t) = A e^{i(kx-\omega t)} + B e^{i(-kx-\omega t)}. \tag{2}$$  I'd expect that the solution is normalized:  $$ \int_{-\infty}^\infty |\psi(x,t)|^2 dx = 1. \tag{3}$$  But   $$|\psi(x,t)|^2 = \psi(x,t)\psi^* (x,t) = A^2 + B^2 + AB ( e^{2ikx} + e^{-2ikx} ), \tag{4}$$  and the integral diverges:  $$ \int_{-\infty}^\infty |\psi(x,t)|^2 dx = \frac{AB}{2ik} (e^{2ikx} - e^{-2ikx})\biggl|_{-\infty}^\infty + (A^2+B^2) x\biggl|_{-\infty}^\infty. \tag{5}$$  What is the reason for this? Can it be corrected?
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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.
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I was wondering if the good old quadratic potential was the only potential with equally spaced eigenvalues. Obviously you can construct others, such as a potential that is infinite in some places and quadratic in others, but that's only trivially different. I am not referring to equally spaced as a limiting behavior either, I mean truly integer spaced.  Any ideas? If not, is there a proof for its uniqueness?  If there are other potentials with equally spaced eigenvalues, can one use them as starting points for a free-field QFT? It would be interesting to know if there is a deeper mathematical relation between all of these potentials and whether they could be used to study interacting systems.