In QM particle on a ring, the Schrodinger in cylindrical coordinates is given by $$\frac{d^2\psi}{d\phi^2}=\frac{-2IE}{\hbar^2}\psi$$ where $I$ is the moment of inertia. Setting $ k^2=2IE/\hbar^2$, the solution is supposed to be $\psi_k(\phi)=e^{ik\phi}/\sqrt{2\pi}, \space k\subseteq\Bbb{Z}\tag{1}$ But the general solution is:$$\psi(\phi)=c_1e^{ik\phi}+c_2e^{-ik\phi}\tag{2}$$ I am unable to derive eq(1) from eq(2) using: Normalization: $\int_0^{2\pi}|\psi|^2d\phi=1$ $\space$and Cyclic boundary condition :$\psi(\phi)=\psi(\phi+2\pi)$. Can anyone guide me with the math? Edit: I think that in , OP asks why is $c_-$ is taken as zero. As far as I know we can't take any constant to be zero. I am looking to derive eq(1) systematically, only from the two conditions stated. I think that by solving the integrals, we will eventually get $c_1=\frac{1}{\sqrt{2\pi}}$ and $c_2=0$.
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
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 $-\frac{\pi}{2}\hbar$.
19824
I am new to QFT, so I may have some of the terminology incorrect.  Many QFT books provide an example of deriving equations of motion for various free theories. One example is for a complex scalar field: $$\mathcal{L}_\text{compl scaclar}=(\partial_\mu\phi^*)(\partial^\mu\phi)-m^2\phi^*\phi.$$ The usual "trick" to obtaining the equations of motion is to treat $\phi$ and $\phi^*$ as separate fields. Even after this trick, authors choose to treat them as separate fields in their terminology. This is done sometimes before imposing second quantization on the commutation relations, so that $\phi$ is not (yet) a field of operators. (In particular, I am following the formulation of QFT )  What is the motivation for this method of treating the two fields as separate? I intuitively want to treat $\phi^*$ as simply the complex conjugate of $\phi,$ not as a separate field, and work exclusively with $\phi$.  Is it simply a shortcut to obtaining the equations of motion $$(\square +m^2)\phi=0\\ (\square + m^2)\phi^*=0~?$$  I also understand that one could write $\phi=\phi_1+i\phi_2$ where the two subscripted fields are real, ; perhaps this addresses my question in a way that I don't understand.
88883
Consider a  $\phi(x)$, and let its  be $$ \frac{1}{p^2-m^2-\Pi(p^2)}=\int_0^\infty\mathrm d\mu^2\ \rho(\mu^2)\ \frac{1}{p^2-\mu^2} $$  We usually have $\Pi(m^2)=0$, which implies that there is a one-particle state with mass $m^2$. In terms of the , this reads $$ \rho(\mu^2)=\delta(\mu^2-m^2)+\sigma(\mu^2) $$ where the support of $\sigma(\mu^2)$ is disconnected from $m^2$.  On the other hand, the continuum contribution usually begins at $(2m)^2$, meaning that $\Pi(k^2)$ has a branch cut from $(2m)^2$ to $\infty$. Again, in terms of the spectral function, this is written as $\text{supp}(\sigma)=[(2m)^2,\infty)$.  My question  I would expect that the first bound state to lie close to $(2m)^2$, but slightly below, something around $(2m)^2-\mathcal O(\alpha^n)$, where $\alpha$ is the coupling constant and $n\in\mathbb N$. For example, in a simplified QED model, the bound state of an  has binding energy $\sim 7\mathrm{eV}\sim m\alpha^2$, and so I would expect the branch point to be somewhere around $(2m)^2-m^2\alpha^4$.  I know the value of the self-energy to one loop, and the branch point is indeed exactly at $(2m)^2$. I don't know where to find higher loop corrections, so I cannot conclude whether   1) at higher loops, the branch point moves a bit closer to $m^2$, or 2) it stays at $(2m)^2$.   I would expect that the correct behaviour is 1), but this contradicts the fact that all books I've read always depict the threshold of pair production as an hyperboloid with its bottom sitting exactly at $(2m)^2$. On the other hand, if it turns out that the correct option is 2), I am lead to ask why is perturbation theory unable to correctly predict the position of the branch point (I know this is a subtle issue because bound states are, to some extent, non-perturbative entities; but AFAIK, perturbative calculations do contain a lot of information about bound states).
89177
When trying to solve the Schrödinger equation for hydrogen, one usually splits up the wave function into two parts: $$\psi(r,\phi,\theta)= R(r)Y_{\ell,m}(\phi,\theta).$$ I understand that the radial part usually has a singularity for the $1s$ state at $r=0$ and this is why you remove it by writing: $$R(r) = \frac{u(r)}{r}$$ But what is the physical meaning of $$R(r=0) = \infty~?$$ Wouldn't this mean that the electron cloud is only at the centre of the atomic nucleus?
114596
I understand that the  is given by  $$F^{\mu\nu}\mapsto\begin{pmatrix}0 &amp; -E_{x} &amp; -E_{y} &amp; -E_{z}\\ E_{x} &amp; 0 &amp; -B_{z} &amp; B_{y}\\ E_{y} &amp; B_{z} &amp; 0 &amp; -B_{x}\\ E_{z} &amp; -B_{y} &amp; B_{x} &amp; 0 \end{pmatrix}$$  where $\mu$, $\nu$ can take the values {0,1,2,3} or {$t,x,y,z$}.  So, for example  $$F^{01}=F^{tx}=-E_x$$  My question is, what would the following expression be?  $$F^{t\rho}=?$$ or $$F^{z\rho}=?$$  where $\rho=\sqrt{x^{2}+y^{2}}$ is the radial coordinate in ?  And more generally, how can we construct the Electromagnetic Tensor in cylindrical coordinates? Where $\mu$, $\nu$ now take the values {$t,\rho,\varphi,z$}.
45781
In the book of Ashok Das, Field theory-path integral approach, he begin the demonstration of the Schwinger-Dyson equation using the fact that the $\delta Z[J]=0$, so  \begin{equation} \delta Z[J]=\int \mathcal{D} \phi \frac{\delta S[\phi,J]}{\delta \phi(x)} e^{iS[\phi,J]}=0, \end{equation} but we already know that  \begin{equation} \frac{\delta S[\phi, J]}{\delta \phi(x)}=F(\phi(x))-J(x), \end{equation} where $F(\phi(x))$ is the equation of motion.   So if we go back to the first equation and use the identification  \begin{equation} \phi(x)\rightarrow -i\frac{\delta}{\delta J(x)}, \end{equation}  we conclude that  $$ \int \mathcal{D}\phi\left(F(\phi(x))-J(x)\right)e^{iS[\phi,J]}=\left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)\int \mathcal{D}\phi e^{iS[\phi,J]} $$ $$ \left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)Z[J]=\left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)e^{iW[J]}=0. $$  But is here where I get lost, how did he pass from the above equation for  $$ e^{-iW[J]}\left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)e^{iW[J]}=F\left(\frac{\delta W[J]}{\delta J(x)}-i\frac{\delta}{\delta J(x)}\right)-J(x)=0. $$  A more important question is: what does this last equation mean at all, mathematically speaking, because the functional derivative now is acting on nothing.
5833
I simply have no idea how this result is found $$I=\hbar  \sqrt{\ell(\ell+1)}.$$ The result seems to just be dumped in textbooks rather than explained. I can get the result that $I_z=\hbar m_j$. Please explain as clearly as possible how this is derived.  Also, any pointers towards any books/videos that may help me learn quantum mechanics would be appreciated.
61285
I'm reading the derivation of the QED generating functional on Mandl&amp;Shaw, "Quantum Field Theory" 2nd ed., 12.5.2..  The authors start from the expression (schematically): $$Z[j]=\frac{\langle 0\vert S[j]\vert 0\rangle}{\langle 0 \vert S\vert 0\rangle}$$and derive the formula: $$Z[j] = \frac{e^{V}Z_0[j]}{e^{V}Z_0[0]},\\V=i\intop \text d^4 x \mathscr L _{\text {int}}(\frac{i\delta}{\delta j(x)}),$$which is precisely what one obtains from the path integral expression: $$Z[j]=\frac{\intop \text D[\phi]e^{iS[\phi]-i(j,\phi)}} {\intop \text D[\phi]e^{iS[\phi]}}.$$ Here $Z_0$ is the generating functional for the free theory. Let me know if my notation is not clear.  In doing so, they completely disregard the normal ordering prescription in the interaction density $$\mathscr L _{\text {int}}\to \colon \mathscr L _{\text {int}}\colon.$$ Now, omitting the normal ordering amounts to include equal time contractions in the application of Wick's theorem. At the end of page 271, the authors state (and prove) that the only diagram which arises from this omission in QED is the tadpole diagram:    which, in QED, is identically zero. Also, in exercise 12.7, they state the in scalar QED, the tadpole diagram leads (in second order) to a simple correction $\delta m^2$ to the self-energy of the charged scalar.  I'm wondering, how does this work in general, that is, what is the effect of including equal time contraction in the perturbative expansion of Green functions? Do they always simply lead to constant (i.e. momentum-independent) corrections to the self-energy, which are incorporated in the physical masses through renormalization?
110404
In reading Weinstock's Calculus of Variations, on  he explains how Schrodinger apparently first derived the Schrodinger equation from variational principles.  Unfortunately I don't think page 262 is showing so I'll explain the gist of it:  "In his initial paper" he considers the reduced Hamilton-Jacobi equation   $$\frac{1}{2m}\left[\left(\frac{\partial S}{\partial x}\right)^2 \ + \ \left(\frac{\partial S}{\partial y}\right)^2 \ + \ \left(\frac{\partial S}{\partial z}\right)^2\right] \ + \ V(x,y,z) \ - \ E \ = \ 0$$   for a single particle of mass $m$ in an arbitrary force field described by a potential $V = V(x,y,z)$.  With a change of variables $S \ = \ K\log(\Psi)$, (where $K$ will turn out to be $\hbar=h/2\pi$) it reduces to  $$\frac{K^2}{2m}\left[\left(\frac{\partial \Psi}{\partial x}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial y}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial z}\right)^2\right] \ + \ (V \ - \ E)\Psi^2 \ = 0.$$  Now instead of solving this he, randomly from my point of view, chose to integrate over space  $$I  = \iiint_\mathcal{V}\left(\frac{K^2}{2m}\left[\left(\frac{\partial \Psi}{\partial x}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial y}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial z}\right)^2\right] + (V - E)\Psi^2\right)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$   then Weinstock extremizes this integral which gives us the Schrodinger equation.  Apparently as the book then claims on page 264 it is only after this derivation that he sought to connect his idea's to deBroglie's wave-particle duality.  Thus I have three questions,  1) What is the justification for Feynman's famous quote:     Where did we get that [Schrödinger's equation] from? It's not possible to derive it from anything you know. It came out of the mind of Schrödinger.   The Feynman Lectures on Physics   in light of the above derivation. I note that all the derivations I've seen of the Schrodinger equation doing something like using operators such as $i\hbar\partial/\partial t = E$ to derive it always mention it's merely heuristic, yet what Schrodinger apparently originally did seems like a roundabout way of solving the Hamilton-Jacobi equation with no heuristic-ness in sight. What subtleties am I missing here? Why would I be a fool to arrogantly 'correct' someone who says Schrodinger is not derivable from anything you know?  Note I've read a ton of the threads on this topic on this forum &amp; none even go near the calculus of variations, in fact the above contradicts  that even refers to Schrodinger's Nobel prize lecture, so hopefully it's not a duplicate.  2) Is the mathematical trick Schrodinger has used something you can use to solve problems?   3) Why can't you use this exact derivation in the relativistic case?  Edit: On the topic of complex numbers:  On page 276, 14 pages after he explains what I posted, &amp; after he goes through how Schrodinger linked his work with DeBroglie's work, it is only then that Weinstock says:     In a more complete study of quantum mechanics than the present one the   admissibility of complex eigenfunctions $\Psi$ is generally shown to be   necessary. If $\Psi$ is complex, the quantity $|\Psi|^2$ is employed as the   position probability-density function inasmuch as $\Psi^2$ is not restricted   to real nonnegative values.   Apparently Schrodinger was able to do what I have posted using real-valued functions &amp; have K as I have defined it, without i. If you're following what Weinstock is saying he shows how the hydrogen atom energy levels are explainable without complex numbers, i.e. he is able to derive a physical interpretation of the eigenvalues (discrete energy levels) of the Schrodinger equation that were in accord with experiment (see Section 11.3 Page 279 on). As far as I understand it it is in trying to find a physical interpretation of the eigenfunctions that one is forced into complex numbers, though apparently, according to the book, it can be shown to be necessary. That is something to take into account!
113328
This relates a little bit to my previous question (), but it's different in a way and more historical.  One of the postulates of quantum mechanics is:     To every observable in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics ().   However, absent in the list of postulates is what these operators are.  So is the mathematical expression for each individual operator also a postulate that's not listed, or are they derivable from other axioms?  Specifically, where did  expressions come from?  Maybe the question can better be answered historically.  If I had to guess, I'd say it began with the wavefunction.  Some scientists thought perhaps the squared modulus of the wavefunction corresponded to charge distribution or something, but Born was the one who got it right.  He essentially made a guess that the squared modulus corresponded to the probability of finding a particle at a certain location.  What's funny is that he initially wrote down the wrong expression in his paper (), and his paper even got rejected by the first journal he submitted it to, but the right expression is in a corrective footnote.    But again, Born simply guessed what the squared modulus meant.  There was no procedure; he just thought of an idea and then experiment verified it to be correct.  Thus:  $$\langle\psi|\hat{X}|\psi\rangle = \langle\psi|x\rangle\langle x|\hat{X}|x\rangle\langle x|\psi\rangle = \int d\vec{x}\ \psi^*(\vec{x})\ \psi(\vec{x})$$  So basically in the position basis, the position operator is simply $\hat{X} = x$, which seems pretty intuitive.  I believe the momentum operator came about next; and it was "derived" by fiddling around with the de Broglie relation and just substituting classical expressions for momentum into the Schrodinger equation (can't remember where I read that).  The fact that plugging these classical ideas into a quantum equation just happens to match experiment for momentum measurements seems like a pure coincident to me (it's along the same lines as trying to make the Schrodinger equation relativistic; one way of shoehorning relativity into it gives you the Dirac equation and a different way gives you the Klein-Gordon equation).  I imagine the spin operator came about by simple observation.  An electron goes up or electron goes down in a magnetic field.  Not hard to guess what the expression for that operator might be.  So anyway, am I on the right track here?  Have all of the operators come about simply by guessing?  And if I'm wrong, how did they actually come about?  How would I come up with a new operator for a new type of measurement?
86806
I have the following problem that I was unable to solve for class, but I had a couple first steps that I started with that I am unable to finish.  I know I can't get this since it's already been turned in, but I would like to see if this was even a viable option for doing this proof in the first place.  ``Given an orthogonal set of states, $\{\phi_{n}\}$, and a Hamiltonian, $\hat{H}$, show that the partition function, $Q_{\beta}$, satisfies the following $$ Q_\beta \geq \sum_{n}\exp\{-\beta \langle \phi_n|\hat{H}|\phi_n\rangle \}$$ with equality holding when the $\phi_n$ states are eigenstates of the Hamiltonian.''  I started by dropping in the identity in the exponential (as eigenstates of the Hamiltonian) $$ \sum_n \exp\{ -\beta\sum_k \langle \phi_n|\psi_k\rangle\langle \psi_k |\hat{H}|\phi_n\rangle\}=\sum_n \exp\{ -\beta \sum_k E_k |c_{nk}|^2\}$$  Then I am left with showing that $$\sum_n \exp\{ -\beta \sum_k E_k |c_{nk}|^2\} \leq \sum_k \exp\{ -\beta E_k\}$$ with equality showing up again the same way, with a Kronecker delta $\delta_{nk}$ collapsing the sum in the exponential.  I realize I didn't make it very far, so this might not be the best way to show this, but it seems manifestly true just by looking at it, but I can't actually show it.  Does anyone have any hints about how to continue with this?
47199
In the theory of free scalar bosons (KG field) it is a basic result that the propagator $\Delta(p)$ has poles at $p^2=m^2$, with residue $1$ (or any other constant, depending on conventions). Thinking of $p^2$ as a variable is acceptable because it can be shown that the propagator is a function of $p^2$, even in the interacting case.  On the other hand, in the theory of free spinor fields (Dirac field), we can show that the propagator is not a function of $p^2$, but of $\not p$, so we can't use the $p^2$ trick. Conventions aside, the propagator is $$ S(p)=\frac{\not p+m}{p^2-m^2+i \epsilon} $$  It is clear that we cannot think of poles as $p^2\to m^2$, because $S(p)$ depends on $p$ in a non-so-trivial way. The usual solution is to think of $S$ as a function of $\not p$, as if $\not p$ was a number instead of a matrix, thus writing $$ S(\not p)=\frac{1}{\not p-m+i\epsilon} $$ and, again, we find a pole at $\not p=m$, with residue $1$. I think this is not mathematics. This is just nonsensical to me (may be I'm too skeptic, and we can give a meaning to this last formula).  In an interacting theory, we define the field strength normalisation $Z$ as the residue of the propagator at the poles: $$ \Delta(p^2)=\frac{Z}{p^2-m^2+i\epsilon}+\int_{M_\text{thresh}^2}^\infty \mathrm d\mu^2\rho(\mu^2)\frac{1}{p^2-\mu^2+i\epsilon} $$ $$ S(\not p)=\frac{Z}{\not p-m+i\epsilon}+\int_{M_\text{thresh}^2}^\infty \mathrm d\mu^2\frac{\not p\rho_1(\mu^2)+\mu \rho_2(\mu^2)}{\not p^2-\mu^2+i\epsilon} $$  Now, I don't understand what's the proper definition of $Z$ as a residue. In what sense is it a residue? What's the variable? I just can't accept it is a residue as $\not p\to m$. Do we really take this "think of $\not p$ as a variable'' seriously? Is it possible to formalise these matters, by thinking of a residue for an actual variable (as $p_0$)?  Perhaps the $\not p$ trick is just that: a trick, which may simplify calculations, but such that it can be shown that the actual result, found by a more standard procedure, is the same. Is this the case? If so, what is the correct procedure?  (I would really appreciate if the answers don't assume that we can boost to the rest frame of the particle, as I'd like it to be as general as posible. I want to take into account the possibility of $m=0$, so please don't boost into $k=(m,\boldsymbol 0)$ if it's not really necessary)
81203
I've been studying the postulates of QM and seeing how to derive important ideas from them. One thing that I haven't been able to derive from them, however, is the identity of the momentum operator.  For simplicity, I'm only thinking about no relativistic effects, no spin, no time-dependent potentials, and one spatial dimension. Also I'm assuming the position operator is simply multiplication by $x$, as in, I'm in position space. So the Hamiltonian operator is  $ H = -\frac{\hbar^2}{2m}\nabla^2+V$.  I know that the momentum operator is $p = -i\hbar \frac{\partial}{\partial x}$.  But how do I get there from the postulates? I know that it makes sense, as it results in the Ehrenfest Theorem, the De Broglie wavelength hypothesis, the Heisenberg Uncertainty Principle (for $x$ and $p$), the momentum operator being the generator of the translation operator, and possibly many other desirable theorems, and correlations with classical momentum.  But none of these are postulates (at least, not in the various formalisms I encountered), so you can't derive $p = -i\hbar \frac{\partial}{\partial x}$ from them. Rather, they are consequences of it. You need to know the operator beforehand to see that they are correct. Yes, this is just semantics, but that is the core issue for me:  Regardless of how much sense it makes, is the identity $p = -i\hbar \frac{\partial}{\partial x}$ (under the assumptions I made) a Postulate, meaning that you can't derive it from other postulates, or can it in fact be obtained from them? And in the latter case, could you show me how?  Note: I know that there are many different and equivalent sets of postulates for QM. But in none that I saw did they name it as a postulate nor properly derived it.
60815
Assuming the eigenvalue of position operator $\hat x$ equal to $k$, can I not write:  $$\begin{align} \langle\psi_n|x|\psi_m\rangle &amp;= \langle x\psi_n|\psi_m\rangle \\ &amp;=\langle k\psi_n|\psi_m\rangle \\ &amp;=k\langle\psi_n|\psi_m\rangle \\ &amp;=k\delta_{nm}  \end{align}$$  But I know that $\langle x \rangle =0$ in case of even potentials (I don't know how that happens) and what I have written above is wrong, at least in case of even potentials.  Taking the example of infinite 1D square well, the states are : $$ \psi_{n} \left(x\right)=A\sin\left(\frac{nx\pi}{L}\right)dx $$ then, $$ \langle\psi_n|x|\psi_m\rangle =A\int_{-\infty}^{\infty}\sin\left(\frac{mx\pi}{L}\right)x\sin\left(\frac{nx\pi}{L}\right)dx $$ If m=n=1, $$ &lt;x&gt;=A\int_{-\infty}^{\infty}x\sin^{2}\left(\frac{x\pi}{L}\right)dx $$ if we apply an even potential then the equation gets reduced to$$ &lt;x&gt;=\frac{1}{L}\int_{-L}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=0 $$ while in case of a potential(neither even nor odd), the equation leads to $$&lt;x&gt;=\frac{2}{L}\int_{0}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=L/2 ~? $$  Here $n=m=1$ but $ &lt;x&gt;=0 $ for even potentials, which is confusing me! It should be $k$ right?
58203
I stumbled on this question while answering the following question:  Given a general wavefunction $\psi(x)$ and an operator $\hat Q$, with eigenvalue spectrum $\{q_r\}$ and corresponding eigenfunctions $\{u_r\left(x\right)\}$, show that the expectation value of $\hat Q$ is given by $$\langle\hat Q\rangle = \int_{-\infty}^\infty{\psi^*\hat Q \psi \,dx}$$  I started of with the ususal definition of the expected value and inserted the identity operator in position representation: $$\langle\hat Q\rangle = \langle \psi|\hat Q|\psi\rangle = \langle \psi|\left(\int dx|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\right)|\psi\rangle$$  Taking $\langle\psi|$ and $|\psi\rangle$ into the integral gives: $$\left(\int dx\langle \psi|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\psi\rangle\right)$$  Pulling the 2nd integral into the first:$$\int dx\int  {dx}'\langle \psi|x\rangle\langle x'|\psi\rangle\langle x| \hat Q|x'\rangle$$  This is the point where I'm stuck. If $\langle x|\hat Q|x'\rangle = \hat Q \delta(x-x')$, then the last expression would evaluate to the answer. However I  don't know why this should be true and even if it is true I wouldn't know how to prove this equality.
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.