Given a harmonic oscillator $\ddot{x}+\omega_0^2x=0$ if we calculate the Wronskian using   we get $W[x_1,x_2](t)=W_0e^{\int_{t_0}^t 0 ds}=W_0$  which means (for our two solutions $x_1$ and $x_2$) that  $$ det\begin{pmatrix}  \dot{x_1} &amp; \dot{x_2} \\ x_1 &amp; x_2  \end{pmatrix}=\dot{x_1}x_2-\dot{x_2}x_1=W_0$$  is some conserved quantity. To me it smells like it should be equal to something of the form $\frac{E}{m}$ (because if we multiply $\dot{x_1}x_2-\dot{x_2}x_1$ by $m$ we get something in units of energy(we don't, I made a mistake, but anyway the question makes sense). Is the conservation of the Wronskian actually the conservation of something energy-ish (to the factor of $m$) for our system or is it something else, if not what is it (physically, what is its physical meaning)?  Ps: I think it probably isn't energy (it definitely isn't energy) because in the case of parametric resonance (aka.  $\ddot{x}+\omega_0^2(t)x=0$) the Wronskian is also conserved but the Energy isn't. I might be wrong :)  pss:  answer states that the conservation of the Wronskian means the conservation od angular moment, however I find it unclear what angular momentum would mean for a 1D system
121476
Noether's principle is the paradigm that symmetries of Hamiltonian and Lagrangian systems correspond to conservation laws of various kinds. Consider a one-dimensional harmonic oscillator $$\tag{*} \ddot{x} + x =0.$$ If $x_1, x_2$ are two solutions, we have the conservation of their Wronskian: $$ \frac d {dt}\left( x_1\dot{x}_2 -x_2\dot{x}_1\right) =0.$$     Does this conservation law correspond to a symmetry of the system? If yes, which one?    A more general question is the following. Consider a Hamiltonian system, that is, the following ODE  $$ \tag{**} \frac{ d}{dt} \begin{bmatrix} q \\ p \end{bmatrix} = J\nabla_{q,p} H(q, p; t), $$  where $J$ is an antisymmetric matrix. Here one has  which states that, if $\Omega(t)$ is a region of phase space that evolves with the flow of $(**)$, then  $$\frac d{dt}\text{Vol }\Omega(t) = \frac d{dt} \iint_{\Omega(t)}dqdp=0.$$ (The aforementioned conservation of the Wronskian is a special case of this theorem, obtained by taking $\Omega(t)$ equal to the parallelogram spanned by $(x_1, \dot{x}_1)$ and $(x_2, \dot{x}_2)$).    Same question as before:      does this conservation law follow from a symmetry, and if yes, which one?    The linked Wikipedia's page suggests that the conservation of volume in phase space follows from time translation invariance. This does not seem to me to be the case, because Liouville's theorem holds even in case of time-dependent Hamiltonians. The simplest example is a time dependent harmonic oscillator $\ddot{x}+b(t) x =0$. Here one still has conservation of the Wronskian: $$\frac d{dt}\left(x_1\dot x_2 - x_2 \dot x_1\right) =-b(t)x_1x_2 +b(t)x_1x_2 =0.$$
46296
In quantum mechanics, when two observables commute, it implies that the two can be measured simultaneously without perturbing each other's measurement results. Or in other words, the uncertainty in their measurements are not coupled.   But in classical mechanics, in an analogous way we have the Poisson brackets instead, where when two functions/variables of the systems have a vanishing Poisson bracket with each other, we say they are in "involution".  I understand that when a variable $j$ has a vanishing Poisson bracket with the Hamiltonian of the system, means that $j$ is a constant of motion, conserved in time. But for the general case:      If the Poisson bracket of $f$ and $g$ vanishes ($\{f,g\} = 0$), then $f$ and $g$ are said to be in involution.    What does the above mean physically? Is it in some sense related to measurement results as it is with commutators in QM, meaning two variables in involution are completely independent of each other? Finally, is there a simple classical example, where one can see 2 observables are in involution, but 2 other (same system) may not be (i.e. counter-example for involution)? I think it would help better understand what in "involution" means.
65260
When I studied QM I'm only working with time independent Hamiltonians. In this case the unitary evolution operator has the form $$\hat{U}=e^{-\frac{i}{\hbar}Ht}$$ that follows from this equation  $$ i\hbar\frac{d}{dt}\hat{U}=H\hat{U}. $$ And in this case, Hamiltonian in Heisenberg picture ($H_{H}$) is just the same as the Hamiltonian in Schrödinger picture ($H_{S}$), i.e. it commutes with $\hat{U}$. Now I have a Hamiltonian that depends explicitly on time. Specifically, $$H_{S}=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega \hat{q}^2-F_0 \sin(\omega_0t)\hat{q}.$$  And in my problem I need to calculate $H_H$ (Hamiltonian in Heisenberg picture).  I've found that differential equation for $\hat{U}$ (I've mentioned about it above.) has generally solution in the form (with $U(0)=1$) $$U(t)=1+\xi\int\limits_0^t H(t')\,dt'+ \xi^2\int\limits_0^t H(t')\,dt'\int\limits_0^t' H(t'')\,dt''+\xi^3\int\limits_0^t H(t')\,dt'\int\limits_0^t' H(t'')\,dt''\int\limits_0^t'' H(t''')\,dt'''+...$$  So my questions are:   Is there other ways to calculate $\hat{U}$, could give a link or tell me about them? If you know form of the solution for my case, please tell me. If you know any articles or papers articles on this topice, please link them to me, too.
75561
I read that $U=q+w$ and $H=u+PV$ so aren't $PV$ and $w$ same? If they are, can't we write $H=q+2w$? Also, $dU=$ change in internal energy wrt change in temperature keeping $V$ constant times $dT+$change in internal energy wrt change in volume keeping $T$ constant times $dV$ Here is the second term $=w$? And, $dH=$ change in enthalpy wrt change in temperature keeping $P$ constant times $dT+$ change in enthalpy wrt change in pressure keeping $T$ constant times $dP$. Here is the second term $=PV$?
103570
Textbooks often introduce the quantum harmonic oscillator with the example of a mass on a spring, giving the Hamiltonian $$H = \frac{1}{2} k x^2 + \frac{1}{2m}p^2 \qquad [x,p] = i \hbar \, .$$ The raising/lowering operators are then introduced as $$a = \sqrt{\frac{m \omega_0}{2 \hbar}} \left( x + \frac{i}{m \omega} p \right) \, ,$$ where $\omega_0$ is the oscillation frequency $\omega_0 = \sqrt{k/m}$. This is rather messy and unenlightening.  Is there a useful physical interpretation of the prefactors in definition of $a$? In other words, is there a way to write the creation/annihilation operators and other useful relations in a way that reveals more physical insight?
62771
It is often said that the classical charge $Q$ becomes the quantum generator $X$ after quantization. Indeed this is certainly the case for simple examples of energy and momentum. But why should this be the case mathematically?  For clarity let's assume we are doing canonical quantization, so that Poisson brackets become commutators. I assume that the reason has something to do with the relationship between classical Hamiltonian mechanics and Schrodinger's equation. Perhaps there's a simple formulation of Noether's theorem in the classical Hamiltonian setting which makes the quantum analogy precisely clear?  Any hints or references would be much appreciated!  Mathematical Background  In classical mechanics a continuous transformation of the Lagrangian which leaves the action invariant is called a symmetry. It yields a conserved charge $Q$ according to Noether's theorem. $Q$ remains unchanged throughout the motion of the system.  In quantum mechanics a continuous transformation is effected through a representation of a Lie group $G$ on a Hilbert space of states. We insist that this representation is unitary or antiunitary so that probabilities are conserved.   A continuous transformation which preserves solutions of Schrodinger's equation is called a symmetry. It is easy to prove that this is equivalent to $[U,H] = 0$ for all $U$ representing the transformation, where $H$ is the Hamiltonian operator.  We can equivalently view a continuous transformation as the conjugation action of a unitary operator on the space of Hermitian observables of the theory  $$A \mapsto UAU^\dagger = g.A$$  where $g \in G$. This immediately yields a representation of the Lie algebra on the space of observables  $$A \mapsto [X,A] = \delta A$$  $$\textrm{where}\ \    X \in \mathfrak{g}\ \ \textrm{and} \ \  e^{iX} = U \ \  \textrm{and} \ \  e^{i\delta A} = g.A$$   $X$ is typically called a generator. Clearly if $U$ describes a symmetry then $X$ will be a conserved quantity in the time evolution of the quantum system.  Edit  I've had a thought that maybe it's related to the 'Hamiltonian vector fields' for functions on a symplectic manifold. Presumably after quantization these can be associated to the Lie algebra generators, acting on wavefunctions on the manifold. Does this sound right to anyone?
114639
I am reading "The undivided universe: an ontological interpretation of quantum theory" and cannot understand this derivation.  From the Schrödinger equation: $$ i\hbar \frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi + V\psi $$ They derive the quantum Hamilton-Jacobi equation $$ \frac{\partial S}{\partial t} + \frac{(\nabla S)^2}{2m} + V + Q = 0 $$  using the polar form $\psi = Re^{iS/\hbar}$, but I don't understand what they do afterwards.
67979
Background of the question (see pp. 161, section 47 in Landau &amp; Lifshitz's quantum mechanics textbook Vol3, 2nd Ed. Pergamon Press). We have the following potential well $$U(x)\leq E \quad\text{for} \quad x \leq a ,$$ $$U(x)&gt;E \quad\text{for} \quad x&gt;a .\tag{47.0}$$ The WKB solutions right and left to the turning point are $$\begin{align}\psi=&amp;\dfrac{C}{2\sqrt{p}}\exp{\left(-\dfrac{1}{\hbar}\left|\int_a^x pdx\right|\right)} \cr &amp;\quad \text{for} \quad x&gt;a, \end{align}  \tag{47.1}$$ $$\begin{align} \psi= &amp; \dfrac{C_1}{\sqrt{p}}\exp{\left(\dfrac{i}{\hbar}\int_a^x pdx\right)}\cr +&amp;\dfrac{C_2}{\sqrt{p}}\exp{\left(-\dfrac{i}{\hbar}\int_a^x p dx\right)}\cr &amp;\quad \text{for} \quad x&lt;a,\end{align} \tag{47.2}$$ respectively. Most quantum mechanics textbooks determine the relation between $C$ and $C_i$'s by finding the exact solution near the turning point. And then let the exact solution match with the WKB solutions. However, in Landau &amp; Lifshitz's quantum mechanics textbooks (vol3, section 47) they let $x$ vary in the complex plane and pass around the turning point $a$ from right to left through a large semicircle in the upper complex plane. Landau claims starting from +$\infty$, when arriving at $-\infty$ (left to $a$), there is phase gain $\pi$ in the denominator of prefactor in Eq.(2). From this we can determine $$C_2=\frac{C}{2}\exp\left(i\frac{\pi}{4}\right).\tag{47.4a}$$ They also claim the first term will exponentially decay along the semicircle in the upper half plane. Question is why? Can we show $$\Im{\left(\int_a^x pdx\right)},$$ where $\Im$ stands for the imaginary part, is positive?
43426
Most sources say that  acts like a joint phase-space distribution in quantum mechanics and this is justified by the formula   $$\int_{\mathbb{R}^6}w(x,p)a(x,p)dxdp= \langle \psi|\hat{a}\psi\rangle$$  as this allows us to compute the expectation value for the operator $\hat{a}$ corresponding to the physical quantity $a$. However, without the knowledge of this formula, how did Wigner come up with this definition:  $$w(x,p)=\frac{1}{(2\pi)^3}\int_{\mathbb{R}^3}\psi\left(x-\frac{v}{2}\right)\psi^*\left(x+\frac{v}{2}\right)e^{iv \cdot p}dv \quad ?$$  I would be greatly indebted for any mathematical motivation that anyone could provide. Similarly, I also wonder at the mathematical motivation for the Weyl quantization.
109997
I recently started studying simple harmonic motion and I came across two equations for the displacement of a particle, as given in my textbook:  \begin{align} y &amp;= a\sin(\omega t +\phi)\\ x &amp;= a\cos(\omega t+ \phi) \, . \end{align}  I really have no clue when to use which one. It would help me a lot if someone could give an explanation on this.
64171
Can someone explain  simply?  I know the math, however I don't understand the relevance.   To explain what I'm asking, what is the significance of it when doing quantum mechanical calculations such as getting Planck length or other such equations.
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 $-\frac{\pi}{2}\hbar$.
50221
Suppose we have a particle of mass $m$ that is in an eigenstate $|\psi\rangle$ of the Hamiltonian $\hat{H}=\hat{T}+\hat{V}$, where $\hat{T}$ is the kinetic energy operator and $\hat{V}=V(\hat{r})$ is the potential energy operator. If the potential has lower bound $V_0$, then is it necessary for the energy eigenvalue $E$ of $|\psi\rangle$ to be greater than $V_0$? Classically this is true, since we regard negative kinetic energy as physically unrealizable/meaningless. However, I don't know if I can necessarily say the same in the quantum case. For example, what I'm tempted to do is write  $$\langle\psi | \hat{T}|\psi \rangle = \langle\psi | E-\hat{V}|\psi\rangle$$  and say "If $E&lt;V_0$, then the RHS is necessarily negative, implying the LHS is as well, which we will regard as physically meaningless. If $E=V_0$, then $|\psi\rangle$ is trivial.", but I'm not sure if that's right.   Confused by that, I then wanted to show that, if $E&lt; V_0$, then a non-trivial $|\psi\rangle$ is non-normalizable. However, I'm not entirely sure how to do this.
90556
The energy level of a simple harmonic oscillator is $E_n=(n+\frac{1}{2})\hbar\omega$. Is there any physical explanation why these levels are equally spaced ($= \hbar\omega$)?  Maybe this  can be helpful.
45792
I read that it makes a difference whether you calculate $\frac{dE(\lambda) }{d \lambda}=0$ or $\frac{dE(\omega)}{d \omega}=0$ in the sense that the maximum energy density with respect to the wavelength does not coincide with the frequency maximum that one would assume to be at $\omega_{max} = \frac{c}{\lambda_{max}}$. Actually, I do not understand why this is so ( Now, I am only interested in a pure physical explanation, mathematically this is clear and due to the chain rule). Does anybody know how to explain this odd thing?  Somehow I feel that the core of the question has not been completely answered. Although I can easily look up wikipedia, where the need for the chain rule is explained, I am rather interested in understanding where the following argument breaks down, which seems to be why this question has caused some confusion in the past(as you can see by googling this question):  So we have $E(\omega)$ the energy radiated at a given frequency. Now this function has a maximum somewhere, so there is a frequency where a maximal amount of energy is emitted. In other words: If you add the sum of the photon's energies at each frequency that are emitted, you will notice that the maximum is reached at this frquency. Now $E(\lambda)$ tells you basically the same for the wavelength, but again: We know where at which frequency the maximal amount of energy is radiated, so we know the corresponding wavelength.   I think there is a need to explain this.   If anything is unclear, please let me know.
46448
It is well known from Noether's Theorem how from continuous symmetries in the Lagrangian one gets a conserved charge which corresponds to linear momentum, angular momentum for translational and rotational symmetries and others.  Is there any elementary argument for why linear or angular momentum specifically (and not other conserved quantities) are conserved which does not require knowledge of Lagrangians? By elementary I mean, "if this is not so, then this unreasonable thing occurs".   Of course, we can say "if we want our laws to be the same at a different point in space then linear conservation must be conserved", but can we derive mathematically the expression for the conserved quantity without using the Lagrangian?  I want to explain to a friend why they are conserved but he doesn't have the background to understand the Lagrangian formalism.
24621
I saw this Lagrangian in notes I have printed:  $$ L\left(x,\frac{dx}{dt}\right) = \frac{m^2}{12}\left(\frac{dx}{dt}\right)^4 + m\left(\frac{dx}{dt}\right)^2\times V(x) -V^2(x). $$ (It appears in the exercises in the first chapter of Goldstein.)  What is it? Is it even physical? It seems like it doesn't have the right units of energy.