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Note: My approach lacks rigour and I'm rusty. I'm sure there are lots of holes, overly-complicated parts, and perhaps even just outright mistakes. However, I think the gist of the argument motivates why two operators which commute leads one to consider solutions of the form you mention. Paraphrased: Why does does the fact that $\hat{A}$ and $\hat{B}$ commute ...

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I'll try to give a slightly more precise view which might be confusing at first but will pay off later. You are minimizing the action $S[\psi,\psi^*]$ where the square brackets indicate that $S$ is a functional: an object which takes a function as argument and spits out a number. Here $\psi$ and $\psi^*$ are still considered completely independent functions. ...

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I'm not 100% confident in my answer as I am new to this topic, but I hope I am of help. Edit: Your lagrangian is a slightly strange, as pointed out by AccidentalTaylorExpansion, so you may want to double check that it is correct. The below should hold for the lagrangian you have given however. The lagrangian you have is a 'lagrangian density'. If the ...

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Yes, the two equations describe quantum particles, i.e., the "first quantization" is already done, and the only further quantization possible is second. (Note how it is different for photons, which are classically already described by a wave equation, see this discussion.)

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The Pöschl-Teller, Morse-Rosen, and Eckart equations can also be solved by ladder operator methods. Barut, Inomata, and Wilson published two papers in 1987 on this subject but [1] [2]. The Lie algebra sorcery in those papers is a little above my head so I started chasing down references; a lot of the math seems to trace back to Infeld and Hull (1951), The ...

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No the states in a finite square well and a simple harmonic oscilator are not the same. The square well is quite simple. Inisde the well it is a trig function. From the edge of the well outward it decays exponentially. The harmonic oscilator is quite a bit more complicated, but still exactly solvable. The solutioins involve Hermite functions. The energy ...

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Consider your one-dimensional Schroedinger equation $$-\frac{d^2\psi}{dx^2} +V(x)\psi=E\psi$$ where $V(x)$ is zero except in a finite interval $[-a,a]$ near the origin. $V$ doe not have to be left-right symmetric. Let $L$ denote the left asymptotic region, $-\infty <x<-a$, and similarly let $R$ denote $\infty>x>a$. For $E=k^2$ and $k>0$ ...

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Can anyone give me an example where a non-normalizable state $\lvert i \rangle$ time evolves to a non-normalizable state $\lvert f \rangle$ through $U$ and that in the end there is a volume $V$ factor that comes out to cancel $\langle i \lvert i \rangle$ and $\langle f \lvert f \rangle$ which are just equal to $V$? What about the simplest example: a plane ...

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In an actual physical measurement one will always have a finite precision, i.e., final volume: e.g., a particle detected in a bubble chamber leaves a trace of finite width. The non-normalizability is important to deal with, in order to be able, e.g., to use Fourier transform and other related mathematical apparatus, but it is just a mathematical artefact. In ...

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Okay I found a little error in my calculation, I basically neglected the product rule in the $\nabla(i(\nabla\lambda)\psi)$ term because I treated the operator as a scalar. So I really should be thinking about simplifying this in terms of operating on the function so I don't make this mistake again: $$(\nabla+i\nabla\lambda)^2\psi=\nabla^2\psi+\nabla(i(\... 0 To elaborate on Qmechanic's answer, and on Leo L. comment : Are they different kinds of "infinity" at 𝑥=0? Yes, there are different kinds of infinity for a potential, and different kinds of zero width. It all comes down to taking a rectangular potential wall of small thickness \epsilon and of large value V(\epsilon) within the thickness, zero ... 5 Well, as I understand your problem, the first step is to find the potential in a power series in r=R-R_0 around the equilibrium point. Because it is an equilibrium point, you know that the first order term proportional to r^1 is zero. If you find it is not, that means you made an error. The next term is proportional to r^2. Just looking at the shape of ... 7 If you want to solve this (almost) exactly, you will need to do so numerically. If you’re happy with an approximation, you can expand V(R) about its minimum, and then first look at the harmonic approximation by comparing the term in R^2 with the harmonic oscillator Hamiltonian. You then then treat the R^3 and R^4 terms in the series expansion as ... 0 We always normalize the wave equation. Any state must be normalized before deductions related to an arbitrary physical observable is made because the only realistically achievable states are those which satisfy the Born's interpretation. The reason for normalization is statistical. We have attached a physical significance to the wave functions by the square ... 0 The Hamiltonian operator for an electron in this model is$$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$$where m is the electron mass. So you have Schrödinger's equation$$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\Psi(x)=E\Psi(x) \tag{1}$$and the periodic condition$$\Psi(x)=\Psi(x+L). \tag{2}$$The solutions of (1) and (2) are the normalized wave functions$$\Psi_n(...

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What the Klein-Gordon equation essentially distinguishes from Schroedinger's equation is the fact that it is more (or even 100%) like a field equation: $$(\Box + m^2)\phi = 0$$ Why is this ? First of all, it is a relativistic equation. In a relativistic theory the concept of a wavefunction no longer works since it is not causal. In Schroedinger's equation ...

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The solution of the Dirac equation for ground-state hydrogen atom (under point nuclei, fixed nuclei position) includes, as quoted in this link, $$r^{\gamma-1},$$ where $$\gamma = \sqrt{k^2 - Z^2 \alpha^2 } = \sqrt{ 1 - 1/137^2} < 1,$$ thus it diverges at the origin. Namely not continuous (not belong to $C^0$, thus cannot be $C^2$), but still within $L^2$...

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$U$ and $U^\dagger$ are mutual inverses so their product is dimensionless. The only dimension comes from the $d/dt$, which has units of s$^{-1}$.

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It's not necessary for a general time-reversible system to have the form of a unitary evolution. This is basically the answer to your question, but to make it more clear I'll give a trivial example: $$\frac{dx}{dt} = 1$$ i.e. $x(t) = t$. This system is clearly time-reversible, with the reverse dynamics given by ${dx}/{dt} = -1$. However the evolution ...

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The Hamiltonian is $$H(q,p)~=~\frac{1}{2}p^2+W(q).\tag{1.2.1}$$ The momentum needs to be rotated oppositely $$p\to \mathrm{e}^{-\mathrm{i}\theta}p$$ in order to preserve the CCR. For the unstable quartic potential (1.3.3) the rotated Hamiltonian $$\tilde{H}(q,p)~:=~H(\mathrm{e}^{\mathrm{i}\theta}q,\mathrm{e}^{-\mathrm{i}\theta}p)\mathrm{e}^{2\mathrm{i}\... 4 As you seem to be assuming that there is only an e^{-cx} for x>a and that c>0. Thus you have already assumed that you are in L^2. This condition on c requires that E<V_0 and you have bound states. The pair of equations for the even case, for example, can by simplified by dividing one by the other to get$$ k\tan ka = c $$which needs ... 5 TL;DR: OP has a point. Given Griffiths' condition (2.91) on p. 52, the solution \psi to the 1D TISE will be a linear combination of exponentially growing and exponentially damped solutions in the asymptotic regions. One needs to impose an extra condition to get rid of the non-normalizable exponentially growing solution. A heavy-handed condition (which ... 0 The Hamiltonian is proportional to the time derivative of the wave function, \hat{H}=i\hbar\partial \psi/\partial t. Suppose you want to find the time derivative of the expectation value of operator \hat{O}.$$d<\hat{O}>/dt=\frac{\partial}{\partial t}\int \ \psi^*\hat{O}\psi dr^3=\int \frac{\partial \psi^*}{\partial t}\hat{O}\psi +\psi^*\frac{\...

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