# Tag Info

4

You do the same thing you always do when applying a function to an observable - you diagonalize it, then apply the function to the eigenvalues, then undo the diagonalizing procedure. The formal underpinning of this procedure is given by Borel functional calculus.

3

The equations of motion for the position determine the accelerations: they are second-order differential equations in time: $$\vec F = m\vec a = m\ddot{\vec x }$$ So the acceleration, the second derivative of the location in time, has to be determined from the state of the physical system in some way. Typically, it is determined using the $F=ma$ formula ...

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The reasoning is that the wavefunction must be normalizable. Equations 0.23--0.25 show that if $j \to \infty$, then the power series becomes a non-normalizable wavefunction. This means that the series must be finite. The reasoning from Equation 0.22 shows that after some $j$, every other $j$ must be zero. Since this doesn't provide any information about ...

2

I assume you have not diagonalized the operator, else you would not be asking the question. The absolute value of the operator $\hat{O}$ is presumably $\sqrt{\hat{O}^\dagger \hat{O}}\equiv \hat{B}$, so the crucial question is what type of algorithm one would choose to evaluate the square root $\hat{B}$ of the operator $\hat{A}= \hat{O}^\dagger \hat{O}=\hat{... 2 To expand on Andrei's answer a bit, start with your equation: $$\ddot x + \omega^2 x = C$$ and rewrite it as: $$\ddot x + \omega^2 \left(x - \frac{C}{\omega^2}\right) = 0$$ Then define a new variable$y$by: $$y = x - \frac{C}{\omega^2}$$ and differentiate twice to get: $$\ddot{y} = \ddot{x}$$ Finally substitute into your original equation to ... 2 Just add a constant say$C$to your solution.$x=A\cos(\omega t)+B\sin(\omega t)+C$Taking the second derivative, the term with$C$will be 0. But you still have$\omega^2C=-V\omega/c$so$C=-V/(\omega c)$. Therefore$x=A\cos(\omega t)+B\sin(\omega t)-V/(\omega c)$1 This might help you out a bit: In what sense is a quantum field an infinite set of harmonic oscillators? From my understanding, most people think it provides a useful way to conceptualize uncoupled quantum fields physically. It doesn't, however, work for coupled quantum fields. The main problem with this seems to be that infinite harmonic oscillators give ... 1 We know $$f_n=\frac1{2\pi}\sqrt{\frac km}$$ We then ask ourselves what is k and m. For this case, mass is$980kg$. Is k stiffness of 1 spring or 4 spring? The 980kg mass is not sitting on 1 spring. So it should be 4 spring. The tricky part is we don't use 4k. Instead we use k for stiffness equivalent to 4 springs. With 80kg, we get 1.2cm deflection. So it ... 1 You need the effective spring constant of the four springs in parallel and then use the whole mass of the car to consider the motion of the car or use the spring constant of one spring and use a quarter of the mass of the car. 1 A simple pendulum performs simple harmonic motion when it is displaced very slightly. You can say that a simple pendulum performs a periodic motion which can be treated as simple harmonic motion in small oscillation Now lets move forward supposing it to be pure SHM. The time period of a simple pendulum does not depend on how much it is displaced( but it ... 1 Why would you consider the amplitude to be equal to the horizontal displacement? There's also a vertical displacement, which is arguably more important, without vertical displacement there would be no oscillation, since the potential energy would remain the same. Or take for example a torsion pendulum: You wouldn't describe the amplitude as$r\sin{\...

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The "cartesian" basis for the 2D Harmonic oscillator is given in terms of the $x$ and $y$ eigenstates of the corresponding number operators $\hat n_x$ and $\hat n_y$, written as $|n_x\rangle \otimes|n_y\rangle$. These number operators have to been understood as $\hat n_x \equiv \hat n_x \otimes 1_y$ and $\hat n_y \equiv 1_y\otimes \hat n_y$, with $1_x$ and $... 1 A simple harmonic motion is one where the acceleration (or restoring force) is directly proportional to the displacement and in the opposite direction of the displacement. For a mass$m$on a spring with spring constant$k$, the differential equation describing the motion becomes:$m\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} = -kx$That equation has as solution: ... 1 I do not believe you can show it starting by the equation, but it becomes clear when you solve the differential equation that the two quantities are independent:$\omega\$ is an arbitrary parameter on the equation of motion, and A is an arbitrary constant that appears in the solution.

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