New answers tagged

1

The number operator does not commute with the position operator. We have $$\hat{H}=\frac{\hat{P}^2}{2m}+\frac{1}{2}m\omega^2\hat{X}^2=\left(a^\dagger a+\frac{1}{2}\right)\hbar\omega=\left(\hat{N}+\frac{1}{2}\right)\hbar\omega$$ Then, $$\hat{N}=\frac{\hat{H}}{\hbar\omega}-\frac{1}{2}$$ where $\hat{H}$ is the Hamiltonian for the harmonic oscillator. Using $[\...


1

$wa\sqrt{1-y^2}$ is the velocity of SHM while $ƒ*Λ$ - velocity of on any wave


-2

I think your definition of the average displacement is missing a normalisation factor of 1/t


1

This isn't the definition of SHM I would go with, but it is equivalent so we'll run with it for now. The point is that the particle moving around the circle at uniform speed is purely fictitious, the real particle is its 'shadow' moving back and forth along the $x$ axis. If you visualise this for a bit, you should see that when the particle is at the points ...


0

If you have a system with a single level of energy $\hbar\omega$ and it is occupied by $n_B$ particles (necessarily bosons, of course), then the energy of the system is $n_B\hbar\omega$, which is equivalent to a single Harmonic oscillator at the $n_B$-excited level.


1

As far as the single Gaussian $x_1$ integration goes, OP is doing the correct thing (up to possible typos) in eq. (7). However since the time-increment $$\epsilon~\ll~ \omega^{-1}\tag{i}$$ is supposed to be small (in order for Feynman's fudge factor $1/A$ to be valid), we have under the square root $$ \frac{\sin (\omega \epsilon)}{2\epsilon^2\omega\cos (\...


2

The action that goes in the path integral is a functional of the path $x(t)$. Namely, $$S[x(t)]=\int_0^T dt \frac{m}{2}\left(\dot{x}^2-\omega^2 x^2\right)$$ which can be discretized very simply $$S=\frac{m}{2}\sum_{i=1}^N\frac{(x_{i+1}-x_i)^2}{\epsilon}-\epsilon\omega^2 x_i^2$$ The $x$ in this action is not necessarily a classical trajectory, but this is ...


3

Your overall Hamiltonian for this system is actually $$ H = H_{1} \otimes \mathbb{I} + \mathcal{I} \otimes H_2 $$ where your overall Hilbert space is $L_{2}(\mathbb{R}_{3}) \otimes \mathbb{C}^2$, and $\mathcal{I}$ is the identity on $L_{2}(\mathbb{R}_{3})$, while $\mathbb{I}$ is the identity on $\mathbb{C}^2$. Let $|\mathbf{n} \rangle \in L_{2}(\mathbb{R}_{...


1

You have $[H_1, H_2] = 0$, so yes, it is possible to diagonalize $H = H_1 + H_2$ using simultaneous eigenkets of $H_1$ and $H_2$. And to the second question yes again, the two states are indeed orthogonal. You can use the reasoning that if one individual component is orthogonal then the whole state must be orthogonal or, since the two tensor products kets ...


1

My 2cents on it is studying harmonic systems was natural because any local or global potential minima can be approximated as a quadratic for modeling purposes. At least for basic quantum mechanics that would be my justification. QFT or particle physics someone else would have to answer the precise why it is powerful. I am aware it is the case, but it is well ...


1

I think instead of $x_i \le E_i$, we have $x_1 \le E_1$ and $\omega x_2 \le E_2$, from which the conclusion readily follows.


0

Assuming that the power series $$\sum_{s=0}^{\infty}g(N,s)t^s $$ converges for $|t|< T$ and some $T>0$, we can derive it under the sign of the series, therefore: \begin{align} \frac{1}{n!}\left( \frac{d}{dt}\right)^n \sum_{s=0}^{\infty}g(N,s)t^s = \sum_{s=0}^{\infty}g(N,s)\frac{1}{n!}\left( \frac{d}{dt}\right)^n t^s = \sum_{s=n}^{\infty}g(N,s)\frac{...


0

Change your equation to: $$ ma + kx + \frac{\rho Av^2C}{2} = 0 $$ and it should work.


2

You forget the direction of the damping force. In the linear equation the $-v$ term always has the opposite sign to $v$, which is correct. But in your quadratic damping equation $-v^2$ is always negative. If $v$ is negative, that force is increasing the speed of the particle, not decreasing it, and the particle shot off to infinity as in your graph. ...


0

After some further offline reflection and discussion, the following understanding was developed. The confusion in the question stems from the fact that the operator $A_\ell$ is better understood as acting only on the radial part of the wave function. Some of the details are explained below. Since the Hamiltonian $H = \frac{\mathbf{P}^2}{2\mu} + \frac{1}{2}m\...


12

First, let's start by completing the square as suggested. The equation will now be of the form $$ \frac{d^2\Psi}{dz^2}+(\nu+\frac{1}{2}-\frac{1}{4}z^2)\Psi=0 $$ with boundary conditions $\Psi(z=L)=\Psi(z=\infty)=0$ and $$ z=\sqrt{\frac{2m\omega}{\hbar}}x+L \\ \nu=\frac{E}{\hbar\omega}+\frac{L^2}{4}-\frac{1}{2} \\ L=-\sqrt{\frac{2mg^2}{\hbar\omega^3}} $$ This ...


3

$F = -kx$ is the definition of a harmonic oscillator. It's not something that's proven. Most real systems can be reduced to a harmonic oscillator for sufficiently small displacements from the equilibrium; the proof of that involves Taylor expansion. There's a $-$ in the formula because the force is restoring - it always attempts to shift the object back to ...


4

No. There are two points of view on this. First, it's a definition of what a simple harmonic oscillator/ideal spring is. Second, it's an empirical formula that is approximately correct in many circumstances. It is literally the same thing as searching for the minimum of a function by setting the derivative of the function equal to zero, and then ...


0

I guess it doesn´t really matter, since the cos(x) function is an even function, and thus cos(x) = cos (-x).


5

Absolutely. The Hermite polynomials $$ H_n(x) = (-1)^n e^{x^2} \partial_x^n e^{-x^2} = \left(2x - \partial_x\right)^n \cdot 1 $$ are orthogonalized by $$ \int_{-\infty}^\infty H_m(x) H_n(x)\, e^{-x^2} \,dx = \sqrt{\pi}\, 2^n n! ~ \delta_{nm} ~, $$ whereas the (nonpolynomial) Hermite functions $$ \psi_n(x) = \left (2^n n! \sqrt{\pi} \right )^{-\frac12} e^{...


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