11

The eigenstates of the operator $N = a^\dagger a$ can be labeled by their eigenvalues, i.e. $N \phi_n = n \phi_n$, where $n$ is an integer. Note that $$n = \langle \phi_n,N\phi_n\rangle = \underbrace{\langle\phi_n,a^\dagger a \phi_n\rangle = \langle a \phi_n,a\phi_n\rangle}_{a \text{ and } a^\dagger\text{ are mutually adjoint}} = \Vert a \phi_n\Vert^2 \geq ...


5

Expanding on J. Murray's answer (which basically is the answer), we can start from the so-called number operator and have $\hat{N} = a^{\dagger}a$. As it was explained Murray's answer, its spectrum is non-negative, because $$ \langle \psi | \hat{N} | \psi \rangle = \langle \psi | a^{\dagger}a | \psi\rangle = || a|\psi\rangle ||^2 \geq 0$$ so every ...


4

You can do this by direct substitution in the position or momentum basis. Here it is for $x$: $$\hat a|0\rangle= [i\sqrt{\frac{m\omega}{2\hbar}}\big(\hat x - \frac i {m\omega}\hat p \big)][\big(\frac{m\omega}{\pi\hbar}\big) e^{-mwx^2/2\hbar}]$$ $$\propto [x+\frac {\hbar}{m\omega}\frac d{dx}]e^{-mwx^2/2\hbar}$$ $$\propto xe^{-mwx^2/2\hbar}+\frac {\hbar}{m\...


2

...you need $\omega = \sqrt{\frac{k}{m}}$ to even show (which the book doesn't) that $x(t) = A\sin(\omega t)$ is a solution. Actually, you don't. You can go through the whole process of deriving the differential equation that governs a simple (or physical) pendulum, approximating it, and solving it without ever using the symbol $\omega$ or talking about a ...


2

The equation for SHM is given by $$ \frac{\partial^2y}{\partial t^2}+ \omega_0^2y = 0\, . $$ Where $y$ is the displacement of the body with respect to a defined origin, and $\omega_0$ is the characteristic frequency of the system. I believe when you are asking whether a wave undergoes SHM, you are effectively asking whether the points on a wave follow a ...


2

The easiest way to derive your solution is probably to convert the first equality into a differential equation: $$ a(t)=e^{iHt}ae^{-iHt}\qquad\Longrightarrow\qquad \frac{d}{dt}a(t)=e^{iHt}(iHa-aiH)e^{-iHt}=ie^{iHt}[H,a]e^{-iHt} $$ where I used $$ \frac{d}{dt}e^{iHt}=iHe^{iHt}=e^{iHt}iH\qquad\qquad \frac{d}{dt}e^{-iHt}=-iHe^{-iHt}=e^{iHt}(-iH) $$ It ...


2

Yes. Simple harmonic motion can be defined by the relationship that acceleration is toward a fixed point and proportional to the displacement from that point. $$\ddot y = -\omega^2 y.$$


2

Physically, it does not matter whether we use $\sin()$ or $\cos()$. It's a matter of convention and you are free to choose either one. Therefore, if somebody defined the amplitude, the frequency and the "initial phase" $\varphi_0 = 0$ of the oscillation, the function $x(t)$ is not (!) well defined. However, if we say, the initial phase is such that the ...


1

In the frame of reference of the wedge, the cylinder will have motion only along the incline (since they are never losing contact) so you can balance forces perpendicular to the incline. I would also recommend trying the same problem for a block first instead of the cylinder as the latter one will get a bit more calculative


1

The force is given by the first derivative of the potential: \begin{equation} F(X) = - \frac{d}{dX}V(X) = -6X^5 -4X^3. \end{equation} At small displacements, i.e. when $X\ll 1$ we have $X^5 \ll X^3$, that is the returning force is proportional to $X^3$ and not to $X$, as is in the case of a harmonic oscillator. The spring constant is indeed zero. You are ...


1

Periodic motion is not the same as being a harmonic oscillator. A harmonic oscillator describes the very particular force configuration F=-kx, where k is the spring constant, and this is not the force for the potential you propose. Therefore, there need not be an equivalent of a spring constant just because the motion is periodic. For example, a particle ...


1

Like the comment says, you know $$ [H,a] = \omega\,\left[a^\dagger a + \tfrac12, a\right] = -\omega\, a\,. $$ Moreover, you also know the $n$fold nested commutator of $H$. Calling $\mathrm{ad}_H(X) \equiv [H,X]$ one has $$ \mathrm{ad}_H^n(a) \equiv\; \stackrel{n\text{ times}}{\overbrace{[H,[H,[\cdots[H}},a]\cdots]]] = (-\omega)^n a\,. $$ This is useful ...


1

First of all, the functions of operators are always to be understood as a Taylor expansion. In your case: \begin{equation} e^{\lambda a_+} = \sum_{n=0}^{+\infty} \frac{(\lambda a_+)^n}{n!}. \end{equation} Let is now consider operators $A$ and $B$ whose commutator is a complex number (i.e. not an operator): $[A, B] = c$. And let us take an infinitely ...


1

It's actually important to recall that this is true only by definition, because it's part and parcel of what refraction means. But refraction is not the only thing that can happen when light passes between two mediums, even though it is probably by far the most common thing that happens in this situation. There are solutions of Maxwell's equations where ...


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