# Tag Info

26

The reason for having two prongs is that they oscillate in antiphase. That is, instead of both moving to the left, then both moving to the right, and so on, they oscillate "in and out" - they move towards each other then move away from each other, then towards, etc. That means that the bit you hold doesn't vibrate at all, even though the prongs do. You ...

13

This question first posed to me by a friend of mine. For the subtleties involved, I love this question. :-) The "flaw" is that you're not counting the dimension carefully. As other answers have pointed out, $\delta$-functions are not valid $\mathcal{L}^2(\mathbb{R})$ functions, so we need to define a kosher function which gives the $\delta$-function as a ...

11

$y(t)=A \sin(\omega t+d)=A \cos(\omega t+d-\left(\frac\pi2\right))$ The two forms are the same except for the phase term, so which one you use depends on when you started your stopwatch. If you started off at max displacement at t=0, then you use the cosine form, if you start out with max velocity at t=0, then you use the sine form. If you are deducing ...

10

You can get a proportionality using one of the most basic techniques in science: dimensional analysis. What you do is look at the dimensions of the various quantities in the problem and try to see how they can fit together. Here you want the period, which is a time. I'll write $[T]$ to indicate that it has dimensions of time. What could it depend on? If ...

10

The Hilbert space ${\cal H}$ of the one-dimensional harmonic oscillator in the position representation is the set $L^{2}(\mathbb{R})$ of square integrable functions $\psi:\mathbb{R}\to\mathbb{C}$ on the real line. The Dirac delta distribution $\delta(x-x_{0})$ is not a function. In particular, it is not square integrable, cf. this Phys.SE post. One may ...

10

Ladder operators are usually constructed to form a Lie algebra (we want them to have specific conmutation relations). The mathematical basis is weight theory. The important thing of Lie algebras is that they are a vector space and their elements, which are called generators obbey this conmutation rule: $$[X_i,X_j]=f_{ijk}X_k$$ Where we have used the ...

10

It depends what you're doing, and indeed most of the quantum optics literature dismisses the term as it does not contribute to the dynamics. However, it is important that beginning students form an intuition for how and where zero-point energies come in, and why they are necessary. Take a look at the eigenfunctions of the harmonic oscillator, in position ...

9

The ground state of the harmonic oscillator $|0\rangle$ obeys $$a|0\rangle = 0$$ which means that the action of $a$ can't be undone: once you act with it on a state, you set to zero the coefficient in front of $|0\rangle$ in the decomposition into occupation eigenstates. Any candidate inverse operator $a^{-1}$ acting on zero will give you zero again; you ...

8

Yes, you are on the right track. The series you have there is called Dyson's series. First note that the $n$'th term looks like $$U_n = (-\frac{i}{\hbar})^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n)$$ The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing ...

8

Although I think user Siva's argument is nice for intuition, I feel that the key mathetmatical point is being obscured; you just need to be careful about what you mean by the "size" of a vector space. The dimension theorem for vector spaces tells us that any two bases for a vector space must have the same cardinality. This allows us to define the dimension ...

8

I assume you have no qualms with the "large $\xi$" approximation - it's fairly obvious that $\xi^2-k^2\approx \xi^2$ for large enough $\xi$. After that you're left with the differential equation $$\frac{d^2\psi}{d\xi^2}\approx-\xi^2\psi.\tag1$$ One way to solve this equation is by the method of divine inspiration: you somehow come up with two linearly ...

7

I) It depends on how abstract OP wants it to be. Say that we discard any reference to 1D geometry, and position and momentum operators $\hat{q}$ and $\hat{p}$. Say that we only know that $$\tag{1}\frac{\hat{H}}{\hbar\omega} ~:=~ \hat{N}+\nu{\bf 1}, \qquad\qquad \nu\in\mathbb{R},$$ $$\tag{2} \hat{N}~:=~\hat{a}^{\dagger}\hat{a},$$ $$\tag{3} ... 7 Why do they have that form and not some other? I suppose one answer is "the form of the Hamiltonian". Because of the form of the Hamiltonian for the QHO, there is a "number" basis for the states. Suppose you don't use the ladder operator algebra to solve for the energy eigenstates of the Hamiltonian. You still find that the energy eigenvalues are of the ... 6 Your solution is correct (multiplication of 1D QHO solutions). Since the potential is radially symmetric - it commutes with with angular momentum operator (L^2 and Lz for instance). Hence you may build a solution of the form |nlm> where n states for the radial state description and lm - the angular. Is it better? Depends on the problem. It's just the other ... 6 The main problem is to determine what corresponds to zero mass of the harmonic oscillator. Remember that a fraction of the spring mass also participates in the motion. By introducing an intercept \beta, your friend takes into account that the true zero of the mass parameter m may be shifted from what you think it is. So an affine model T^2=Cm+\beta is ... 6 This is all about potential; it is common that a particle movement is described by a following ODE: m\ddot{\vec{x}}=-\nabla V(\vec{x}), where V is some function; usually one is interested in minima of V (they correspond to some stable equilibrium states). Now, however complex V generally is, its minima locally looks pretty much like some quadratic ... 6 When one uses complex variables in this way one never multiplies two variables because the whole system is linear: if z is the oscillating variable and you choose to represent it by a complex number, then things like z\times z don't arise in a linear equation, so you don't get the kind of contradiction you astutely and clearly pointed out above. If the ... 5 Just to add to gigacyan's answer, the harmonic oscillator Hamiltonian may be written in terms of raising and lowering operators: \begin{eqnarray} \hat{H}\psi&=&-\frac{1}{2m}\frac{\partial^2\psi}{\partial x^2}+\frac{1}{2}m\omega^2x^2\psi\nonumber\\ &=&(a^{\dagger}a+\frac{1}{2})\omega\psi \end{eqnarray} where ... 5 I think perhaps what you're missing is in the "skipping through the commutator" part. Do you understand where we get this equation (try computing it yourself, if not):$$a_{-}a_{+} = \frac{1}{2 \hbar m \omega}[p^{2} + (m\omega x)^{2}] - \frac{i}{2\hbar}[x,p]$$Now, the canonical commutator, I'm sure you noticed (as it's boxed on the same page in Griffiths) ... 5 I think of 'quantize' as a verb that refers to converting the classical to the quantum picture. The Hamiltonian you wrote down is, after all, the classical one, which is not 'quantized'. Once you solve it by (1) writing down annihilation/creation operators, (2) finding that only particular wavefunctions are allowed, (3) numerically evaluated the energy ... 5 Addressing just the physics part (go to stack overflow for the programming), and using the equation that you've been given:$$ x(t) = A \cos \left( \omega t + \delta \right) $$Let's look at the form of the solution. It is sinusoidal The curve will have a maximum value of A (because cosine has a maximum value of 1) When \delta is 0, \pm 2\pi, \pm ... 5 Well, the reflection of a wave at the end happens always. One can picture this by imagining the succesive atoms being pushed off the equilibrium position as the wave propagates. Since the endpoint is fixed, it has nowhere to be pushed but the few atoms near it (I am considering idealized linear chain for simplicity) that have already being perturbed will, ... 5 1) There are many inequivalent quantum systems that have the same classical limit \hbar\to 0. 2) For instance, assume for simplicity that the quantum system is described by a single pair of creation and annilation operators,$$[\hat{a},\hat{a}^{\dagger}] ~=~\hbar {\bf 1}, \qquad\qquad ...

5

Yes, of course, they're the squeezed states. If your change involves a simple change of the frequency only, you may simply "rescale" all the energy eigenstates by $x\to Cx$ and $p\to p/C$. This is achieved by the operation $$|n\rangle \to |n'\rangle = \exp[\frac {i\ln(C)}{2\hbar}(xp+px)] |n\rangle$$ I inserted the Hermitian part of the operator $xp$ ...

5

If the total force $F$ on a mass $m$ follows Hooke's law, $$F~=~-kx,$$ then one can use Newton's 2nd law $$F=ma,$$ to infer that the motion is a simple harmonic motion $$a =-\omega^2x, \qquad\qquad \frac{2\pi}{T}~=~\omega~=~ \sqrt{\frac{k}{m}}~,$$ cf. OP's correct belief. Now it only remains to solve the ODE $$... 5 The second solution is there to allow for arbitrary start and stop times. Using standard trig identities you can convert an arbitrary linear combination of \sin and \cos into a time-displaced sinusoidal function:$$A\sin(\omega t)+B\cos(\omega t)=R\cos(\omega(t-t_0)),$$where R=\sqrt{A^2+B^2} and \tan(\omega t_0)=A/B. 5 You may recall from high school algebra that x^2 + y^2 = (x + iy)(x - iy). Because the way the adjoint operator works, you could define an operator \hat a = x + iy, and its adjoint becomes \hat a^\dagger = x - iy. The hamiltonian for the quantum oscillator is just this relation with some constants. You have to be careful because the ladder operators ... 5 The reason that creation and destruction operators don't commute is that, on top of 'moving a state up and down energy levels', they multiply it by a number in the process, and this number depends on where you are in the ladder. More specifically,$$\begin{cases} \hat{a}|n\rangle&=\sqrt{n}|n-1\rangle,\text{ while}\\ ...

5

They almost are. Clearly you replace $q_j$ by $x_j$ since the canonical commutation relationships are the same between these two and the $p_j$. Your QM course equation is then a special case of the one in the paper: if you then expand the last term in your QM course equation, you have equivalence if $A_{11} = A_{NN} = k+K$, $A_{jj} = k+2 K,\;j\neq 1, N$ and ...

5

The wavefunction $\psi(x)$ will satisfy the Schrödinger equation for the harmonic oscillator on an interval $x\in (-\frac L2 , \frac L2)$. We could write it as $$\psi '' +\left(\frac{2E}{\hbar \omega} - \xi^2\right) \psi =0,\tag{1}$$ where $\xi = \sqrt{\frac{m\omega}\hbar} x$ is the rescaled $x$-coordinate and dash denotes differentiation w.r.t $\xi$. ...

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