The energy level of a simple harmonic oscillator is $E_n=(n+\frac{1}{2})\hbar\omega$. Is there any physical explanation why these levels are equally spaced ($= \hbar\omega$)?
Maybe this link can be helpful.
[Physics] Why are the energy levels of a simple harmonic oscillator equally spaced
eigenvalueenergyharmonic-oscillatorquantum mechanics
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In Newtonian mechanics, one may derive a virial theorem, which says that the (time-averaged) kinetic and potential energies are related as
$$ \tag{1} 2\langle T \rangle ~=~p\langle V \rangle ,$$
if the potential $V(r) \propto r^p $ is a power law. Thus for the classical harmonic oscillator (HO)
$$ \tag{2} \langle T \rangle ~=~\langle V \rangle.$$
In QM the averaging procedure is of a different nature, but even so, it turns out that there are no quantum mechanical corrections to (2) in the HO case.
I believe my answer here to the question "How does one determine ladder operators systematically?" gives at least a partial answer to your question. It is a partial answer because I assume a little more than your bare question, but then, as we see by looking carefully at yuggib's answer observes you can obviously write down a hamiltonian with equally spaced eigenvalues and then characterise the whole family of such hamiltonians. It becomes clear that you need to talk about more than simply the Hamiltonian to answer your question: we need to define other observables and how they behave to define something akin to a "potential".
Let's look at yuggib's answer. You can obviously write down a hamiltonian with equally spaced eigenvalues. Then, as you counter, in your comment:
"Does this really qualify as a proof? Sure you can make operators with any set of discrete eigenvalues, but how do you know that they are not all just equivalent to each other and the harmonic potential? Also, I should say that I really mean a potential, because obviously you can write a Hamiltonian as a matrix and give it equally spaced eigenvalues, but how could you know what potential gave rise to it?"
For this question, I choose energy eigenstates bounded from below. Otherwise, you could get arbitrarily negative energy states and there would be no quantum ground state. This may or may not be more than what you want to assume, but I think it is physically reasonable. As I said, I'm giving a partial answer. So now our index set $I$ becomes in fact the set of semi positive integers $\mathbb{N}$. So, in the notation of yuggib's answer, choose an orthonormal basis $\left\{Y_j\right\}_{j=0}^\infty$ for our assumed separable (this is yet another assumption we must bring to bear) Hilbert space of quantum states with $P_j Y_k = \delta_{j\,k} Y_k$, where $\delta$ is clearly the Kronecker delta and $P_j$ are the projection operators onto the basis vectors. Then, the most general Hamiltonian with equispaced eigenvalues is as written in yuggib's answer with:
$$\lambda_i = E_0 + \sigma(i) \Delta$$
where $E_0$ is the ground state energy, $\Delta$ the energy spacing and $\sigma:\mathbb{N}\to\mathbb{N}$ a bijection between $\mathbb{N}$ and itself. So there are an infinite number of Hamiltonians with equispaced energy levels. All members of each family of such Hamiltonians defined by the family's ground state energy $E_0$ and spacing $\Delta$ are unitarily equivelent to one another: two members $\hat{H}_1,\,\hat{H}_2$ are equivalent by $\hat{H}_1 = U\,\hat{H}_2\,U^\dagger$ with $U$ some unitary operator.
So now, how to work this into something like a "potential"? My solution is then to abstractly define position and momentum observables $\hat{X},\,\hat{P}$ and we make our final three assumptions:
They fulfill the canonical commutation relationship $\hat{X}\,\hat{P} - \hat{P}\,\hat{X}=i\,\hbar\,\mathrm{id}$;
Measurements by these observables vary sinusoidally with time;
Our observables are Hermitian operators.
So now we Writing a general quantum state as:
$$\psi = \sum\limits_{j\in\mathbb{N}} \psi_j e^{-i\,\left(j\,\omega_0+\frac{E_0}{\hbar}\right)\,t}$$
so that the mean of a general observable $\hat{A}$ is:
$$\left<\left.\psi\right|\right.\hat{A}\left.\left|\psi\right.\right> = \sum\limits_{j=0}^\infty a_{j,j}|\psi_j|^2 + 2\,{\rm Re}\left(\sum\limits_{j=0}^\infty\sum\limits_{k=j+1}^\infty a_{j,k} \psi_j \psi_k^* \exp(i\,\omega_0\,(k-j)\,t)\,\right)$$
we can readily see that observables with sinusoidally varying measurement means must have two symmetrically placed, complex conjugate off-leading-diagonal diagonal stripes. Moreover the displacement from the leading diagonal must be the same for both $\hat{X},\,\hat{P}$ if they are to fulfill the CCR. The simplest case is when the two stripes are immediately above and below the leading diagonal. In this case, the means of the observables will vary like $\cos(\omega_0\, t + \phi_0)+const.$: if the two stripes are displaced $N$ steps either side of the leading diagonal, then we have variation varies like $\cos(N\,\omega_0\, t + \phi_0)+const.$. The case with the stripes displaced $N$ from the leading diagonal yield analyses that are essentially the same as the below as discussed in my other answer: they essentially pertain to a quantum oscillator with $N$ times the energy spacing of the one we talk about here.
So now, without loss of generality, in our assumed basis we can write the Hermitian observables as:
$$\begin{array}{ll}\hat{X} = \sqrt{\frac{\hbar}{2}}\left(\tilde{X} + \tilde{X}^\dagger\right) \\ \hat{P} = \sqrt{\frac{\hbar}{2}}\left(\tilde{P} + \tilde{P}^\dagger\right)\end{array}$$
where:
$$\tilde{X} = \left(\begin{array}{ccccc}0&x_1&0&0&\cdots\\0&0&x_2&0&\cdots\\0&0&0&x_3&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{array}\right)\quad\tilde{P}=\left(\begin{array}{ccccc}0&p_1&0&0&\cdots\\0&0&p_2&0&\cdots\\0&0&0&p_3&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{array}\right)$$
are both arbitrary lone-striped upper triangular matrices. You'll need to look at my other answer that I referenced above for full details, but by writing down the CCR we find that:
$$\begin{array}{ll}\hat{X} = \sqrt{\frac{\hbar}{2\,m\,\omega_0\,\cos\chi}}\left(a^\dagger\,e^{-i\,\xi} + a\,e^{i\,\xi}\right) \\ \hat{P} = i\,\sqrt{\frac{\hbar\,m\,\omega_0}{2\,\cos\chi}}\left(a^\dagger\,e^{-i\,(\xi+\chi)} - a\,e^{i\,(\xi+\chi)}\right) \end{array}$$
where we have defined the arbitrary complex constant $\alpha = -i\,m\,\omega_0\,e^{i\,\chi}$ by writing it in terms of a second real positive magnitude $m$ with the dimensions of mass and an arbitrary phase factor $\chi$ and where we have also defined:
$$a = \left(\begin{array}{ccccc}0&\sqrt{1}&0&0&\cdots\\0&0&\sqrt{2}&0&\cdots\\0&0&0&\sqrt{3}&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{array}\right)$$
and its Hermitian conjugate as the wonted ladder operators.
One can quite straighforwardly show that our derived $\hat{X},\,\hat{P}$ must have continuous spectrums. So, if we change our co-ordinates so that we work in position co-ordinates, i.e. where $\hat{X}$ becomes the diagonal (multiplication) operator $\hat{X}f(x) = x f(x)$ for $f(x)\in \mathcal{H} = \mathbf{L}^2(\mathbb{R}^3)$, then we can argue as I do in my answer here that there is needfully such a co-ordinate system wherein $\hat{X}f(x) = x f(x)$ and $\hat{P} f(x) = -i\,\hbar\,\nabla f(x)$. So now the Hamiltonian in these co-ordinates is:
$$\begin{array}{lcl}\hat{H} &=& \hbar\,\omega_0 \left(a^\dagger\,a + \frac{1}{2}I\right) + \left(E_0 - \frac{\hbar\,\omega_0}{2}\right)I\\ &=& \frac{1}{2\,m\,\cos\chi} \hat{P}^2 + \frac{1}{2\,\cos\chi}\,m\,\omega_0^2\,\hat{X}^2 - \frac{\omega_0\,\tan\chi}{2}(\hat{X}\hat{P} + \hat{P}\hat{X}) +\left(E_0 - \frac{\hbar\,\omega_0}{2}\right)I\end{array}$$
so that the Schrödinger equation in these co-ordinates is:
$$i\hbar\partial_t \psi = -\frac{1}{2\,m\,\cos\chi} \nabla^2 \psi + \frac{1}{2\,\cos\chi} m\,\omega_0^2 |\vec{x}|^2 \psi + i\,\hbar\,\omega_0\,\tan\chi\,\vec{x}\cdot\nabla \psi + \left(E_0 +i\,\frac{\tan\chi\,\hbar\,\omega_0}{2} - \frac{\hbar\,\omega_0}{2}\right)\psi$$
and more "traditional" Schrödinger equation is recovered when $\chi = 0$ and $E_0 = \hbar\,\omega_0/2$:
$$i\hbar\partial_t \psi = -\frac{1}{2\,m} \nabla^2 \psi + \frac{1}{2} m\,\omega_0^2 |\vec{x}|^2 \psi$$
and we see that we must have a quadratic potential. So, in summary, let's list the assumptions that lead to this conclusion:
- Separable quantum state space;
- Equispaced energy levels bounded from below;
- The existence of Hermitian position and momentum observables fulfilling the CCR and
- Measurements from the observables vary sinusoidally with time.
I have not yet analysed the case where we relax the assumption 4. From the above, we see that equispaced energy levels implies periodic variations of measurements with time, and it seems to me that this relaxation would likely yield a much more general potential in position co-ordinates.
Best Answer
Yes. There is a simple explanation.
The classical harmonic oscillator has a well defined frequency $\omega$, independent of initial conditions. This can only happen if the quantum system has precisely equally spaced energies with gap $\hbar \omega$.
The reasons for this is that motion in a quantum system can only happen if more than one energy level is occupied. The frequency of the motion is then set by the energy difference of the different occupied energy levels. Thus if every classical trajectory has the same frequency $\omega$, then any combination of quantum states must also have frequency differences that are multiples of $\hbar\omega$.
The argument can be made even more refined by thinking about the existence of coherent states and semi-classical superpositions in the quantum system, but the basic argument is as above.
The reverse argument can also be used to conclude that the hydrogen atom will not have equally spaced energy levels, since the frequency for motion in a Coloumb potential depends on the initial conditions.