In one-dimensional quantum mechanics, it seems that the only kind of potential able to produce an "equidistant" spectrum, i.e. with $E_{n+1}-E_{n}=\text{constant}$, is the harmonic oscillator.  Why is that? And is there a way to prove it?
101913
I was wondering if the good old quadratic potential was the only potential with equally spaced eigenvalues. Obviously you can construct others, such as a potential that is infinite in some places and quadratic in others, but that's only trivially different. I am not referring to equally spaced as a limiting behavior either, I mean truly integer spaced.  Any ideas? If not, is there a proof for its uniqueness?  If there are other potentials with equally spaced eigenvalues, can one use them as starting points for a free-field QFT? It would be interesting to know if there is a deeper mathematical relation between all of these potentials and whether they could be used to study interacting systems.
43480
Consider a normal operator on a complex Hilbert space with its spectrum contained in the real line. Show that the operator is hermitian without using the spectral theorem.  What I have tried- Since the spectrum is inside the real line, the boundary of the spectrum as a subset of the complex numbers is the spectrum itself. Hence the spectrum is equal to the approximate point spectrum. I have been unable to proceed further.
73213
Consider standard quantum harmonic oscillator, $H = \frac{1}{2m}P^2 + \frac{1}{2}m\omega^2Q^2$.  We can solve this problem by defining the ladder operators $a$ and $a^{\dagger}$. One can show that there is a unique "ground state" eigenvector $\psi_0$ with $H\psi_0 = \frac{1}{2}\hbar\omega\psi_0$ and furthermore that given any eigenvector $\psi$ of $H$ with eigenvalue $E$, the vector $a^{\dagger}\psi$ is also an eigenvector of $H$ with eigenvalue $E + \hbar\omega$.  However, it is usually stated that we now have all eigenvectors of $H$ by considering all vectors of the form $(a^{\dagger})^n\psi_0$.   How do we know that we have not missed any eigenvectors by this process? e.g. how do we know that eigenvalues are only of the form $E_n = (n+\frac{1}{2})\hbar\omega$?  Also a slightly more technical question, how do we know that the continuous spectrum of $H$ is empty?  The technical details I am operating with are that $\mathcal{H} = L^2(\mathbb{R})$ and all operators ($H, P, Q$) are defined on Schwartz space, so that they are essentially self-adjoint with their unique self-adjoint extensions corresponding to the actual observables.
58791
In the  for quantum mechanics, the first postulate states that a quantum state is a vector in a separable Hilbert space. It means it is assumed the Hilbert space has a basis with at most infinite countable elements (cardinality). In other words, it states that the energy of an arbitrary system is discrete. Is it always true? If not, can you make an specific example in nature that its eigen-energies are uncountable?
97189
I'm having a discussion with someone. I said that it is -even theoretically- impossible to reach $0$ K, because that would imply that all molecules in the substance would stand perfectly still.  He said that this isn't true, because my theory violates energy-time uncertainty principle. He also told me to look up the Schrödinger equation and solve it for an oscillator approximating a molecule. See that it's lowest energy state is still non-zero.  Is he right in saying this and if so, can you explain me a bit better what he is talking about.
107284
suppose we have a potential that's independent of time $V(x,t) = V(x)$ so in Schrödinger equation we get: $$i\hbar \frac{\partial \Psi (x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi (x,t)}{\partial x^2}+V(x)\Psi(x,t)$$ since the LHS involves a variation of $\Psi$ with $x$ and the LHS involves a variation of $\Psi$ with $t$ can we say that a basis for the solutions will be with the form $\Psi(x,t)=\psi(x)T(t)$?
117672
There seems to be something very strange about the relationship between quantum field theory and quantum mechanics.  It is bothering me; perhaps somebody can help.  I'll consider a free Klein-Gordon field.  In standard treatments (e.g. Peskin &amp; Schroeder and Schwartz) the one-particle momentum eigenstates $| \vec{k} \rangle$ are normalized so that  $$ \langle \vec{p} | \vec{k} \rangle = 2 \omega_{\vec{p}} (2\pi)^3 \delta^{(3)}(\vec{p}-\vec{k}), \qquad 1 = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 \omega_{\vec{p}}}| \vec{p} \rangle \langle \vec{p} |. $$ Now, assuming $\langle \vec{x}' | \vec{x} \rangle = \delta^{(3)}(x-x')$ as usual, it follows that $$ \langle \vec{x} | \vec{p} \rangle = \sqrt{2 \omega_{\vec{p}}} e^{i \vec{p} \cdot \vec{x}}. $$ Now, one can compute (here in the Schrodinger picture; see Schwartz 2.76 or P&amp;S 2.42) that $$ \langle 0 | \phi(\vec{x}) | \vec{p} \rangle = e^{i \vec{p} \cdot \vec{x}}. $$ This is supposed to mean that $\phi$ creates a particle localized at position $\vec{x}$.  P&amp;S are a little cautious about the details, but Schwartz claims that the calculation implies  $$ \phi(\vec{x}) |0 \rangle = | \vec{x} \rangle. $$ But this is false because $\langle \vec{x} | \vec{p} \rangle \neq e^{i \vec{p} \cdot \vec{x}}$ with the normalization conventions used.  I suppose it could be true with some weird normalization of $| \vec{x} \rangle$, but I can't see what that might be (and at the very least this is not spelled out on the text).    Even if this works out, it seems extremely strange for there to be a relative normalization between the one-particle states of field theory and the states of one-particle relativistic quantum mechanics.  One ought to be able to redo the correspondence to make the normalization work out, but I don't see how.  (Note that the normalizations can be easily made to agree in the non-relativistic limit $\omega \approx m$, but that's besides the point.  Even if fully relativistic quantum mechanics is inconsistent [as some texts claim without reference], at the very least the perturbative corrections for $v \ll 1$ should be recoverable from field theory.)  [Edit: This seems to go beyond normalization.  We can get a feel for what kind of state $\phi(\vec{x})|0\rangle$ is by computing its wavefunction as a function of $\vec{x}'$, $$ \langle \vec{x}' | \phi(\vec{x}) |0 \rangle = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec{p}}}} e^{i\vec{p} \cdot (\vec{x}-\vec{x}')}. $$ This wavefunction is peaked (I think divergent) at $\vec{x}'=\vec{x}$, so in some sense the particle is centered at $\vec{x}$, but it seems to be quite a stretch to say that it is at $\vec{x}$ (as the books do).  I would go so far as to say the claim is incorrect, since in quantum mechanics saying the particle is at particular position means the wave function is a delta-function there. I guess the same language is used in the Heisenberg picture, when two-point functions are called amplitudes for particles to propagate from one spacetime point to another.  This similarly seems false by the conventional meaning of amplitude as the overlap between two localized states.  Words of wisdom would be appreciated.]
55118
I'm reading  and I'm trying to understand the proof under "Degeneracy in One Dimension".  Here's what it says:     Considering a one-dimensional quantum system in a potential $V(x)$ with degenerate states $|\psi_1\rangle$ and $|\psi_2\rangle$ corresponding to the same energy eigenvalue $E$, writing the time-independent Schrödinger equation for the system:       $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_1}{ \partial x^2} + V\psi_1 =E\psi_1$$       $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_2}{ \partial x^2} + V\psi_2 =E\psi_2$$   Multiplying the first equation by $\psi_2$ and the second by $\psi_1$ and subtracting one from the other, we get:       $$\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$$   Integrating both sides       $$\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$$   In case of well-defined and normalizable wave functions, the above constant vanishes, provided both the wave functions vanish at at least one point, and we find: $\psi_1(x)=c\psi_2(x)$ where $c$ is, in general, a complex constant. So, the two degenerate wave functions are not linearly independent. For bound state eigenfunctions, the wave function approaches zero in the limit $x\to\infty$ or $x\to-\infty$, so that the above constant is zero and we have no degeneracy.   Here are the parts I don't understand:   How does integrating $\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$ yield $\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$.  Without having explicit formulas for $\psi_1(x)$ and $\psi_2(x)$, how do we just apparently pull them both out of the integral? I don't understand how the constant in the last equation vanishes just because $\psi_1$ and $\psi_2$ vanish at infinity. Even if it does vanish, how does that imply $\psi_1=c\psi_2$?
8845
Srednicki's Quantum Field Theory mentions the following at the end of the unit on path integrals in non-relativistic quantum mechanics:   Assume that the total Hamiltonian is of the form,     $$ H = H_0 + H_1, $$   where $H_0$ is the exactly solvable part and $H_1$ is treated as a perturbation.  Without the presence of the perturbation, the ground-state t0 ground-state transition amplitude in the presence of external forces/classical sources is:     $$ \langle 0 | 0 \rangle_{f,h} = \int \mathcal{D}p \; \mathcal{D}q \; \exp \Bigg [i \int_{-\infty}^{\infty} dt \; (p\dot{q} - (1- i\epsilon)H_{0} + fq + h p \Bigg ]. \tag{6.21} $$   In the presence of the perurbation, the amplitude is now given by:    Questions:  1) To get equation $(6.22)$, how can we take the exponential inside the integral, make the necessary manipulations and then take the exponential outside the integral. This seems very ad-hoc to me. How and when do we know that the these steps are valid? Doesn't this required justification from mathematical analysis?  2) More importantly, I don't get the first term involving the perturbing Hamiltonian. Why do we have functional derivatives in its argument? How did it come out to be? Why don't we similarly have functional derivatives in the argument of $H_0$ as well? What's the distinction?  Note: The factor of $(1 - i \epsilon)$ has been suppressed in $(6.22)$.
118416
Yesterday I asked this  concerning the existence of terms. This question is on the same topic. I have been looking at some tables (in Yang and Hamilton (2010), pg 199) at the terms that exist for two electrons in the same configuration. I noticed that for example if the state $^1S$ exists the state $^3S$ would not and vice versa. Is this a general rule? I.e. can we say:     For two electrons in the configuration $nl^2$ if the term $^1L$ exists then the term $^3L$ will not and vice versa.   If so can it be proved easily and if not why not?
81219
Consider a particle in a potential well. Let’s assume it’s a simple harmonic oscillator potential and the particle is in its ground state with energy E0 = (1/2) &#8463;&omega;0. We measure its position (measurement-1) with a high degree of accuracy which localises the particle, corresponding to a superposition of momentum (and therefore energy) states.  Now we measure the particle’s energy (measurement-2) and happen to find that it’s E10 = (21/2) &#8463;&omega;0. Where did the extra energy come from?  In the textbooks it’s claimed that the extra energy comes from the act of observation but I wonder how that could work. Measurement-1 which probed the position of the particle can’t have delivered to it a precise amount of energy, while measurement-2 might just have been passive. No doubt there is entanglement here between the particle state and the measuring device but where, and which measurement?
101363
I am starting to learn about quantum mechanics and was wondering about the solutions to a question about a particle in an infinite potential well of length $L$.  The solutions are: $$\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi n x}{L} \right)$$  And I was wondering what is the meaning of $n$. Isn't there supposed to be only one wave function? In that case, $\Psi = \sum_{n=1} ^\infty a_n \psi_n$, but what are the $a_n$? Or maybe it is probabilistic, and we don't know which $\psi_n$ the particle is currently in?
56133
Every now and then a question comes up about the status of the momentum operator in the infinite square well, and while we have two good answers on the topic  and , I'm generally not satisfied by their level of detail and by how easy (not very) it is to find them when you need them. I therefore posit that it is high time we settle things and build a canonical Q&amp;A thread for this, so, with that in mind:  What's the deal with momentum in the infinite square well?   Is the momentum operator $\hat p=-i\hbar \frac{\mathrm d}{\mathrm dx}$  when restricted to the compact interval of the well? Are there any subtleties in its definition, via its domain or similar, that are not present for the real-line version? Is the momentum operator $\hat p=-i\hbar \frac{\mathrm d}{\mathrm dx}$  in these conditions? If not, why not, and what are the consequences in terms of the things we normally care about when doing one-dimensional QM? If it is not self-adjoint, does it admit a ? If it does, is that extension unique? If the extension is not unique, what are the different possible choices, and what are their differences? Do those differences carry physical meaning / associations / consequences? And what is a self-adjoint extension anyways, and where can I read up about them? What is the spectrum and eigenvectors of the momentum operator and its extensions? How do they differ from each other? Is there such a thing as a momentum representation in this setting? If not, why not? What is the relationship between the momentum operator (and its possible extensions) and the hamiltonian? Do they commute? Do they share a basis? If not, why not? Do these problems have counterparts or explanations in classical mechanics? ()   And, more importantly: what are good, complete and readable references where one can go and get more information about this?
117901
In quantum mechanics we know that the canonical position $\hat x$ and momentum operator $\hat p$ satisfying  \begin{align} [\hat x,\hat p] = i \quad (\hbar = 1) \end{align} have continuous spectrum.   We also know what the spectrum of the operator \begin{align} H = \frac{1}{2}\left(\hat p^2 + \hat x^2 \right), \end{align} is discrete, nondegenerate,  and it has equal spacing between the eigenvalues.  Is something known about the spectrum and eigenkets of the operator \begin{align} \hat O = \hat x \hat p + \hat p \hat x~? \end{align}
20700
The way I've understood it is that eigenfunction of an operator are the different states which the actual wavefunction can take when the property/observable corresponding to the given operator is being observed.   For the Hamiltonian operator, the eigenfunctions are forms the wavefunction will take when energy is being measured. The probability that we get a specific state corresponds to the coefficient from fourier decomposition of the wavefunction.   I've also read/interpreted/misinterpreted that upon measurement of position, a wavefunction will collapse to a single spike with a specific value of position (determinate state of the position observable).  With all of the above in mind, is it correct to think that the eigenstates of the position operator are continuous, because the probability density function is continuous and we can get a spike anywhere?  Also, does this mean that the wavefunction takes the form of a spike when position is being observed, but a sine wave when the energy is being observed?  Edit: by continuous eigenstates I want to say that there is an infinite number of them.
81369
I was recently reading Griffiths' Introduction to Quantum Mechanics, and I stuck upon a following sentence:     but $\Psi$ must go to zero as $x$ goes to $\pm\infty$ - otherwise the wave function would not be normalizable.   The author also added a footer: "A good mathematician can supply you with pathological counterexamples, but they do not arise in physics (...)".  Can anybody give such a counterexample?
104439
In de Broglie's explanation for Bohr's quantization condition, why do we not use $2\pi r = n\lambda/2$, why do we use $2\pi r = n\lambda$?  We know that in waves on strings, for standing waves creation in a string between two rigid supports, we use "Length of String" $ = n \lambda/2$. Why do we not use a similar concept here?