In a first course on quantum mechanics, everybody learns some version of the following statement:     Postulate: To every classical observable $A$ of a physical system, there corresponds a Hermitian operator $\hat{A}$ such that a measurement of $A$ performed on  a system in state $|\psi\rangle$ is expected (in the probabilistic sense) to return $\langle \psi | A\psi \rangle$. Possible outcomes of the measurement correspond to eigenvalues of $A$...   One then learns that $\hat{x} = x$ is the operator corresponding to position, $\hat{p} = -i\hbar\frac{\partial}{\partial x}$ is the operator corresponding to momentum, and from these, one can construct the operators for the kinetic and potential energies, angular momenta, etc.  But what happens if we want to find the operator corresponding to some more complicated classical observable, say, $$ A = \frac{e^{\sin(x^2 p)}}{x+xp} + \cos\left(\frac{p}{x}\right)? $$ (Assume here that we've scaled $x$ and $p$ by some characteristic length and momentum, so that they become dimensionless.) Granted, this bizarre quantity might not have a compelling physical interpretation, but it is still in principle a classical observable, that can be computed with knowledge of $x$ and $p$. It seems to me that if we are to take this postulate seriously, there ought to be a well-defined procedure that associates to any function of $(x,p)$ a Hermitian operator acting on the appropriate Hilbert space. What is this procedure?  More generally, if we have some classical system $S$ with configuration manifold $M$, it seems to me that any real-valued function $f: T^*\!M \to \mathbb{R}$ on the phase space of $S$ (i.e., cotangent bundle of $M$) defines (in principle) a classically observable quantity. (Or does $f$ need to be continuous/smooth?) Quantum mechanics ought to prescribe a mechanism  that associates to any such $f$ some Hermitian operator $\hat{f}: L^2(M) \to L^2(M)$. What is this mechanism, and how does it work in general?
122753
Okay, I know that in quantum mechanics the quantum observable is obtained from the classical observable by the prescription   $$ X \rightarrow x,\quad P \rightarrow -i\hbar\frac{\partial}{\partial x} $$   in the position basis. Now my questions are:    What if $x$ or $p$ appears in the denominator in a classical expression?  How to promote this to a quantum expression? What would be the meaning of division by an operator?   My expression likely contains a mixture of $x$ and $p$. For e.g., it could contain terms like $$\frac{p}{x^2}$$ or $$\frac{xp}{(x^2 + a^2)^{3/2}}.$$    How to resolve products of non-commuting operators like $x$, $p$ in a satisfactory way?
46558
A typical problem in quantum mechanics is to calculate the spectrum that corresponds to a given potential.    Is there a one to one correspondence between the potential and its spectrum?  If the answer to the previous question is yes, then given the spectrum, is there a systematic way to calculate the corresponding potential?
111787
I know that there are differences in measurement due to quantisation, probability, and collapse in quantum physics, but I am having trouble explaining how these ideas relate to measurement in an understandable manner.  In what ways does measurement in quantum mechanics differ from measurement in classical mechanics?
89599
The  is defined as $$ \rho_\psi ~:=~ \frac{\lvert\psi(t)\rangle  \langle \psi(t)\vert}{ \langle \psi(t) |\psi(t)\rangle }$$ in the Schrödinger picture. $\rho_\psi$ is obviously a time dependent projector, and the equation of motion on these projectors become: $$   i\hbar\frac{d}{dt} \rho_\psi ~=~ [H,\rho_\psi]   \tag{S} $$ but my book also reports that the  on the operators/observables is:  $$   \mathrm{i}\hbar\frac{d}{dt} A ~=~ [A,H] .  \tag{H} $$ Why are the signs in eqs. (S) and (H) opposite?   Isn't $A$ an operator like  $ \rho_\psi$, although time independent? They belong to the same operator space, so I don't think I can apply duality, but I know that $A$ operate on the states to give us the expectation value through the relation  $$  \mathrm{Tr}( \rho_\psi A)$$ so it should be in the dual space of the observables.
43272
Position, momentum, energy and other observables yield real-valued measurements. The Hilbert-space formalism accounts for this physical fact by associating observables with Hermitian ('self-adjoint') operators. The eigenvalues of the operator are the allowed values of the observable. Since Hermitian operators have a real spectrum, all is well.  However, there are non-Hermitian operators with real eigenvalues, too. Consider the real triangular matrix:  $$ \left( \begin{array}{ccc} 1 &amp; 0 &amp; 0 \\ 8 &amp; 4 &amp; 0 \\ 5 &amp; 9 &amp; 3 \end{array} \right)  $$  Obviously this matrix isn't Hermitian, but it does have real eigenvalues, as can be easily verified.  Why can't this matrix represent an observable in QM? What other properties do Hermitian matrices have, which (for example) triangular matrices lack, that makes them desirable for this purpose?
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?
56020
This question is about non-normalisable states in quantum mechanics, e.g. the eigenstates of the position operator $|x\rangle$ which are defined by the eigenvalue equation  $$\tag{1} \hat{X}|x\rangle = x|x\rangle.$$  Since the spectrum of the operator is continuous, the states are not square normalisable. Instead one has an equation such as  $$\tag{2} \langle x'|x\rangle = \delta(x-x')$$  which gives the overlap meaning as a . However it is clear that there is still at least an ambiguity in choosing a constant, i.e. why no have  $$\langle x'|x\rangle = a \times \delta(x-x')$$  where $a$ is $a \in \mathbb{R} $.    So I went to see how Equation (2) is derived. In Schleich's textbook on quantum optics in phase space I found the following, very short and in my opinion not clear/complete derivation.  From Equation (1) which can be taken to be the definition of the position operator and the assumption that the operator is hermitian which is a fundamental postulate of quantum mechanics one obtains:  $$\tag{3} \left( x - x' \right) \langle x'|x\rangle = 0.$$  That was easy so far, but then Schleich jumps immediately to equation (2). But the solutions for $\langle x'|x\rangle$ of (3) are all functions/generalised functions that are non-zero everywhere except at $x=x'$. So my question is: Why is (2) the unique solution? Where does the additional information come from?  My suspicion is that we impose another requirement, which is probably the completeness of the states, which would make (2) clear since it is the identity element in the position representation.
101251
This is a question I've been asked several times by students and I tend to have a hard time phrasing it in terms they can understand. This is a natural question to ask and it is not usually well covered in textbooks, so I would like to know of various perspectives and explanations that I can use when teaching.  The question comes up naturally in what is usually students' second course in quantum physics / quantum mechanics. At that stage one is fairly comfortable with the concept of wavefunctions and with the Schrödinger equation, and has had some limited exposure to operators. One common case, for example, is to explain that some operators commute and that this means the corresponding observables are 'compatible' and that there exists a mutual eigenbasis; the commutation relation is usually expressed as $[A,B]=0$ but no more is said about that object.  This naturally leaves students wondering     what is, exactly, the physical significance of the object $[A,B]$ itself?   and this is not an easy question. I would like answers to address this directly, ideally at a variety of levels of abstraction and required background. Note also that I'm much more interested in the object $[A,B]$ itself than what the consequences and interpretations are when it is zero, as those are far easier and explored in much more depth in most resources.    One reason this is a hard question (and that commutators are such confusing objects for students) is that they serve a variety of purposes, with only thin connecting threads between them (at least as seen from the bottom-up perspective).   Commutation relations are usually expressed in the form $[A,B]=0$ even though, a priori, there appears to be little motivation for the introduction of such terminology. A lot of stock is placed behind the canonical commutation relation $[x,p]=i \hbar$, though it is not always clear what it means.  (In my view, the fundamental principle that this encodes is essentially de Broglie's relation $\lambda=h/p$; this is made rigorous by the  but that's quite a bit to expect a student to grasp at a first go.) From this there is a natural extension to the Heisenberg Uncertainty Principle, which in its general form includes a commutator (and an anticommutator, to make things worse). Canonically-conjugate pairs of observables are often introduced, and this is often aided by observations on commutators. (On the other hand, the energy-time and angle-angular momentum conjugacy relations cannot be expressed in terms of commutators, making things even fuzzier.) Commutators are used very frequently, for example, when studying the angular momentum algebra of quantum mechanics. It is clear they play a big role in encoding symmetries in quantum mechanics but it is hardly made clear how and why, and particularly why the combination $AB-BA$ should be important for symmetry considerations.  This becomes even more important in more rigorous treatments of quantum mechanics, where the specifics of the Hilbert space become less important and the algebra of observable operators takes centre stage. The commutator is the central operation of that algebra, but again it's not very clear why that combination should be special. An analogy is occasionally made to the Poisson brackets of hamiltonian mechanics, but this hardly helps - Poisson brackets are equally mysterious. This also ties the commutator in with time evolution, both on the classical side and via the Heisenberg equation of motion.   I can't think of any more at the moment but they are a huge number of opposing directions which can make everything very confusing, and there is rarely a uniting thread. So: what are commutators, exactly, and why are they so important?
23488
In the particle in a box, harmonic oscillator and in Hydrogen Atom, we can safely assume $$\Psi(x,t) = \psi(x)e^{-i\omega t}.$$ So why not make it a postulate to consider the wave function to be always in the form $$\psi(x)e^{-i\omega t}$$ We can still explain all the foundational experiments, so my question is why not make this amendment to the theory so that we can avoid a lot of nightmares and some peace of mind. Why unnecessarily consider so general thing when we can do away with simple things most of the time?  Oh my actual question is where do we run into problems (practical physical situations/experiments) if we consider such a thing? If we run into problem anywhere, I hope to make any other simple changes to fix them but without going back on this one.
27861
Hm, this just occurred to me while answering another question:   If I write the Hamiltonian for a harmonic oscillator as $$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ then wouldn't one set of possible basis states be the set of $\delta$-functions  $\psi_x = \delta(x)$, and that indicates that the size of my Hilbert space is that of $\mathbb{R}$.  On the other hand, we all know that we can diagonalize $H$ by going to the occupation number states, so the Hilbert space would be $|n\rangle, n \in \mathbb{N}_0$, so now the size of my Hilbert space is that of $\mathbb{N}$ instead.  Clearly they can't both be right, so where is the flaw in my logic?
89742
Quantum Mechanics is very successful in determining the overall statistical distribution of many measurements of the same process.  On the other hand, it is completely clueless in determining the outcome of a single measurement. It can only describe it as having a "random" outcome within the predicted distribution.  Where does this randomness come from? Has physics "given up" on the existence of microscopic physical laws by saying that single measurements are not bound to a physical law?  As a side note: repeating the same measurement over and over with the same apparatus makes the successive measurements non-independent, statistically speaking. There could be a hidden "stateful" mechanism influencing the results. Has any study of fundamental QM features been performed taking this into account? What was the outcome?    Edit: since 2 out of 3 questions seem to me not to answer my original question, maybe a clarification on the question itself will improve the quality of the page :-)  The question is about why single measurements have the values they have. Out of the, say, 1000 measure that make a successful QM experiment, why do the single measurements happen in that particular order? Why does the wave function collapse to a specific eigenvalue and not another? It's undeniable that this collapse (or projection) happens. Is this random? What is the source of this randomness?  In other words: what is the mechanism of choice?    Edit 2: More in particular you can refer to chapter 29 of "The road to reality" by Penrose, and with special interest page 809 where the Everett interpretation is discussed - including why it is, if not wrong, quite incomplete.
101367
Given a problem, how does one know whether to use quantum mechanics or classical mechanics?   Take for example electron scattering from a nucleus. The electrons are given a wavefunction in this case. Why not use the classical approach?
54908
Is it logically correct to assert that the expectation of the momentum $$\langle \hat p \rangle=0$$ for any bound state because it is bound to some finite region? What is the physical interpretation of the fact that $$\langle \hat p \rangle=0$$ in an energy eigenstate $\psi_n(x,t)$  but $$\langle \hat p \rangle\neq0$$ in some superposition state $$\psi(x,t)=c_m\psi_m(x,t)+c_n\psi_n(x,t)~?$$ Here $\psi_n(x,t)$ the eigenstates of the Hamiltonian, for example, in the problem of particle in a box (say).
58203
I stumbled on this question while answering the following question:  Given a general wavefunction $\psi(x)$ and an operator $\hat Q$, with eigenvalue spectrum $\{q_r\}$ and corresponding eigenfunctions $\{u_r\left(x\right)\}$, show that the expectation value of $\hat Q$ is given by $$\langle\hat Q\rangle = \int_{-\infty}^\infty{\psi^*\hat Q \psi \,dx}$$  I started of with the ususal definition of the expected value and inserted the identity operator in position representation: $$\langle\hat Q\rangle = \langle \psi|\hat Q|\psi\rangle = \langle \psi|\left(\int dx|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\right)|\psi\rangle$$  Taking $\langle\psi|$ and $|\psi\rangle$ into the integral gives: $$\left(\int dx\langle \psi|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\psi\rangle\right)$$  Pulling the 2nd integral into the first:$$\int dx\int  {dx}'\langle \psi|x\rangle\langle x'|\psi\rangle\langle x| \hat Q|x'\rangle$$  This is the point where I'm stuck. If $\langle x|\hat Q|x'\rangle = \hat Q \delta(x-x')$, then the last expression would evaluate to the answer. However I  don't know why this should be true and even if it is true I wouldn't know how to prove this equality.
54295
Does there exist any geometrical description of canonical commutation relation (CCR) of quantum mechanics $$[\hat{x},\hat{p}] = i \hbar \, ,$$ e.g. in phase space?  The commutator, along with the normalization condition, leads to the discretization of energy of a bound system, so does there exist some kind of phase space representation as in  which would tell me why and how exactly using the CCR actually leads to the discretization of energy? I would like to avoid going for the Wigner representation (WR) because, if I remember correctly, the RW has commuting functions in phase space. While we're at it, does the phase space of a quantum system has to do anything with connes noncommutative geometry since $\hat{x}$ and $\hat{p}$ do not commute so we cannot do something like  as we do in classical mechanics?
62243
Alastair Rae states that there are 4 postulates of Quantum Mechanics in his text on the subject matter.  The first part of his second postulate can be stated as:      Every dynamical variable may be represented by a Hermitian operator whose eigenvalues represent the result of carrying out a measurement of the value of the dynamical variable...   My question is: just how much can we deduce from saying that the some of the operators involved in Quantum Mechanics are Hermitian?  When I use the word Hermitian, I am referring to the property that $A=A^+$ where $A$ is an operator and $A^+$ is the adjoint of $A$.  Does an operator being Hermitian automatically imply that there will exist eigenfunctions and eigenvalues corresponding to this operator?  Does being Hermitian imply that, should there exist eigenfunctions corresponding to this operator, they will form a complete and orthonormal set?  That any wavefunction can be expanded in terms of these eigenfunction?  Just how much is hidden in the word "Hermitian"?
61693
Let me remind you about the following classical  examples in quantum mechanics.     Example 1. Bound states in 1-dim potential V(x).   Let $V(x)$ be a symmetric potential i.e.   $$V(x) = V(-x)$$   Let us introduce the parity operator $\hat\Pi$ in the following way:   $$\hat\Pi f(x) = f(-x).$$   It is obvious that    $$[\hat H,\hat\Pi] = 0.$$   Therefore, for any eigenfunction of $\hat H$ we have:   $$\hat H|\psi_E(x)\rangle = E|\psi_E(x)\rangle = E\hat\Pi|\psi_E(x)\rangle,$$   i.e. state $\hat\Pi|\psi_E(x)\rangle$ is eigenfunction with the same eigenvalue. Is $E$ a degenerate level? No, because of linear dependence of $|\psi\rangle$ and $\hat\Pi|\psi\rangle.$   Consider the second example.     Example 2. Bound states in 3-dim a potential $V(r)$. Where $V(r)$ possesses central symmetry, i.e. depends only on distance to center.   In that potential we can choose eigenfunction of     angular momentum $\hat L^2$ for basis    $$|l,m\rangle,$$   where $l$ is total angular momentum and $m$ - its projection on chosen axis (usually $z$). Because of isotropy eigenfunction with different $m$ but the same $l$ correspond to one energy level and linearly independent. Therefore, $E_l$ is a degenerate level.      My question is if there is some connection between symmetries and degeneracy of energy levels. Two cases are possible at the first sight:   Existence of symmetry $\Rightarrow$ Existence of degeneracy Existence of degeneracy $\Rightarrow$ Existence of symmetry     It seems like the first case is not always fulfilled as shown in the first example. I think case 1 may be fulfilled if there is continuous symmetry. I think the second case is always true.