44

Observables don't commute if they can't be simultaneously diagonalized, i.e. if they don't share an eigenvector basis. If you look at this condition the right way, the resulting uncertainty principle becomes very intuitive. As an example, consider the two-dimensional Hilbert space describing the polarization of a photon moving along the $z$ axis. Its ...


37

Self adjoint operators enter QM, described in complex Hilbert spaces, through two logically distinct ways. This leads to a corresponding pair of meanings of the commutator. The former way is in common with the two other possible Hilbert space formulations (real and quaternionic one): Self-adjoint operators describe observables. Two observables can be ...


32

Poisson brackets play more or less the same role in classical mechanics that commutators do in quantum mechanics. For example, Hamilton's equation in classical mechanics is analogous to the Heisenberg equation in quantum mechanics: $$\begin{align}\frac{\mathrm{d}f}{\mathrm{d}t} &= \{f,H\} + \frac{\partial f}{\partial t} & \frac{\mathrm{d}\hat f}{\...


31

Recall that commuting observables in quantum mechanics are simultaneously observable. If I have observables A and B, and they commute, I can measure A and then B and the results will be the same as if I measured B and then A (if you insist on being precise, then by the same I mean in a statistical sense where I take averages over many identical experiments). ...


22

If OP wants to evaluate $[A,e^B]$ in terms of $[A,B]$, there is a formula $$\tag{1} [A,e^B] ~=~\int_0^1 \! ds~ e^{(1-s)B} [A,B] e^{sB}. $$ Proof of eq.(1): The identity (1) follows by setting $t=1$ in the following identity $$\tag{2} e^{-tB} [A,e^{tB}] ~=~ \int_0^t\!ds~e^{-sB}[A,B]e^{sB} .$$ To prove equation (2), first note that (2) is trivially true ...


20

Short answer: complexifications facilitate representation theory. In physics, we typically want to find representations of a Lie algebra $\mathfrak g$, and often times determining the representations of its complexification $\mathfrak g_\mathbb C$ is easier. Moreover, we have the following theorem (see ref 1. Proposition 4.6) which tells us that ...


20

The classical poisson bracket with the generator of any transformation gives the infinitesimal evolution with respect to that transformation. The familiar $$ \partial_t f = \{H,f\}$$ means nothing else than the time evolution of any observable is given by its Poisson bracket with the Hamiltonian, which is the generator of time translation. More generally, ...


17

According to the topic of deformation quantization, the first few entries in the dictionary between $$ \text{Quantum Mechanics}\quad\longleftrightarrow\quad\text{Classical Mechanics}\tag{0}$$ read $$ \text{Operator}\quad\hat{f}\quad\longleftrightarrow\quad\text{Function/Symbol}\quad f,\tag{1}$$ $$ \text{Composition}\quad\hat{f}\circ\hat{g} \quad\...


17

I) Here we will work in the Heisenberg picture with time-dependent self-adjoint operators $\hat{Q}(t)$ and $\hat{P}(t)$ that satisfy the canonical equal-time commutation relation $$\tag{1} [\hat{Q}(t),\hat{P}(t)]~=~ i\hbar~{\bf 1}, $$ and two complete sets of instantaneous eigenstates$^1$ $\mid q,t \rangle $ and $\mid p,t \rangle $, which satisfy $$\tag{...


17

In a context like this, a symmetry is a transformation that converts solutions of the equation(s) of motion to other solutions of the equation(s) of motion. In this case, the equation of motion is the Schrödinger equation $$ i\hbar\frac{d}{dt}\psi=H\psi. \tag{1} $$ We can multiply both sides of equation (1) by $U$ to get $$ Ui\hbar\frac{d}{dt}\psi=UH\psi. ...


16

There's a trick to proving this result which you would certainly be forgiven for not spotting! Consider the quantity $\exp(\mu X) \exp(\mu Y)$ which appears on the right hand side. Now differentiate this with respect to $\mu$. Say what!? Yeah, bear with me, just try it: $$\frac{d}{d \mu} \exp(\mu X) \exp(\mu Y) = X\exp(\mu X) \exp(\mu Y) + \exp(\mu X) Y \...


15

In the position representation, the matrix elements (wavefunction) of a momentum eigenstate are $$\langle x | p\rangle = \psi_p(x) = e^{ipx}$$ The wavefunction shifted by a constant finite translation $a$ is $$\psi(x+a)$$ Now the momentum operator is the thing which, acting on the momentum eigenstates, returns the value of the momentum in these states, ...


15

Actually your result doesn't quite follow. Something slightly more general does. Everything begins and ends with the canonical commutation relationship (CCR): $$[\hat{\mathbf{X}},\,\hat{\mathbf{P}}]=\hat{\mathbf{X}}\,\hat{\mathbf{P}} - \hat{\mathbf{P}}\,\hat{\mathbf{X}} = i\,\hbar\,\mathbf{I}\quad\quad\quad\quad(1)$$ where $\hat{\mathbf{X}}$ and $\hat{\...


15

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}\\ \hat{a}^\dagger|n\rangle&=\sqrt{n+...


15

I'd like to expand a little bit on the interpretation of commutators as a measure of disturbance (related to incompatibility, as touched on in the other answers). My interpretation of the commutator is that $[A,B]$ quantifies the extent to which the action of $B$ changes the value of the dynamical variable $A$, and vice versa. Let's assume that $A$ is a ...


15

They're really the same $\hbar$, and they come from the exact same place. To see this, it's best to work in Heisenberg picture, where the Schrodinger equation becomes $$\frac{dA}{dt} = \frac{1}{i \hbar} [A(t), H]$$ for any quantity $A(t)$. Now compare this to the equation of motion in the Hamiltonian formalism of classical mechanics, $$\frac{df}{dt} = \{f, ...


15

The answer is basically the same as the answer to this question. It's a very subtle point about infinite-dimensional Hilbert spaces. Note that the fact that your commutator is nonzero means that the operators must act on an infinite-dimensional Hilbert space (or else you could take the trace and would automatically get 0). So you need to be careful about the ...


14

You have already got "practical" answers, so I intend to answer form another point of view. There is a quite famous theorem due to Stone and von Neumann, later improved by Mackay, and finally by Dixmier and Nelson, roughly speaking establishing the following result within the most elementary version. (Another version of the theorem focuses on the unitary ...


14

It seems that the question (v1) is caused by the fact that there are two different notions of the commutator: One for group theory: $$\tag{1} [A,B] ~:=~ ABA^{-1}B^{-1}$$ (or sometimes $[A,B] := A^{-1}B^{-1}AB$, depending on convention), which is relatively seldom used in physics. One for rings/associative algebras: $$\tag{2} [A,B]:=AB-BA,$$ which is ...


14

Although the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}$ is the most famous, there is an expression for the commutator: $$[\gamma^\mu,\gamma^\nu] = 2\gamma^\mu \gamma^\nu - 2 \eta^{\mu\nu}$$ The matrix defined by $[\gamma^\mu,\gamma^\nu]$ actually has a purpose: it forms a representation of the Lorentz algebra. If we define $S^{\mu\nu}$ as $1/4$ the ...


13

Those things are surely not enough to find the inner product $\langle q|p\rangle$ uniquely. For example, starting with the conventional $Q,P$, you may redefine them by a canonical transformation, for example by $$ Q\to Q'=Q, \quad P\to P'= P + Q^3 $$ Then $P', Q'$ obey all the four conditions in the same way as $P,Q$. They also have eigenstates and $|p'\...


13

No. You can add an arbitrary constant shift (or an arbitrary operator commuting with $x$) without affecting the CCR. For 1-dimensional QM, the general solution of the CCR with $\hat x$ represented as multiplication by $x$ on wave functions with argument $x$ is $\hat p=\hat p_0-A(\hat x)~~$, where $\hat p_0$ is the canonical momentum operator , and $A(x)$ is ...


13

Commutativity is not a transitive relation: If operator $A$ commutes with $B$ and $C$, $$AB=BA \quad\text{and}\quad AC=CA,$$ then there is no reason that $B$ and $C$ should commute. Example: Take $A=y$, $B=x$, and $C=p_x$. In particular, if commuting selfadjoint operators $A$ and $B$ have a common basis of orthonormal eigenvectors, and if commuting ...


13

Does this mean that the operator $\hat O$ (an observable) is special in some way? I believe it means there is no such $\hat O$. If $\hat O$ corresponds to an observable, we require the eigenvalues to be real. Let $|o\rangle$ be an eigenket of $\hat O$ with real eigenvalue $o$: $$\hat O |o\rangle = o |o\rangle$$ Now consider the following $$\hat O \...


12

Both the commutator (of matrices) and the Poisson bracket satisfy the Jacobi identity, $[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0$. This is why Dirac was inspired by Heisenberg's use of commutators to develop a Hamilton-Jacobi dynamics style of Quantum Mechanics which provided the first real unification of Heisenberg's matrix mechanics with Schroedinger's wave ...


12

So, why can't the uncertainty relations be violated in such a case, if I could, say, measure the position of the object with this wave function That's the catch. You can't. Or rather, you can measure the position, but the result you get will vary from one measurement to the next, because the wavefunction $\exp(x^2/2i - cx)$ is not an eigenstate of position....


12

No. In full generality, the super-commutator of two fundamental fields is identical to the Dirac bracket of the corresponding classical variables (modulo the standard obstructions). If the system is unconstrained, then the Dirac bracket agrees with the Poisson bracket, which means that two independent fields super-commute. But if there are non-trivial ...


11

Your running into circles will stop once you commit yourself to a choice. What to regard as postulate is always a matter of choice (by you or by whoever writes an exposition of the basics). One starts from a point where the development is in some sense simplest. And one may motivate the postulates by analogies or whatever. The CCR are a simple coordinate-...


11

I) The proofs of both the first (algebraic) Bianchi identity and the second (differential) Bianchi identity crucially use that the connection $\nabla$ is torsionfree, so they are not entirely consequences of the Jacobi identity. Proofs of the Bianchi identities are e.g. given in Ref. 1. II) The second Bianchi identity may be formulated not only for a ...


11

This is a basic example of a BCH formula. There are many ways to prove it. For example, write the exponential as $$ \exp(\mu X + \mu Y) = \lim_{N\to \infty} \left(1 + \frac {\mu X+ \mu Y}N\right)^N = \dots $$ Because the deviations from $1$ scale like $1/N$, it is equal to $$ = \lim_{N\to \infty} \left[\left(1 + \frac {\mu X}N\right)\left(1 + \frac {\mu Y}...


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