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1

You are on the right track. Pushing one more step to the final answer may leave you disappointed: $\sigma_x \sigma_K$ can equal zero! To see this, I find it more helpful to think just in terms of $x$ and $K$ as linear operators satisfying certain commutation relations, rather than thinking explicitly in terms of integrals of wavefunctions. Specifically, we ...


2

The 2nd way $$\langle p\rangle = \int_{-\infty}^{\infty}\frac{\hbar}{i}\frac{d}{dx}|\Psi|^2 dx$$ will produce a complex result in general (in the example above it is will simply be zero), not having a physical measurement analog. The operator operates on some vector (either $\Psi$ or $\bar{\Psi}$), whereas the $|\Psi|^2$ is a simple real number.


11

So I believe it's standard to place the operator inbetween the conjugate of the wavefunction and the wavefunction itself. For instance, $$\langle p\rangle = \int_{-\infty}^{\infty}\Psi * \frac{\hbar}{i}\frac{d}{dx}\Psi dx$$ Yes, that is correct, and Is it wrong to do this? $$\langle p\rangle = ...


2

Since $\hat{p}$ is a Hermitian operator, one can always expand the wave function $|\psi\rangle$ as a linear combination of the eigenstates of $\hat{p}$, $$|\psi\rangle=\sum_{p}\psi(p)|p\rangle,$$ where the eigenstate $|p\rangle$ satisfies the equation $\hat{p}|p\rangle=p|p\rangle$. With this setup, we can first show $\langle\psi|\hat{p}|\psi\rangle=\langle ...


0

If I understand correctly the question (not sure I do...), The question is basically a linear algebra one. Consider an operator $\hat{A}$, that has a eigenvalue $a$, the eigenfunctions/eigenvectors of $\hat{A}$ are denoted by $|a\rangle$ such that: $$ \hat{A}|a\rangle=a|a\rangle $$ Now consider a composite of $\hat{A}\circ\hat{A}$, that operates on ...


1

However then I say: Why do you make things so complicated? Suppose you want to calculate $\exp(A)$. Why don't you define $$\exp(A)~:=~1+A+1/2 A^2 + \ldots $$ and require convergence with respect to the operator norm. An example: Consider the vectorspace spanned by the monomials $1,x,x^2,\ldots$ and let $A=d/dx$. Then you can perfectly define ...


5

Consider the general case that we want to calculate $$ \langle p |F(r) |p'\rangle.$$ By inserting the resolution of the identity $\int d^3r\, |r\rangle\langle r|$ we find that we need to compute $$\tilde{F}(q = p-p') = \int d^3 r \, e^{i(p-p')r} F(r). \tag{1}$$ This integral will converge if $\int dr\, |F(r)|$ is finite. Such a function is said to be $L^1$. ...


0

The only crucial point is the degeneracy of eigenspaces. Consider the finite dimensional Hilbert space $\cal H$ (the extension to the infinite dimensional case is more difficult also because a part of continuous spectrum may appear) and a pair of commuting Hermitian operators $A$ and $B$ on that space such that the following requirement is satisfied. R.: ...


2

Consider self-adjoint operators $A$ and $B$, on a Hilbert space $\mathscr{H}$. Roughly speaking (forgetting about domains of definition), it is possible only if your Hilbert space can be written as $\mathscr{H}=\mathscr{H}_1\otimes \mathscr{H}_2$; where $\mathscr{H}_1$ and $\mathscr{H}_2$ are Hilbert spaces such that: $A=A_1\otimes 1$ and $B=1\otimes B_2$, ...


2

A very explicit argument: $$\phi(\xi)=(\xi-c_1)\dots (\xi-c_{r-1})(\xi-c_r)(\xi-c_{r+1})\dots(\xi-c_n)$$ where $r\leq n$ and $c_k> c_l$ whenever $k>l$. We can order the $c_i$ like this because the $c_i$ are real. Now, $$X_r(\xi)=(\xi-c_1)\dots (\xi-c_{r-1})(\xi-c_{r+1})\dots(\xi-c_n) $$ Clearly, the set of zeros of $X_r(\xi)$ is $\{c_i\ \bigl|\ ...


2

If all the c's are different, then since $X_r(\xi)$ is the quotient when $\phi(\xi)$ is divided by $(\xi-c_r)$, $(\xi-c_r)$ cannot be a factor of $X_r(\xi)$ else two of the c's would equal $c_r$. $X_r(c_r)$ is, by the remainder theorem (I imagine this has some other name), the remainder when $X_r(\xi)$ is divided by $(\xi-c_r)$. Since $(\xi-c_r)$ is not a ...


2

Here is another version of the same proof. If $\langle \psi | A \psi \rangle \in \mathbb R$ for all $\psi \in \cal H$, then $\langle \psi | A \psi \rangle^* = \langle \psi | A \psi \rangle$ for all $\psi \in \cal H$. Since $\langle \psi | \phi \rangle^* = \langle \phi| \psi\rangle$ we have that $\langle \psi | A \psi \rangle = \langle A\psi | \psi ...


2

The assumption that $\mathcal{D}$ is invariant under $\phi(f)$ for each $f\in \mathcal{S}(\mathbb{R^4})$, the Schwartz space of functions of rapid decrease is one of the Wightman axioms. Its main use is for the vacuum expextation values $(\psi_0,\phi(f_1)...\phi(f_n)\psi_0)$ to make sense (where $\psi_0$ is the vacuum state), what would not happen in general ...


6

The sort of trick involved in removing the $|P\rangle$ on both sides to get the conjugate imaginary equation $$\langle P|\xi|P\rangle = \langle P|a|P\rangle \tag1 $$ is quite common but it is indeed nontrivial to grasp the first time. In essence, you leverage the fact that in an equation of the form $$ ⟨\psi|\hat A|\phi⟩=⟨\psi|\hat B|\phi⟩\tag2 ...


1

Think of $\lvert \psi \rangle$ as being written $a \lvert n \rangle + b \lvert n{+}1 \rangle$ -- it is just a linear combination of $\lvert n \rangle$ and $\lvert n{+}1 \rangle$ with (possibly complex) coefficients $a$ and $b$. Converting from a ket to a bra (i.e., finding the dual) distributes over addition and scalar multiplication, and it ...


1

First, note that a unitary transformation can not modify the commutation relations.. $$AB-BA=C$$ Use the fact that $U^\dagger U=U U^\dagger=1$ to get, $$AU U^\dagger B-BU U^\dagger A=C$$ and then multiply by the conjugate transpose from the left and $U$ from the right, $$ U^\dagger AU U^\dagger B U^\dagger- U^\dagger BU U^\dagger AU^\dagger= U^\dagger C ...


3

If $A^\prime$ and $B^\prime$ commute then there exists a set of mutual eigenvectors of $A^\prime$ and $B^\prime$. For any eigenbasis of $A^\prime$ there exists a unitary transformation $W$ which takes that basis to the mutual eigenbasis of $A^\prime$ and $B^\prime$. Consequently if there is a unitary operation such that $ |\langle \psi | b \rangle |^2 = ...


0

What you want to ask is whether the position operators of a system with two harmonic oscillators , labeled as $A$ and $B$, commute. If the system is not entangled, so operator $\hat{q_a}$ act only on $A$, $\hat{q_b}$ act on $B$ individually. Assume that the state of the system is $\psi$ , $\psi=\psi_a\psi_b$. So ...


0

Doesnt the system collapse into ζ1 given we know this is the state at time t=0? True, but after the measurement at $t_0$ the system follows the unitary dynamics given by $H$. As you can easily check, $\zeta_1$ is not an eigenstate of the Hamiltonian, iff $E_1\neq E_2$. Hence the system will not remain in the state $\zeta_1$ during time-evolution. If ...


0

A really cool example came up when a friend of mine and me discussed a spin-1/2 system. I had some trouble and it turns out that exactly the same misunderstanding that started this topic lead me to really strange results: I'll write $S_x, S_y$ and $S_z$ for the canonical components of the spin operator and $|S_i; \pm\rangle$ for its eigenvectors, i.e. $S_y ...


0

It doesn't matter that the ladder operators are Hermitian or not. If they commute with some other operator, then you can have simultaneous eigenvalues. The relation between simultaneous eigenstates and commutation relations is a result of mathematics, not physics.


1

It is a distinction corresponding to different types of spectral measures. The absolutely continuous spectrum corresponds to absolutely continuous measures, singular spectrum to continuous singular measures (both with respect to Lebesgue measure). Refer e.g. to Reed-Simon Chapter VII for a more detailed description.


1

Besides the truncated BCH formula (which is also proven in this Phys.SE post and mentioned in Steven Mathey's answer), in practice one often wants to normal order the differential operators, i.e. putting all the $x$'s to the left of all the $\partial_x$. To this end the formula $$\tag{1} e^{a\partial_x}f(x) ~=~ f(x+a)e^{a\partial_x}. $$ is useful. Hence ...


3

Suresh is right, the Baker-Campbell-Hausdorff formula will save you. If $\left[A,B\right]$ commutes with $A$ and $B$ you have, $$ \exp\left(A\right)\exp\left(B\right) = \exp\left(A+B+\frac{1}{2}\left[A,B\right]\right) \, .$$ Then the answer to your question is $$ \exp\left(\partial_x\right)\exp\left(x\right) = 2 \exp\left(x+\partial_x\right) \sinh(1/2) \, ...


7

Calculus method The Taylor series of a function of $d$ variables is as follows: $$f(\mathbf{x} + \mathbf{y}) = \sum_{n_1=0}^{\infty}\ldots\sum_{n_d=0}^{\infty} \frac{(y_1-x_1)^{n_1} \ldots (y_d - x_d)^{n_d}}{n_1!\ldots n_d!} \left. \left( \partial_1^{n_1}\ldots \partial_d^{n_d} \right) f \right|_{\mathbf{x}}.$$ where $\partial_i^{n_i} f$ means "the ...


2

Compute the time dependence of $a$ in the interaction picture first (no hats here, because it's too much typing). Let $a_I$ denote the lowering operator in the interaction picture. Define the following symbols: $H_i \equiv \hbar \omega \left( a_i^{\dagger}a_i + 1/2 \right)$ $H_0 \equiv \sum_i H_i$ $K_i \equiv H_i/\hbar$. $K \equiv H_0/\hbar$ $\bar{a}_i ...


2

Hints: Assume that $H$ is a complex Hilbert space. Assume that $A:H\to H$ is a normal operator$^1$. Then a version of the Spectral Theorem says that $A$ is orthonormally diagonalizable. Let $(\lambda_i)_{i\in I}$ denote the set of different eigenvalues of $A$ with corresponding multiplicities $(m_i)_{i\in I}$. Let $P_i$ be the orthogonal projection ...


0

$\langle \phi_{k'} \rvert$ is not an eigenfunction of $G_k$, hence the second line of your derivation is not correct. The eigenfunctions are $\langle \psi_{k} \rvert$, and that is exactly what we usually solve Lippmann-Schwinger equation for.


2

The "wavefunction" way of talking about things is a special case of the more abstract "Hilbert space" formulation. The abstract formulation says that states live in a Hilbert space, that is a complex vector space with an inner product (plus some technical assumption about completeness). The Hamiltonian is then a linear operator on that space. The way you ...


5

This may be a bit more abstract than what you're looking for, but this is just a special case of a more general construction; determining the infinitesimal generators of a representation of a Lie Group acting on a vector space. In the case at hand, the Lie group is $G=\mathbb R^3$, the group of spatial translations, and the vector space is the Hilbert space ...


1

The action of an operator is not equivalent to performing a measurement. For example, for a spin-1/2 particle in the eigenstate "spin up along z", measuring spin along y produces either the eigenstate "spin up along y" or "spin down along y". But the action of $\hat{S}_y$ on "spin up along z" creates a superposition of the two spin y eigenstates. I'll ...


1

The relation $$\langle a|b\rangle\propto\delta(a-b)$$ is nothing unusual, it is simply an orthogonality condition. If the proportionality was an equality, and in addition we had completeness, the set of states would form an orthonormal basis. The reason why the delta function shows up is that you assume your operator to have a continuous spectrum of ...


1

The steps that you wrote down till eq. 1 is in fact a simple proof of the following theorem (which can be looked up in elementary text books of quantum mechanics): Eigen functions (of a Hermitian operator or more generally a symmetric operator on a separable Hilbert space) belonging to distinct eigenvalues are orthogonal. This is always true for separable ...


2

While i was typing, two good answers were posted. Since I don't want to delete everything, I'll leave this here nontheless. Without appealing to Lie theory, one might argue by physical reasoning. The unitary operators your book has in mind depend on a continous parameter $\alpha$. They describe continous transformations of the quantum mechanical state ...


3

Well, quantum mechanics is famous for not being intuitive for earthlings like us, but the following couple of facts might help: Observables in quantum mechanics are Hermitian/selfadjoint operators. The spectrum ${\rm Spec}(\hat{A}) \subseteq \mathbb{R}$ of a Hermitian/self-adjoint operator $\hat{A}$ belongs to the real axis $\mathbb{R}\subseteq ...


11

There's no escaping the Lie theory if you want to understand what is going on mathematically. I'll try to provide some intuitive pictures for what is going on in footnotes, I'm not sure if it will be what you are looking for, though: On any (finite-dimensional, for simplicity) vector space, the group of unitary operators is the Lie group $\mathrm{U}(N)$, ...


3

The momentum and spin operators do commute. Since $H_s$ is a sum of products of commuting Hermitian operators, it is Hermitian (assuming $\alpha$ is real). The matrix you have written to represent $H_s$ is correct and Hermitian. But you're right that it appears to be anti-Hermitian. To see what's wrong, consider the simpler $1\times 1$ matrix $$ ...


3

The relation (4) literally switches the states, adds an overall complex conjugate, and removes a hermitian bar over the operator. (Actually, no one uses bars anymore to denote hermitian conjugates, they use daggers instead. And because stacked bars get ugly, I'll use stars for complex conjugation of plain complex numbers.) Thus we have $$ \langle B \mid ...


1

Note that the spectrum ${\rm Spec}(\hat{A}) \subseteq \mathbb{R}$ of a Hermitian/self-adjoint operator $\hat{A}$ belongs to the real axis $\mathbb{R}\subseteq \mathbb{C}$, cf. e.g. this Phys.SE post. It is therefore not surprising that a reality condition for a classical field naturally translates into a Hermiticity/self-adjointness condition for the ...


1

As other answers mention, it was originally (in QED) about getting a neutral vacuum. It is useful to go back to Schwinger's old version of QED, before Dyson's approach became accepted. See Pauli: Selected topics in field quantization. Pauli presents both ways of looking at it: 1) define the electric current as sum of two terms (p.20 [6.4]), such that the ...


2

First of all, note that the radial operator ordering ${\cal R}$ is implicitly implied in many textbooks of CFT (e.g. Ref. 1). For instance, eq. (2.2.7) on p. 39 in Ref. 1 is discussing Wick's theorem between two operator ordering prescriptions. In this case between normal ordering $:~:$ and radial ordering ${\cal R}$. See also e.g. this Phys.SE post. The ...


2

The choice of normal ordering prescription $:~:$ is typically adjusted to the choice of vacuum state $|\Omega\rangle$ so that the bra-ket-sandwich of normal-ordered operators $$\langle \Omega|:\hat{\cal O}_1\ldots \hat{\cal O}_n : |\Omega\rangle~=~0 $$ vanishes. The relation of normal ordering prescription to Wick theorem and other operator ordering ...


2

In classical physics, quantities are ordinary, commuting $c$-numbers. The order in which we write terms in expressions is of no consequence. In quantum field theory (QFT), on the other hand, quantities are described by operators that, in general, don't commute. Classical physics is a low-energy approximation of quantum physics - the road from quantum to ...



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