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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 $\mathrm{U}(N)$, ...


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 = ...


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 ...


6

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 $L^1$. ...


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 ...


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 ...


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 ...


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) \, ...


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 ...


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

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 = ...


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 ...


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 ...


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 ...


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

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 ...


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

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

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

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

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.


2

The problem here is how to quantize systems whose classical hamiltonian involves factors of the form (for example) $p^nx^m$, because these cannot be unambigously represented in a formalism where $p$ and $x$ do not commute. As such there are many alternatives (all of which are classically equivalent) but only one is quantum-mechanicaly relevant. In most ...


2

Yes, this is possible - but only for states with zero total angular momentum. To see why, the first step is seeing that if $\Delta\sigma_x^2=\Delta\sigma_y^2=0$ on state $\psi$, then $\psi$ is an eigenstate of both $\hat\sigma_x$ and $\hat\sigma_y$: $$ ...


2

I don't know how elementary you consider a simple position dependent mass, but due to ordering ambiguity in the kinetic term $\hat{p}^2/2m(\hat{r})$ such a system will have a quantum Hamiltonian different from the classical one. For example: Analytic results in the position-dependent mass Schrodinger problem Position-dependent effective masses in ...


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 ...


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 ...


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 ...


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 ...


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 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 ...



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