8

Your lecturer is trying to express some useful intuition, but in a vague way. To make it as simple as possible, let's go back to the simple harmonic oscillator. Recall that $$[a, a^\dagger] = 1, \quad [H, a^\dagger] = a^\dagger$$ and let's work throughout with units $k = m = 1$. Now suppose that some state has energy $n$, $$H |\psi\rangle = n |\psi \rangle.$...


7

Let me put on my math hat by providing a counterexample: Let's say $$ A = a_1, \\B= a_2, \\C= a_3, \\D = a_4 $$ where $a_i$ are regular bosonic anhiliation operators, therefore $$ [a_i, a_j] \equiv 0 $$ If OP's proposition works: $$ \{ A B , \, C D \} \equiv A B C D + C D A B = 2a_1a_2a_3a_4 $$ will be reduced to zero, since any combination of $[a_i, a_j] ...


5

If you want to be formal, the function $\psi : \mathbb{R}\to \mathcal{H}, t\mapsto \lvert\psi(t)\rangle$ needs to be understood as a function between Banach spaces (every Hilbert space is in particular a Banach space). The correct notion of derivative is then the Fréchet derivative. Note that this vector-valued function is much easier to differentiate ...


5

You are asking about the "physical meaning" of the celebrated Weierstrass transform, which is used routinely in physics, of course: $$ \bbox[yellow]{ e^{\partial_x^2}f(x) =\frac{1}{\sqrt{4\pi}} \int_{-\infty}^\infty f(x-y)~ e^{-y^2/4}\;dy}~. $$ In your case, $$\langle x |\exp(-a {\hat{p}}^2)| \psi \rangle = \exp(a \hbar^2 \partial_x^2)~\langle x|\psi \...


5

Yes it acts on the $A_x$: $$ (-i\hbar \partial_x A_x) \psi= ( -i\hbar A_x\partial_x) \psi+ (-i\hbar \psi \partial_x) A $$


4

$\hat{D}_x(x)=e^{-i\frac{x}{\hbar}\hat{p}_x}$: the spatial displacement operator, moves the wave function $\psi$ along the x coordinate, $\hat{p_{x}}$ is the momentum operator which generates the displacement. First let me say that using $x$ in the definition of $\hat{D}_x(x)$ is really confusing, because $x$ is also used as the spatial coordinate of ...


4

You are indeed missing some pieces. You can immediately see that there is no way your formula works in general since the full expansion reads $$ \begin{aligned} e^{ix\cdot P} K_\mu e^{-ix\cdot P} &= \sum_{n,m=0}^\infty \frac{i^{n-m}}{n!m!} (x\cdot P)^n\,K_\mu\,(x\cdot P)^m\,. \end{aligned}\tag{1}\label{ini} $$ The problem is that this is a mess because ...


4

It is just how the operators work in the position basis. I will show how it works for the position operator. The momentum operator can be handled similarly First, we know that matrix elements of the position operator in the position basis are given by $\langle x|\hat X|x'\rangle=x'\delta(x'-x)$. Second, we know the position basis vectors form a complete ...


4

All of my work is based on the assumption that I can write the following:$$\Pi|\vec{r}\rangle = -|\vec{r}\rangle$$ This is wrong. $|{-\vec r}\rangle$ is an eigenvector of the position operator $\hat{\vec x}$ with eigenvalue $-\vec r$. $-|{\vec r}\rangle$ is an eigenvector of the position operator $\hat{\vec x}$ with eigenvalue $+\vec r$, since $$ \hat{\vec ...


3

Yes, indeed, you simply need to calculate the matrix elements of the Hamiltonian in the new basis: \begin{array} \hat{H}_{11}' = \langle \phi_1'|\hat{H}|\phi_1'\rangle = \frac{1}{2}(\langle\phi_1| + \langle\phi_2|)\hat{H}(|\phi_1\rangle + |\phi_2\rangle) = \\ \frac{1}{2}(\langle \phi_1|\hat{H}|\phi_1\rangle + \langle \phi_1|\hat{H}|\phi_2\rangle + \langle \...


3

Okay, so I cleaned up the latex a bit (Use \langle and \rangle for the braket notation. Though I wish they'd install the braket package, which would make it even easier). First, you have a mistake in the first line, and it doesn't quite make sense when going to second quantization anyway: There should be no sum over individual particles any more! In your ...


3

If you had many systems in the state $|\alpha\rangle$ and you were to make measurements of the photon number for each of those systems, then $\langle\alpha|\hat n|\alpha\rangle$ is the expectation value of those measurements. This can be calculated using inner products if you can directly calculate $\hat n|\alpha\rangle$ and $\langle\alpha|\left(\hat n|\...


3

The previous answer gives the meaning, here is the calculation. The coherent state $\mid \alpha\rangle$ is an eigenvectior of the annihilation operator $\mathsf{a}$, with $$\mathsf{a}\mid \alpha\rangle= \alpha \mid \alpha\rangle .$$ The Hermitian conjugate of this equation is: $$\big(\mathsf{a}\mid \alpha\rangle\big)^+ = \langle\alpha\mid\mathsf{a}^+ = \...


3

A nice way to see what's happening is to construct an approximate "toy" state $\newcommand{\ket}[1]{\left|#1\right>} \ket r $. For instance you might construct a wavefunction that's "at" a location ten units to the right of your origin by $$ \ket{10}=\left\{ \begin{array}{cl} 1 & \text{where } 9<x<11 \\ 0 & \text{elsewhere} \end{array} \...


3

Like all proofs about properties of "maximal" sets in any context, this one too proceeds by assuming that we have a set that lacks the property and then constructing something we can add to the set, showing it was not maximal: Assume we have a set of $n$ operators $A_i$ and that there is a degenerate common eigenvector, i.e. an (w.l.o.g.) two -dimensional ...


2

Expressing the momentum operator as spatial derivative gives us: $$ \hat D_z=\exp\left(-i\frac{z}{\hbar}(-i\hbar\partial_x)\right)=\exp(z\partial_x) $$ The result of $\left.\hat D_z\psi\middle|x\right>$ can be expressed as: $$ \left.\hat D_z\psi\middle|x\right> = \psi(x)+z\psi'(x)+\frac{z^2}2\psi''(x)+\frac{z^3}6\psi'''(x)\ldots $$ which can be seen ...


2

In order for your formula to be valid, the states $|i\rangle$ must be eigenvectors of $H$ with the energy $e_i$. Otherwise you will not get $$\langle j | G_r | i \rangle = \langle j| \frac{1}{E_f-e_i+i\gamma} |i\rangle$$ as what you did was explicitly act with $H$ on the state to the right. While $|j\rangle$ might be any basis of states whatsoever, it will ...


2

The trace of any matrix/operator will be same regardless of what basis you use, provided they are complete. So it doesn't matter whether you choose eigenvectors of $A$ or $H$, but you must be consistent and use the complete basis. If you decide to use the eigenbasis of $A$, then you can't simply substitute the scalar energy $E_i$ for $H$, you must keep $H$ ...


2

Because mass is supposed to be a quantity which is to remain constant for all eternity in quantum mechanics. Mass is energy in the rest frame of the of the particle. The reason why you don’t see this in non-relativistic quantum mechanics is that you can add an arbitrary constant to the Hamiltonian. In relativistic quantum mechanics, that offset is fixed with ...


2

You have $$\sum_l \bar U_{lj} a_l = \sum_l U^\dagger_{jl} a_ l \neq \sum_l \big(U^\dagger_{jl} a_l\big)^\dagger$$ and if you had started from $\{\bar a_i, \bar a_j^\dagger\} $, you'd have $$ \{\bar a_i, \bar a_j^\dagger\} = \big\{\sum_k \bar U_{ki} a_k, (\sum_l \bar U_{lj} a_l)^\dagger\big\} = \sum_k \sum_l \big\{\bar U_{ki} a_k, (\bar U_{lj} a_l)^\dagger\...


2

The conditions, $[A,[A,B]]=[B,[A,B]]=0$, are needed for commuting operators to satisfy the identity that $[A,F(B)]=[A,B]\frac{\partial F}{\partial B}$. It turns out that these conditions are not needed if one works with anticommuting operators A and B (i.e., $\{A,A\}=\{B,B\}=0$) instead. To show this, we note that any function $F(B)$ constructed from an ...


2

The easiest way to derive your solution is probably to convert the first equality into a differential equation: $$ a(t)=e^{iHt}ae^{-iHt}\qquad\Longrightarrow\qquad \frac{d}{dt}a(t)=e^{iHt}(iHa-aiH)e^{-iHt}=ie^{iHt}[H,a]e^{-iHt} $$ where I used $$ \frac{d}{dt}e^{iHt}=iHe^{iHt}=e^{iHt}iH\qquad\qquad \frac{d}{dt}e^{-iHt}=-iHe^{-iHt}=e^{iHt}(-iH) $$ It ...


2

A Hilbert space is, by definition, an inner product space (meaning that it is also a normed space) and it is Cauchy Complete. Completeness means that all the normal definitions of limiting procedures go through as usual (with no more than minor notational changes).


2

Short answer: Intuitively, for the displacement operator, the exponential accumulates an infinite number of infinitesimal displacements, and this gives rise to an overall macroscopic finite displacement. The same principle holds for the rotation operator, i.e., accumulation of many small rotations. Since the momentum operator generates a displacement via ...


2

First of all, the phrasing of the question might be a bit misleading, since you are obviously talking not about any state, but about any eigenstate of $\mathbf{\hat{A}}$, since it satisfies the property \begin{equation} \mathbf{\hat{A}}|\psi\rangle = a|\psi\rangle. \end{equation} Note, btw, that $\mathbf{\hat{A}}$ is not necessarily Hermitian (otherwise $a$ ...


2

This is not an assumption, it is a requirement for consistency. The symmetry transformation acts on operators and states, it does not act on numbers. So the equation $A\lvert \psi_n \rangle = a_n\lvert \psi_n\rangle$ simply becomes $A'\lvert \psi_n'\rangle = a_n\lvert \psi_n'\rangle$ after applying the transformation. This equation must be true for any ...


2

It is a different application of the word determinate. Griffiths simply means that the result of a measurement is determinate if the state is an eigenfunction. That is not the same as determinism in time evolution.


2

In the Schrödinger picture, the Hilbert space $\mathcal{H}$ is physically the set of states at a given time. A function like $\psi(x,t)$ is not a state, but a time evolution of a state. Operators are also a priori not time dependent: they take functions of $x$ and return functions of $x$. A time dependent operator is really an operator valued function; you ...


2

A physical way to think about this is that, if you have a degeneracy then you have a symmetry. Namely, if $|1\rangle,|2\rangle$ are states with the same set of eigenvalues $\lambda_1,\ldots,\lambda_n$ under $A_1,\ldots,A_n$, then there exists an operator that rotates $|1\rangle$ and $|2\rangle$ into each other and leaves all other observables untouched $$ U(\...


1

I'll write down the whole calculation for completion. Firstly we define $$A = C-\langle C\rangle \qquad B = D-\langle D \rangle$$ We firstly evaluate the following $$\frac{\langle v|A^2|v\rangle}{\langle v | v \rangle} = \frac{\langle v|C^2-2C\langle C\rangle+\langle C\rangle^2|v\rangle}{\langle v | v \rangle} = \frac{\langle v|C^2|v\rangle}{\langle v | v ...


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