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1

Your Hilbert space is finite-dimensional (specifically, 4-dimensional). It means that you need no fancy way of deriving the spectrum, just proceed with the standard approach: Write your Hamiltonian in the matrix form and solve the characteristic equation $ \det ( H - \lambda \cdot 1_{4 \times 4} ) = 0 $ with respect to $\lambda$, which would give you ...


3

He's doing a linear approximation. Suppose $\Delta x$ is very small. Then $\langle x - \Delta x | \alpha \rangle$ is almost equal to $\langle x | \alpha \rangle$, but not quite, because $\Delta x$ isn't zero. So we do a first order approximation: Let's write $\langle x | \alpha \rangle$ as $f(x)$. Then $f(x - \Delta x) \approx f(x) - \Delta x ...


0

Notice that in $[\partial_xO(x),O(x')]$, the partial derivative only acts on $x$, not on $x'$. So we can pull the partial derivative operator out of the bracket and get $\partial_x[O(x),O(x')]$.


0

It boils down to a matter of convention. Nothing stops you from choosing the annihilation operator to be $a^\dagger$. Still, in quantum field theory, you decompose e.g. a scalar quantity in plane waves $$ \Phi(x) = \int \frac{d^4 k}{4 \pi} \left( a(k) e^{ik\cdot x} + a^*(k) e^{-ik\cdot x} \right)$$ where obviousely the second part is the c.c. of the first ...


0

I came here before asking myself the same question and, as I figured it out, I'd like to answer just for the record. Respecting the former answer, in the Schrödinger picture, if $\phi(x)=\sum_i\left(\alpha_i(x){U}+b_i(x){V}\right)$, the condition $[\phi(x),\phi(y)]=0$ implies $[U,V]=0$ and thus also $[\partial_i\phi(x),\phi(y)]=0$. Now, in the Heisenberg ...


0

As was said in the comments, my attempt to a solution is to decompose $\hat{x}$ as a linear combination $\alpha \hat{A} + \beta \hat{B}$. This seems to be general, because if $\hat{p}$ is a linear combination of $\hat{A}$ and $\hat{B}$, then the position will be too, since it's the inverse Fourier transform of $\hat{p}$ and such transformation is linear. I'd ...


3

Yes, operators in quantum mechanics can be understood basically as infinite matrices, $|r\rangle$ as basis vectors and $\psi(r)\equiv \langle\psi|r\rangle \sim \psi_r \sim "\psi_i"$ as components of the state vector numbered by a continuous index $r$. $\langle r|F|r'\rangle$ are indeed just matrix components of the operator $F$. Generally $\langle ...


3

In a certain sense, what you said that every operator might be represented in position representation as a integral operator may be true for many operators only if you allow distributions to be used, and even though that's not always the case. You are kind of confusing things when you say about the dual of distributions. What is a distribution is the ...


3

Unfortunately, I am not so sure what ⟨r|F|r′⟩ is ... in order to evaluate this expectation value it's not an expectation, it is a matrix element; think of it as the components of the operator $F$ on the position basis. If the operator is 'diagonal' on the position basis then $\langle r|F|r' \rangle$ is zero except when $r = r'$. Thus, for example, ...


0

The operator $\vec s\cdot \vec n$ is just the inner product $$ \vec s\cdot \vec n = s_x n_x + s_y n_y + s_z n_z $$ Here, you substitute Pauli matrices for $s_x,s_y,s_z$ (they should be called $\sigma$, not $s$, but I adopted your conventions) and $$ (n_x,n_y,n_z) = (\sin\beta\cos\alpha,\sin\beta\sin\alpha,\cos\beta)$$ for the unit vector in the desired ...


3

There is no way to represent Grassmann variables using matrices ! Actually, this is the big obstacle that hinders the use of the so-called quantum state diffusion approach for systems placed in Fermionic baths. You may find many papers on this by googling this topic. Also, individual Grassmann variable has no physical meaning. It is something invented mostly ...


0

Since there still isn't an answer but the question has attracted a few upvotes, let me elaborate on my comment. This is more maths, than physics, but anyway. Writing $\sum_{jk} \mathrm{tr}(E_j^{\dagger}E_i)|j\rangle\langle k|$ doesn't give you anything. This is indeed a rank-one decomposition, but the theorem does not tell you that ANY rank-1 decomposition ...


0

A measurement is not a primitive in physics. Rather, a measurement is a physical process that takes place according to the same laws of physics as any other physical process. Those same laws apply to the measurement apparatus, to the person doing the measurement and to the records he makes of the measurement. What distinguishes a measurement from any other ...


0

What you are asking is called, in mathematical terms, spectral theorem. I don't know how much you are interested in details, but any self-adjoint operator $A$ (linear partial differential operator on a Hilbert space) can be written as $$A=\int_{\sigma(A)}\lambda dP_\lambda\;,$$ where $\sigma(A)$ is the spectrum of $A$ and $dP_\lambda$ the spectral measure ...


2

It is usually very difficult to give a characterization of the domain of self-adjointness of an operator. However, the Harmonic oscillator is a well-known operator. Unluckily, this does not mean there is a completely explicit form of its domain. Anyways, I will give you what in my opinion is the best shot at explicitness: As you may know there are ...


0

The measurement procedure described by the measurement is as follows: You have a quantum state and do some measurement on it. Doing a measurement means you get one of a set of results. For example, if you measure the component of the spin of an electron in a certain direction, you have two possible results: Spin up, or spin down. The state of the electron ...


3

First of all, if you focus on proper functions, instead of elements of $L^2$, the domain is much more tough than your candidate ($L$ extended to our domain is again simply essentially self adjoint but not self-adjoint). The self-adjointness domain contains functions which are nowhere differentiable. A trivial example: If you consider the simpler operator: ...


1

You found $$x = e^{ik} + e^{-ik} - 2 = 2(\cos(k) - 1) \leq 0 . $$ If the question is whether $g\ge 0$ then the answer is "no" in the sense that it's eigenvalues are non-positive as we have just showed. A bit of intuition We recognize $$\phi(n+1) + \phi(n-1) -2\phi(n)$$ as some kind of second derivative. Consider the first derivative of a function $f$ ...


8

Qmechanic explained a way in which something with the word "commutator" in it doesn't vanish when applied to two of the same operator. However, I feel it is necessary to point out that plain commutators, as seen in a quantum mechanics course, really, honestly, always, and without fail satisfy $[Q,Q] = 0$ for any operator $Q$. This is because $[A,B]$ is ...


7

I) Yes, they are probably referring to that a Grassmann-odd operator needs not (super)commute with itself. Take e.g. the 1st order Grassmann-odd differential operator $$\tag{1} D~:=~\frac{d}{d\theta}+ \theta\frac{d}{dt}. $$ In eq. (1) $t$ is a Grassmann-even variable and $\theta$ is a Grassmann-odd variable, which (super)commute $$\tag{2} ...


3

Choose the momentum representation, $$x_i = i \hbar \frac{\partial}{\partial p_i}$$ distribute $i \hbar$ and act the commutator on vector $\psi$, $$[x_i, F(\mathbf p)] \psi = i \hbar \left(\frac{\partial}{\partial p_i}(F(\mathbf{p}) \space \psi) -F(\mathbf p) \frac{\partial }{\partial p_i} \psi \right)$$ and apply the product rule: $$= i \hbar ...


3

The commutation of two variables, in some cases, can be related to Poisson Bracket via $$ \left[\hat A,\,\hat B\right]=i\hbar\left\{\hat A,\,\hat B\right\} $$ Thus, $$ \left[\hat A,\,\hat B\right]=i\hbar\sum_i\left(\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i}-\frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i}\right)\tag{1} $$ ...


3

As @Qmechanic pointed out in a comment, we are free to use any operator representation. In momentum space, $\hat{\bf x} = + i \hbar \ \partial/\partial {\bf p} $ and $\hat{\bf p} = {\bf p}$, so $$ \begin{eqnarray} \left[\hat{x}_i,F\left(\hat{\bf p}\right)\right] &=& \left[i \hbar \frac{\partial}{\partial p_i},F\left({\bf p}\right)\right] \\ ...


3

Let me expand a bit on the intuition part and write down an example. This is all essentially already covered by yuggib's answer. Your confusion about positive operator valued measures, as also pointed out, is that they are not to be confused with measurement outcomes. The problem with measurement outcomes is that they are rather arbitrary. Often, they rely ...


3

To expand the comment, spectral measures, or projection valued measures are introduced to characterize self-adjoint operators. They are families of orthogonal projections on the Hilbert space that, when acting on vectors suitably, define a measure. If you denote by $\{P_\lambda\}_{\lambda\in\mathbb{R}}$ this family, a self adjoint operator $A$ corresponding ...


3

I assume that $n =1,2,\ldots$ and I indicate by $\psi_n$ the unit vector $|n\rangle$. A generic vector in the Hilbert space can therefore be written as $$\psi = \sum_{n=1}^{+\infty} c_n \psi_n$$ where $\sum_n |c_n|^2 < +\infty$. The action of $R$ and $L$ on that vector respectively is: $$R \psi = \sum_{n=1}^{+\infty} c_n \psi_{n+1}$$ and $$L \psi = ...


2

The translational operator works as $$ \tau(a)\lvert x\rangle = \lvert x+a \rangle $$ on position eigenstates. This implies through $$ \langle x \rvert \hat x + a\lvert x \rangle = \langle x \rvert \hat x \lvert x \rangle + \langle x \rvert a \lvert x \rangle = x + a = \langle x + a \rvert \hat x \lvert x + a \rangle = \langle x \rvert ...


0

Consider the kinetic energy operator $$\hat{K}=\frac{1}{2m}\hat{p}^{2}.$$ Then $[r^{-1}, \hat{K}]$ is ambiguous depending on the coordinate system. Moreover, in Cartesian coordinates, it is quite peculiar compared to its Poisson bracket... (At least, if I did my math correctly, which is possible considering how sloppy/quick it was done, as I am ...


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

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

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 ...


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 ...


9

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 = ...



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