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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 = ...


9

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 ...


8

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} ...


6

This issue is a bit confused in textbooks, however the statement of the professor is physically wrong (mathematically all the procedure can be rigorously justified using the theory of distributions). The point is that the claimed position operator is not the position operator because it is not even self-adjoint (nor Hermitian) in the relevant Hilbert space ...


4

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

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 ...


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, ...


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

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 ...


3

Yes, it is possible to represent the fermionic operators as matrices with the caveat that the fermionic Fock space of states is a super vector space, and the matrices are super matrices. If we have two creation operators $\hat{c}^{\dagger}_{\sigma}$, $\sigma\in \{\uparrow,\downarrow\}$, then there are: 2 bosonic states (1 vacuum state ...


3

Really a good question (unfortunately Ī came too late to actually help). All this stuff perfectly makes mathematical sense, but its fundamentals lie in abstract algebra, and rarely are explained to students, and hence remain concealed from those who lack appropriate imagination. To understand firmly bra and ket, not just multiply rows by columns and ...


3

Although we can define the momentum as a self-adjoint operator in $L^2[0,1]$ as you proposed, I think it's rather artificial to think about it as having relation to momentum in the case of $L^2(\Bbb{R})$. Realize that the operator $p_1$ with domain $D(p_1)=\{\psi\in\mathcal{H}^1[0,1]\,|\,\psi(1)=\psi(0)\}$, is related to spatial translations via the unitary ...


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: ...


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 = ...


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

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

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

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

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 ...


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 ...


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

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 ...


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

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 ...


2

Let $$|\psi'\rangle = e^{i\theta}|\psi\rangle.$$ The probability to be in the $n$:th eigenstate of the observable $\hat O$ is given by the expectation value of $P_n$ where $P_n$ is the projector onto the $n$:th eigenstate. Since the expectation value in the state $|\psi'\rangle$ is $$\langle \psi'|P_n|\psi'\rangle = \langle \psi|e^{-i\theta} P_n e^{i\theta} ...


2

Here is the answer (I will not consider the constants on the denominator of your Fourier transform for simplicity, however they are there ;-) ). When you write the operator $\hat{\phi}$ you have to be careful. I will drop the hats, because it will be clearer I think (maybe here the hat stands for an operator and not for the fourier transform). Your operator ...


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 ...


1

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 ...



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