# Tag Info

1

There is some restrictions on previous answer. Hamiltonian must be time-independent to use $U = e^{-iHt}$ rule. For time-dependent hamiltonian, time-evolution in form $\psi(t) = U(t,t_0) \psi(t_0)$ takes U in more general form $U(t,t_0) = \mathcal{T}\,\exp(-i \int _{t_0} ^ t H(\tau) d\tau)$. Of course, if your potential is constant over $t$ to $t_0$ period, ...

4

Neuneck's answer is the pithiest description of how you get normalisable states as superpositions of non-normalisable states, but the following is more of a "why" these states happen. Hopefully, you should see that this discussion is independent of the number of dimensions. Practically speaking, the reason why there are always such states it is because ...

3

I supposed you are in a context of bound states, with normalized eigenfunctions $\psi_n(x,t) = \phi_n(x) e ^{iE_nt}$. Of course, if you calculate $\langle x(t)\rangle_{\psi_n} = \int dx \bar \psi_n x \psi_n$, you will find a position expectation value which does not depend on time. Now, this is not the general case, if you take a linear combination of the ...

2

The scattered states are indeed non-normalizable. This is because a plane wave is an unphscial state (which you can for example see by trying to calculate the Heisenberg uncertainty, which will read $\Delta x \cdot \Delta p = \infty \cdot 0 = ??$). In order to create a physical state, you need to specify boundary conditions, i.e. a physical wavefunction at ...

1

The answer to your question "Would it be valid to..." is "yes." Whenever your Hamiltonian is time-independent over a time interval beginning at some reference time $t_0$, then Schrodinger's equation takes the form $$|\psi(t)\rangle= \hat{U}\,|\psi(t_0)\rangle$$ where $$\hat{U} = e^{-i H t/\hbar}$$ is the unitary time evolution operator. It doesn't ...

2

First of all, don't underestimate the idea of "trial and error" as a valid explanation. For every polished result that makes things seem to fit smoothly into place, there are likely tens of other attempted results in the notebooks (or waste paper bins) of the researcher who published it and likely thousands of like tries by other researchers who didn't get ...

4

I've always found Griffith's method of 'deriving' the ladder operators to be simple to understand. And, if I forget their precise form, I can always recover it this way. He first introduces ladder operators for the 1-dimensional harmonic oscillator. Here's his basic method: For the harmonic oscillator, ...

5

The term canonical gives it away. The canonical ensemble density matrix $\rho$ is defined as follows in terms of the Hamiltonian $H$ and inverse temperature $\beta = 1/kT$: \begin{align} \rho(\beta) = \frac{1}{Z(\beta)}e^{-\beta H}, \qquad Z(\beta) = \mathrm{tr}(e^{-\beta H}) \end{align} Then the canonical ensemble average of any observable $O$ is given ...

1

The mathematical concept behind the ladder operators is the "root system" of a Lie algebra and things related to it. So I think the most systematic approach to ladder operators is the mathematical one. If you are interested, you can find some explanations in the following post: Link (in the answer that contains the word "roots"). The mathematics and the ...

3

Sorry it is impossible that, if both $\psi, \phi$ belong to the domain of a self-adjoint operator $A$ the identity $$\langle \psi| A \phi \rangle = \langle A\psi| \phi \rangle$$ fails. The point is that your function $T\psi$, where $\psi({\bf x}):= e^{-r}$, does not belong in turn to the domain of the self-adjoint operator $T$ so that: $$\langle T\psi| ... 3 In those notes you linked to, the |u_i\rangle form an orthonormal basis. They are not necessarily the eigenkets of the operator O. By choosing different bases \{u_i\}, the "representation" of the operator O (e.g.., how you write it out in a matrix) will change. You are correct that if one chooses the set of kets that are eigenkets of the operator, ... 4 This is just the potential of a standard harmonic oscillator. The presence of the linear term is just due to the fact that you're using a coordinate system where the minimum of the potential is not at x=0. You can re-write it as$$V(x)=\frac{1}{2}mx'^2+c$$where x'=x+\frac{\lambda}{m} is the coordinate centered at the minimum of the potential and ... 3 It is possible to say something more precise than Martin's answer (that is correct however). The key-point is that self-adjoint operators are closed operators. An operator A: D(A) \to H, with D(A) \subset H a linear subspace of the Hilbert space H is said to be closed if, for every sequence of vectors f_n\in D(A) such that (1) f_n \to f \in ... 4 But can the eigenstates of the position observable be individually thought of as delta functions? Yes they can in a sense but it is rather inaccurate. First, kets and functions(distributions) are somewhat different things although they share most of their properties. If the ket |x'\rangle  satisfies$$ \hat{x} |x'\rangle = x' |x'\rangle, $$then we ... 1 Okay, so a general observable acting on |x\rangle won't give you x' |x\rangle. Only the position operator, acting on the state |x'\rangle will give us x'|x'\rangle, where the x' is a label for the state, think of it as a number, not a variable. Just because the state |x'\rangle is an eigenstate of the position operator, it does not mean that it ... 1 One way to treat Quantum mechanics is that writing the operators in terms of matrices. One can think of operators \hat{X}, \hat{P_x} and \hat{P_x}^{-1} in terms of matrix quantum mechanics. In the |x \rangle basis (where \hat{X} |x \rangle=x|x \rangle), we can discretize the space up to some overall constant as a unit of spacing, and write the ... 14 The operator$$\frac{1}{P_x}$$produces a divergent result if it acts on a wave function with P_x=0. The superpositions of a P_x=0 states with the eigenstates associated with nonzero values of P_x are still singular. That's why a well-defined result of (1/P_x)|\psi\rangle is only obtained if |\psi\rangle contains no P_x=0 admixture in the ... 1 I accepted lionelbrits' answer because I've already figured it out with his/her hint to "slide the operator past". I will write now a more detailed version for others who are interested in the identy but can't figure it out. At first, I advice the reader to read Galindo & Pascual, Volume I, p. 70ff (2012) to see how the time-ordering is introduced and ... 2 Hints to the question (v1): Recall that the operator time ordering is symmetric$$\tag{1}T[A_1(t_1)\ldots A_n(t_n)]~=~T[A_{\pi(1)}(t_{\pi(1)})\ldots A_{\pi(n)}(t_{\pi(n)})], where \pi\in S_n is a permutation. (Here we assume for simplicity that all operators are Grassmann-even. Else there will be additional sign factors.) Recall that if t_1> ... 2 Hint \int_0^t \int_0^{t_1} dt_1 dt_2 \, a(t_1) a(t_2) = \frac{1}{2!} \int_0^t\int_0^t dt_1 dt_2\, \mathcal{T}\{ a(t_1) a(t_2) \} and so forth. You can see this by noting that the (square) integration region in the second integral can be split up into two triangular integration regions like in the first integral. This is one way to define the ... 1 Yes. Also note that in the momentum representation, x = i\hbar \frac{d}{dp}, which is what your commutation relation proved as a special case. You could use this shortcut right off the bat. 2 The problem is that the physical states have positive occupation numbers n_1, n_2,.... With your operators, you have, for instance :  c_\alpha| n_1, n_2, ..., 0, ... \rangle = | n_1, n_2, ..., -1, ... \rangle. This gives you a totally unphysical state, so you would have to add by-hand constraints like n_1 \geq 0, n_2 \geq 0,..., . With the ... 4 Here are three properties that would make your definitions awkward. You can think of a^\dagger\,a as the LU (lower triangular, upper triangular or Cholesky) decomposition of the number observable. Actually, it's not the unique Cholesky factorisation but it is the one found by the outer product version of the algorithm. Your definition would not have this ... 4 Here's a simple argument showing that a and a^\dagger cannot have a common eigenvector using only the commutation relation between them. Suppose,by way of contradiction, there existed a vector that were a common eigenvector of both, namely a nonzero vector |\psi\rangle such that \begin{align} a|\psi\rangle &= \alpha|\psi\rangle, \\ ... 4 An eigenfunction of the creation operator would be a state that satisfies a^\dagger |\psi\rangle \propto |\psi\rangle. Think about the consequences of this for a creation operator. EDIT: Since this got downvoted, let me be more specific. Suppose |\psi\rangle = \sum_{n=0}^\infty c_n |n\rangle, and we require that a^\dagger |\psi\rangle =\lambda ... 0 Dirac argues from symmetry in his Principles of QM: In a 1-D system, \hat{q} and \hat{p} are both observables, with eigenvalues extending from -\infty to +\infty, and are connected by the commutation relation [\hat{q},\hat{p}]=i \hbar. Since one can interchange \hat{q} and \hat{p} in these equations if i is replaced by -i, it follows ... 2 Say that we are handed a 1D quantum mechanical system, which satisfies the canonical commutation relation\tag{1} [\hat{Q},\hat{P}]~=~ i\hbar~{\bf 1}, $$and handed some choice of eigenstates |q\rangle and |p\rangle for every value of q,p\in\mathbb{R}. The eigenstates satisfy$$\tag{2} \hat{Q} \mid q \rangle ~=~q\mid q \rangle, \qquad ...

3

Since I'm not an expert on spectral theory, this will only be a partial answer, however, I believe that this question, is mathematically much more involved than you think. First of all, let's review the finite dimensional case: We have two Hermitian matrices $A,B\in\mathcal{M}_d$ and they commute if and only if their spectral projections commute, i.e. they ...

2

The is a theorem that says Hermitian operator are associated to real eigenvalues. Since eigenvalues correspond with our measure values and they are real, this means that it makes sense to have hermitian operators as observables. Also, what type of observables are we talking about here? Particles? Observables are magnitudes we can measure: position, ...

0

Hints to the question (v1): Let $\hat{Z}^I$ be operators that satisfies a Heisenberg algebra $$\tag{1} [\hat{Z}^I,\hat{Z}^J]~=~i\hbar ~\omega^{IJ} ~{\bf 1},$$ where $\omega^{IJ}=-\omega^{JI}$ is an antisymmetric real matrix. The important property will be that the commutators $[\hat{Z}^I,\hat{Z}^J]$ belong to the center of the Heisenberg algebra, i.e. ...

2

You start from this $[p,F(x)]\psi=(pF(x)-F(x)p)\psi$ knowing that $p=-i\hbar\frac{\partial}{\partial x}$ you'll get $[p,F(x)]\psi=-i\hbar\frac{\partial}{\partial x}(F(x)\psi)+i\hbar F(x)\frac{\partial }{\partial x}\psi=-i\hbar\psi\frac{\partial}{\partial x}F(x)-i\hbar F(x)\frac{\partial}{\partial x}\psi+i\hbar F(x)\frac{\partial }{\partial x}\psi$ from ...

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