# Tag Info

1

I can answer (1) and (2). The answer is: NO. Passing form classical mechanics to quantum one requires, in general, to add more information. There is no rigorous machinery allowing one to write the quantum corresponding of a classical object. Physically speaking, this is because quantum structures are more fundamental in Nature than classical ones. ...

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 ...

1

I'd like to add a really elementary level illustration of some of David Bar Moshe's answer when he says: ...They give two major examples of non separable Hilbert spaces: 1) The infinite tensor product of harmonic oscillators, ... The example below is clearly well below the level the OP is looking for, but hopefully it shows to a wider audience what a ...

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 ...

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, ...

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 ...

7

From the viewpoint of the Wightman axioms, the separability assumption on the Hilbert space can be actually derived from a few of the other axioms if you adopt a formulation using $n$-point functions. The reasoning goes as follows. A (say, scalar) quantum field theory on $\mathbb{R}^d$ can be thought of as being specified by a sequence of $n$-point ...

11

First, let me state that the notion of a Hilbert space is not fundamental in quantum theory. One can realize the same quantum physical system using different Hilbert spaces. This is because quantum states (which are really the objects which physically matter) are only weakly connected to vectors on a Hilbert space. It is true that pure states correspond to ...

1

I) We interpret OP's question (v2) as follows: Why not normalize $$\tag{1} \langle x_1 | x_2\rangle~=~\delta_{x_1,x_2}~:=~\left\{ \begin{array}{ccl} 1 & \text{for} & x_1= x_2, \\ 0& \text{for} & x_1\neq x_2. \end{array} \right.$$ via a continuous Kronecker delta function rather than a Dirac delta distribution $$\tag{2} \langle ... 1 Any good basis should be complete. If the set of all |x> is complete, any other vector |\psi> in the Hilbert space of your system should be writable as |\psi>=\sum_{x} |x><x|\psi>. This sum does not make sense for continuous variables x, hence the need to redefine the completeness relation with an integral (as Jan's answer ... 0 Why cant these basis vectors be normalized to one, only to the delta function? Because that would make those continuously indexed vectors unsuitable for the role of a "continuous basis" for normalizable functions. Here is the explanation. Suppose some function \psi(\mathbf r) is expressed as the integral$$ \psi(\mathbf r) = \int c(k) \phi_k(\mathbf ...

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 ... 3 It's a matter of definition. According to Hirzebruch/Scharlau, Einführung in die Funktionalanalysis (1971), Definition 21.10: An orthonormal system$\{x_i\}_{i\in I}$that meets the [proven equivalent] conditions of this theorem is called a Hilbert basis or just basis of$X$. Wikipedia calls it an orthonormal basis, which is a Schauder basis if your ... 4 It is a basis. The trouble is that "basis" is actually vague if removed from specific context, and what we have here is a Schauder basis. The definition you learned in basic linear algebra that requires a finite linear combination is a Hamel basis. 1$\lambda$stands for the eigenvalue. Eigenvalue equation is:$S_xX=\lambda XS_xX-\lambda X=0(S_x-\lambda I)X=0$Since X is eigenfunction, we seek solutions for$det(S_x-\lambda I)=0\begin{align} (S_x-\lambda I)= \begin{bmatrix} 0 & \frac{\hbar}{2} \\ \frac{\hbar}{2} &0 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & ... 2 Take case for ann\times n$matrix$A$. To find its eigenvalues, first you write the eigenvalue equation for it. $$Au=\lambda u$$ where$u$are its eigenvectors. This can be rewritten in the following way $$Au-\lambda u=(A-\lambda I)u=0$$ with$I$the identity matrix. Let$A-\lambda I=B$, and we know that the equation$Bu=0$has a non zero solution$u$... 2 So now I'm wondering the following: is this the definition of a wavefunction or is there a wavefunction for every observable, being this wavefuntion just a map from the possible values of the observable to their probability amplitudes? There is indeed a wavefunction for every observable. The state$|\psi\rangle$is a vector in a complex Hilbert space. ... 3 Really, there is one wavefunction that runs over all compatible observables, and it's time evolution is governed by the Schrodinger equation with some Hamiltonian. But often times certain observables don't interact much, so you can just treat them as a system in isolation. You can imagine the Hamiltonian as being a matrix acting on some vector space, and if ... 3 You don't need to integrate anything here. The wavefunction is represented as a two component spinor here, which represents the probability of observing a certain spin along some axis. The components of the spinor refer to how the spin of the particle is oriented, they do not represent$x$and$y$components of a vector. So the inner product defined for such ... 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\ranglesuch that \begin{align} a|\psi\rangle &= \alpha|\psi\rangle, \\ ... 4 An eigenfunction of the creation operator would be a state that satisfiesa^\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 ...

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Actually, what your professor isn't telling you - what we always gloss over in intro quantum mechanics for simplicity, but I might as well give it away now because everything will make so much more sense once you get this - is that kets aren't wavefunctions at all. Forget that you ever learned about wavefunctions for a minute. Kets are a form of notation ...

1

To test whether $|0\rangle$ is physical, you would apply $\partial^\mu A_\mu^+$ to it. So $|0\rangle \in V$. Note that 7.48 does not mean that $\partial_\mu A^\mu = 0$ as an operator identity in this space. It means that all it's matrix elements in this space are zero. You have noticed that the state you created, $\partial_\mu A^\mu |0\rangle$, lives not ...

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 ...

1

As you and Christoph already pointed out, the difference comes from the fact that these contiuous "bases" do not belong to the respective Hilbert space themselves. This is why they are not actually bases at all in the general sense. Rather, they are useful mathematical tools to expand any states actually belonging to the Hilbert space. As they obey certain ...

0

Very intuitive. No maths. There is an excited state with a symmetrical probability distribution and no e/m dipole moment. There is a ground state (or less excited state) also with no dipole moment. There is a tiny probability that the excited state electron will be in the ground state that allows both states to be present at the same time producing a finite ...

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