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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 = ...


10

Call $u_1, u_2, u_3, u_4$ the eigenvectors described by you, respectively. Your claims are all right, but realize that both $u_1$ and $u_2$ share the same eigenvalue, that is $1$, i.e., $Pu_1=u_1$ and $Pu_2=u_2$. Hence, any linear combination of $u_1$ and $u_2$ will also be eigenvectors with the same eigenvalue $1$. Try to find eigenvectors of $H$ of the ...


8

Hint: When an eigenvalue for an operator $P$ is degenerate, there are more than one way to chose a set of eigenvectors. If the other commuting operator $H$ lifts that degeneracy, there will be a preferred choice of common eigenvectors. More generally, a set of diagonalizable operators commutes if and only if the set is simultaneously diagonalizable.$^1$ ...


7

Actually, all that is quite known from the foundational work by von Neumann and Birkhoff. In this formulation of QM (and in the subsequent evolution of this research area) one constructs the quantum theory theory starting from the lattice of elementary "YES-NO" observables (see my answer on quantum probabilities for more details) or "elemetary propositions" ...


6

The bound state is defined such that the probability density average will be finite at some particular space region when time passes. While for unbounded states, as time passes, the probability density will tends to zero. See Landau Quantum Mechanics section 10. This can be understand as this, if the state is bounded, i.e. it is exist only within some ...


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 ...


5

Why do electrons in an atom occupy only the stationary states? This isn't true. An electron in an atom can be in any superposition of states. This is one of the basic postulates of quantum mechanics: linearity. For example, say an atom has a ground state 1 and an excited state 2, and let's say we're able to prepare it in a pure state 2. It will decay ...


5

It doesn't matter what sign you choose. Notice that since $|A|^2 = \frac{2}{a}$, you could even pick $A = \sqrt{\frac{2}{a}} e^{i\phi}$, so $A$ doesn't have to be real. The reason is that a wavefunction is only defined up to a global phase. The reason is that we calculate probabilites with $|\psi|^2$ and mean values of operators with $\int \psi^* \hat{O} ...


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 ...


4

Yes, this is perfectly possible. The simplest example is the hydrogen atom, which has an infinite sequence of discrete eigenstates $|nlm\rangle$ with negative energy, and an energy continuum at $E>0$, which are known as Coulomb waves. It is perfectly possible for a state to be in a superposition of those: First of all, it is required by the ...


4

(1) We could very well write $0|\psi\rangle=0$ but we must keep in mind that the first $0$ is a scalar, the one that belongs to the field over which the vector space (where $|\psi\rangle$ lives) is defined, while the second $0$ is the zero vector of that v.s. (we always use the same symbol to denote both, provided we always are careful when writing these ...


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

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

I don't find this proof a good one, since the notation is messy and not very clear (not to say wrong). One proof can be given in a similar way to the one you posted in the link. Suppose the spectrum of $H$ is discrete and the set of eigenstates $\{|\phi_n\rangle\}$ constitutes an orthonormal basis with eigenvalues $E_n$, such that $E_0\leq E_1\leq ...


3

Ok, to expand my comment into an answer: There are only two finite-dimensional division rings (that admit division) containing the real numbers as a finite subring: the complex numbers and the quaternions (application of Frobenius theorem). Also, a vector space (and Hilbert spaces are vector spaces) is usually defined over a field, that is a non-zero ...


3

Since states which are not eigenstates of the Hamiltonian are also not eigenstates of the time evolution, it does not make sense to talk about "bound states" for these states, as they are continually changing into other states. For energy eigenstates, it makes sense to speak of "a bound state", since that state will stay the same forever unless acted upon.


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

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: ...


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

If eigenkets are defined up to arbitrary constants, it is possible to write the sum without any coefficients.


2

One reason we focus on energy eigenstates is that atoms spend almost all of their time in an energy eigenstate, and their spectrum is a result of transitions between them. Another reason is pedagogical: to peel back the onion one layer at a time. But before too long, many courses do include examples of systems that are not in an energy eigenstate. One ...


2

Your understanding of reducible and irreducible representations is a little bit muddled. Let me try to clarify this a bit: A reducible representation $D:G\to \text{GL}(V)$ is one that has a nontrivial invariant subspace $W$. That is, there exists a nonzero $W<V$ such that for all $g\in G$ and all $w\in W$, the action $D(g)w\in W$ remains in the ...


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

Typically, the Hilbert spaces one considers in quantum mechanics are $L^2$ spaces. The elements of these spaces are equivalence classes of functions which differ only on a null set of points, i.e. whose distance in terms of the $L^2$ norm is zero, $\|n-\tilde n\|_{L^2}=0$. That is, you are right, but it's the $L^2$ norm that matters, not the Sobolev norm. ...


2

The integral without delta square would converge to 1, but squaring the function somehow breaks it? Yes. Recall the sifting property: $$\int_{-\infty}^{\infty}f(x)\delta(x - a)dx = \int_{-\infty}^{\infty}f(a)\delta(x - a)dx = f(a)$$ Then, it follows formally that $$\int_{-\infty}^{\infty}\delta^2(x - a)dx = \int_{-\infty}^{\infty}\delta(x - ...


2

What happens when you integrate a function multiplied by a delta function? You get: $$\int f(x) \delta (x-y) dx=f(y)$$ (Because the delta function is zero everywhere, except at zero where its integral gives 1.) So, when integrating over $\delta^2 (x-y)$ we get: $$\int \delta^2 (x-y) dx=\int \delta (x-y) \delta (x-y)dx = \delta (y-y) = \infty$$


2

The Dirac delta is defined by the equation \begin{equation} \int_a^b\mathrm{d}x \,\delta(x)f(x) = f(0) \end{equation} for $a < 0 < b$. By direct application of this definition we get \begin{equation} \int \mathrm{d}x\, \delta(x - y) \delta(x - y^\prime) = \delta(y - y^\prime) \end{equation} If we let $a < c < 0$ we can write \begin{equation} ...


2

The $\delta$ function has the following property: $$ \int \text{d}x\; f(x)\delta(a-x)=f(a) $$ This actually answers both of your questions. First, the non-square-integrability: $$ \int \text{d}x\;\delta(x-y)\delta(x-y)=\delta(y-y)=\delta(0)=\infty $$ according to the rule above if you choose one of the $\delta$'s to be the $f$. Your second question is the ...


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 ...



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