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21

That the eigenfunctions of the free Hamiltonian $H\propto p^2$ are not actually normalizable due to its completely continuous spectrum and therefore cannot be actual quantum states is well-known, although rarely suitably emphasized. (See e.g. Why are eigenfunctions which correspond to discrete/continuous eigenvalue spectra guaranteed to be ...


10

This is indeed possible only in some situations, e.g. when the continuous spectrum is absent (it may also consist of a single point, see Valter Moretti's comment below). A sufficient condition for that to be true is that either the Hamiltonian is compact or it has compact resolvent. Sadly, very few interesting Hamiltonians satisfy that property (an example ...


7

Not every tensor is a simple tensor. The simple tensors of the form $a\otimes b\in A\otimes B$ span the tensor product, which means that a general element $t$ of the tensor product is a linear combination of these simple tensors, i.e. $$ t = \sum c_{ij} a_i\otimes b_j$$ for some basis $a_i,b_j$ of $A$ and $B$ respectively. If $A$ itself is a space of ...


6

No. Counterexample: choose a basis in 2D quantum space, so the basis states are written $\left(\begin{array}{c}1\\0\end{array}\right)$ and $\left(\begin{array}{c}0\\1\end{array}\right)$. Now think of a mixture, with a state's having probability $p$ to be in state 1. The density matrix is then $\mathrm{diag}(p,\,1-p)$, which is not singular for ...


5

Phenomena in quantum mechanics may be expressed using any basis. It doesn't mean that all bases are equally useful for a given situation. In particular, a fundamental postulate of quantum mechanics says that right after every measurement, the system is found in one of the eigenstates of the observable that was just measured. That's why the basis of the ...


5

The answer to your question is NO. The simplest counter-example is the identity matrix. There is no solution to $|a\rangle\langle b| = I$. For example, try to solve $\begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} z & t \end{bmatrix} = \begin{bmatrix} 1&0 \\0&1 \end{bmatrix}$. Since $xt=0$ we know that $x=0$ or $t=0$. But clearly $x$ ...


4

The basis is still $\{|\boldsymbol r\rangle\}$. The abstract Schrödinger equation is $$ i\frac{\mathrm d}{\mathrm dt}|\psi\rangle=H|\psi\rangle $$ where $|\psi\rangle$ is a set of four kets, (with a slight abuse of notation) $$ |\psi\rangle=\begin{pmatrix}|\psi_1\rangle\\|\psi_2\rangle\\|\psi_3\rangle\\|\psi_4\rangle\end{pmatrix} $$ Time is still a ...


4

Historically, you probably want to start with the de Broglie relations (i.e. $p = \hbar k$), which are just a wild guess. This immediately pops out the form of $p$ as an operator if the wavefunction is a plane wave. Mathematically, $p$ should be defined as the generator of translations (or equivalently the conserved quantity corresponding to translational ...


3

OP explicitly asks whether a material object (i.e. not a state of light) has been placed in the superposition $|\psi⟩=|\psi_1⟩+|\psi_2⟩$, where $|\psi_n⟩$ is the $n$th Fock state of a harmonic oscillator. Perhaps the clearest example of this is the achievement of precisely that superposition in the quantum state of a cantilever microwave resonator ...


3

Similarly to AccidentalFourierTransform I am not sure to understand well your issue. However there is a crucial missed point in your argument, usually absent in many textbooks on these topics. It is true that decomposing $H$ as $H= \hbar\omega( a^\dagger a + \frac{1}{2})$ and taking the relations (following from CCR) $[a, a^\dagger] =I$ into account one ...


3

Well, I'm not sure I understood your question so I'm going to write what I think and let's see if it's useful :-) The algebra $[a,a^\dagger]=1$ is all you need to diagonalise $H$, but this is because what $H$ looks like: $$ H=\omega a^\dagger a $$ The important observables, namely $H,P,X$, can be written as polynomials in $a,a^\dagger$: \begin{aligned} ...


3

The annihilation operator is a linear operator. A linear operator can be applied to ANY state. And yes, it returns zero when applied to the ground state. You can really take this as a definition. The definition of the momentum operator $\hat{p}$ is the operator such that $\langle x|\hat{p}|\psi\rangle=-i\hbar \psi'(x)$. One could write this as $\langle ...


3

If $|\pm\rangle$ are the eigenvectors of ${\hat \sigma}_x$, ${\hat \sigma}_x |\pm\rangle = \pm |\pm\rangle$, then a rotating $x$-basis is defined as $$ |+\rangle(t) = \exp\left(-i\frac{\omega t}{2} {\hat \sigma}_x\right)|+\rangle = e^{-i\omega t/2 } |+\rangle $$ $$ |-\rangle(t) = \exp\left(-i\frac{\omega t}{2} {\hat \sigma}_x\right)|-\rangle = e^{i\omega t/2 ...


2

The point is that the domain $D(P)$ of $P$ must be such that $P$ is (essentially) self-adjoint thereon. Otherwise it does not represent an observable. I am assuming that $D(X)= L^2([0,L],dx)$ instead, where $X$ is automatically self-adjoint. The vector $\psi$ you use to prove Heisenberg inequality has to belong to $D(PX) \cap D(XP)$ as you see by direct ...


2

A crucial hypothesis is missed in your construction. Each $\phi_i$ must also satisfy $\phi_i \not \perp \psi_i$, otherwise $\langle \psi_i |E_i \psi_i\rangle >0$ is false. This point provides an answer to your last question as well. If $\psi$ is an added further vector, linearly dependent on the vectors $\psi_i$, the construction you made cannot be ...


2

The answer is superposition. Let's take one classical bit. It can be in two states: $|0\rangle$ or $|1\rangle$. Now let's consider a qubit: its general state will be $$a |0\rangle + b |1\rangle$$ where $a,b$ are complex numbers with the constraint $$|a|^2+|b|^2=1$$ Now you should be able to see the difference. A state like $$\frac{|0\rangle + ...


2

Your 2x2 unitary is mostly determined by its action on the state vector $\left(\begin{array}{c}1 \\0\end{array}\right)$. This is because 1) in a 2-d Hilbert space, for any given vector there is a single other orthogonal vector (up to a phase factor), and 2) a unitary map preserves orthogonality. Once $|u\rangle = U\left(\begin{array}{c}1 ...


2

It's a good question, and the answer is surprisingly simple and physical. There is indeed no fundamental objection to having a superposition of particles of different charge. But it turns out this is not stable. This is basically due to wavefunction collapse, or more sophisticatedly, due to decoherence. Imagine having a single particle in a superposition ...


2

Both states $\Psi_{k,\sigma}$ and $\Psi_{k',\sigma'}$ are meant to be states of the same particle species i.e. they have the same values of the squared mass $k^2$. The inner product of one-particle states from different species $s$ is zero which one might indicate by additional $s,s'$ labels and a Kronecker symbol $\delta_{s,s'}$. Weinberg claims about the ...


2

If you ask about what plays the role of the state of the system at some given time than the answer is: nothing. You talk only about the initial state and what you get in the measurements (expectation values or probabilities of outcomes for observables). The Heisenberg picture is very "Copenhagen" in its spirit and abstracts itself from what's happening with ...


2

Every complex number can be written in the form $re^{i\theta}$ for a real number $r$. We call $e^{i\theta}$ the phase. For example, if $$|\psi \rangle = \frac{1}{\sqrt{2}} ( |0 \rangle + i |1 \rangle)$$ then the phases of the $|0 \rangle$ and $|1 \rangle$ components are $1$ and $i$, and their relative phase is $i$. Now consider $$|\psi' \rangle = ...


2

Quick answer My question is how does the presence of nonzero $J(x)$ results in a non-trivial spacetime dependent value of $\langle 0|\phi(x)|0\rangle$? The equation $\phi(x)=\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}$ works both for $J=0$ and $J\neq 0$. Therefore, $$ \langle \phi(x)\rangle_J= {}_J\langle 0|\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}|0\rangle_J ...


2

You know that the normalisation of the inner product is 1, that is, $$ \langle n\, l\, m\ |\ n\, l\, m\rangle = 1 $$ you can use this information to find the value of $\langle n_x=1\, n_y=0\, n_z=0\ |\ n=1\, l=1\, m=1\rangle$ as, $$ \langle 1\,1\,1|1\,1\,1\rangle= 1 $$ leaving some of the algebra for you as part of the exercise*, you will obtain, $$ 1 = ...


1

Momentum and position are conjugate variables in classical mechanucs, which means they satisfy the Poisson bracket relationship. When quantum mechanics was invented the Poison bracket relation was replaced by the operator commutation relationship which results in the relation under consideration.


1

Wavefunctions combine trough tensor products, which is not the addition that one would expect naively. The reason for this is that a wavefunction contains the description of all possible futures of the system at once, so if there are multiple subsystems, then the wavefuntion of the entire system has to describe all possible futures for each part ...


1

The physical states in the Heisenberg picture are frozen in time, and can be made to coincide with the Schrodinger-picture state at any given time $t_0$. In other words, $$|\psi(t_0)\rangle^S=|\psi\rangle^H$$ which doesn't evolve with time.


1

If you want to do this rigorously use the Cauchy-Schwarz inequality. If we call your two states $\psi$ and $\chi$, then the Cauchy-Schwartz inequality tells us that: $$\lvert\langle \psi \vert \chi \rangle\rvert^2 \leq \langle \psi\vert\psi\rangle\langle\chi\vert\chi\rangle $$ And in particular we get equality if and only if $\chi$ and $\psi$ are linearly ...


1

Here, $V_0(k_2)$ could have been replaced by $e^{k_2 \alpha_{-1}} e^{-k_2 \alpha_1}$ while the factors from $\alpha_{\pm n}$ for $n\gt 1$ could have been neglected because $\alpha_n$ annihilates everything that appears in the matrix element on the right side from $V_0$ (because it ultimately annihilates $|0\rangle$), and similarly for $\alpha_{-n}$ that ...


1

I know that the usual interpretation of the wavefunction in QM is that it´s a probability distribution of measurable quantities. It is a postulate that is necessary to choose the subset of mathematical sets that can be useful to modeling nature. The postulate was chosen because observations were fitted by the hypothesis. Not a deterministic ...


1

The electric charges of the states $\:|uuu⟩,|ddd⟩,|sss⟩\:$ are $\:+2,−1,−1\:$ respectively. More exactly these baryon states are the baryons $\:Δ++,Δ−,Ω−\:$. If $\:|X⟩\:$ or $\:|Y⟩\:$ would represent a baryon what would be the electric charge of this particle ? And electric charge is one of many quantum numbers. This problem is pointed out by @Cosmas Zachos ...



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