New answers tagged

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 = ...


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 ...


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.


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Ab initio the momentum operators can be constructed using de Broglie Plane waves In one dimension, using the plane wave solution of the Schrodinger equation,the wave function Psi = exp. i (kx -wt) , if one takes the partial derivative w.r. to x of the wave function delta/delta x (Psi) = ik. Psi and using de-Broglie relation p = hbar . ...


-1

Momentum is the generator of spacial translations, even in classical physics. Anyway, you can find a derivation here or in Sakurai's book Modern Quantum Mechanics. They are more or less the same and go like this: The translation operator is the operator $T( a)$ such that $$T( a) \mid x \rangle = \mid x+a\rangle$$ From the definition it follows that the ...


0

That $\hat{P} = - i \hbar \partial_x$ generates translations comes from a straight-forward computation: if $\psi$ is continuously differentiable, and $\Psi$ as well as its derivative are square integrable, then you can prove that \begin{align} i \frac{\mathrm{d}}{\mathrm{d} y} \bigl ( \psi(x - y) \bigr ) \big \vert_{y = 0} = - i \partial_x \psi(x) ...


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 ...


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 ...


0

If a particle is a wave function describing a probability amplitude distributed through space, what happens when two wave functions meet? Well, I just try to think allowed and one can not portray a 'wave function' picture unless a physical system is at hand to use the picturization or description. Let us think about a hole in the reactor and neutron ...


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 ...


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 + ...


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Lets not use technical terms and understand this bit by bit, by example. In classical terms if a cat is kept with a radioactive poison inside a box, without any type of knowledge of what is going inside. A person with common sense would say that the cat is either alive "OR" dead. The word OR here is emphasized, because of some reason which we will come to ...


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 ...


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.


0

An eigenstate $ |\psi > $ with eigenvalue $\lambda$ has the property that $H|\psi> = \lambda |\psi>$. Since the dimension of your Hilbert space is 2, you are looking for two eigenstates. In this case, it's a matter of finding the right linear combinations of the given basis states. Since $H|1> = |1> + |-1> = H|-1>$, you are looking for ...


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} ...


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 ...


0

There are explanations for why the probability of a measurement of a particular observable is the square amplitude of the relevant eigenstates. The square amplitude is the only quantity that fits the calculus or of probability, and the rules of decision theory, and quantum physics: http://arxiv.org/abs/quant-ph/9906015 http://arxiv.org/abs/0906.2718. The ...


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 ...


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

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 ...


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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 ...


0

The spin state is given as a linear combination of spin up and spin down states. So $$\psi=c_+| z +\rangle+c_-| z -\rangle$$ The square of the modulus of the complex coefficients $c_+$ and $c_-$ represent the probabilities of each associated base kets. When you multiply $\psi$ by a constant, the state will not change as a ket and a constant times that ...


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 ...


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 ...


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 ...


0

The author shows that since $UN$ equivalent to $UNU^\dagger U$, that it is possible to change the order of the operations, if $N$ is replaced by $M=UNU^\dagger $. In that case, $UN$ = $MU$. The remaining logic follows from this. Note that $ M $ is just $ N $ in the $ U $ basis; the choice of $ U $ uniquely determines how $ N $ changes to $ M $ for this ...


0

Concerning point 2: Operators do not always come through each other cleanly, but there are some very basic rules that always apply, which can be turned into less tedious rules that apply in special cases. Often the latter are taught first, causing mass confusion. General rule: Operators can be expressed as (Sum over a in the set of eigenvectors ) |a > ...


1

The left hand: rotate the state $|JM\rangle$ by applying a rotation $R$ on it. Right hand side: insert completeness condition $\sum_{M'} |JM'\rangle\langle JM'|$ $D$ is the matrix representation of rotation matrix $R$ in basis ${|JM\rangle}$. The rotated state is expanded in terms of basis ${|JM\rangle}$ with coefficient $D$ in terms of rotation matrix.


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 ...


0

You can think, on the left is a short hand notation for a (2J+1) x (2J+1) matrix R applied to a (2J+1) component vector |J,M> with the components labelled by M. On the right, the matrix elements are explicitly shown, and the sum over M' is the matrix multiplication. Actually on the left is an abstract rotation operator R that will rotate any J. When ...


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 ...


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 ...


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 ...


0

a Stern Garlach apparatus does not rotate the state of the particle, what it does is to split a beam or, if you have a single particle, it ``chooses" a state in the desired direction. What you might be looking for is how to write the eigenstate in terms of Z basis. For a 1/2 spin it is going to be: $$|\pm \rangle_y = \frac{1}{\sqrt{2}}\left( |+\rangle_z ...


2

Schmidt decomposition is in general a singular value decomposition (SVD) and it is applied on wave vectors and not on density matrices. While dealing with bi-partite wave vectors we use SVD because there is no restriction that the size of the two systems in question are the same. So the matrix of the wave vector coefficients can be rectangular and SVD can ...


1

The system can be separated, but not necessarily in nice form. For instance, the time derivative of the first eq. reads $$ i\hbar {\ddot c}_1 = - B {\dot c}_1 - {\dot V}c_2 - V {\dot c}_2 $$ Now remove $c_2$ using again the first eq., $$ c_2 = -\frac{i\hbar}{V} {\dot c}_1 - \frac{B}{V} c_1 $$ and ${\dot c}_2$ using the second eq., ${\dot c_2} = ...


1

Hints: The starting point is the 2-point relation $$T(\phi(x)\phi(y)) ~-~:\phi(x)\phi(y): ~=~ C(x,y)~{\bf 1}, \qquad C(x,y)~\equiv~\langle 0 | T(\phi(x)\phi(y))|0\rangle,\tag{1} $$ cf. this Phys.SE post. The relevant Wick's theorem is a nested Wick's theorem $$ T(:\phi(x)^n::\phi(y)^m:)~=~\exp\left( ...


2

Everything you write is correct and there is no inconsistency. When you write $\left<x\right|Hf(x)g(x′)\left|x′\right>$ you just, indeed, put the numbers on the right. But numbers commute so there is nothign wrong with it. Note that you have really no conclusion from it. You can't even transform back to operators on the form ...


2

The delta function is not really a function, it is a distribution, In the strict sense both $\delta (x)$ and $e^{ikx}$ are not normalizable when $n=m$ One way to prove your equations is to use fourier transforms Using Placherels theorem the fourier transform $F([f(x)]k)$ for the function $f(x)$ is given by ...


0

The problem here is that you're assuming that $|x' \rangle$ and $\langle x|$ are autokets of g(X) and f(X), respectively.


-1

Particles are simply high momentum wave states interacting weakly with matter, i.e. their "existence" is observer dependent. Trying to derive them from some form of free field equation is therefor useless and so is the assumption that they are a general phenomenon. They are a highly likely phenomenon for energies that are much higher than the typical em ...


4

The way to do this is using the Wigner-Eckart theorem. The way it is applied to your problem is as follows: $$ \left\langle nlm |\vec{r}| n'l'm'\right\rangle = \left\langle nl ||\vec{r}|| n'l'\right\rangle \left\langle l' m' 1 q | l m\right\rangle $$ where the second factor is a Clebsch-Gordan coefficient and $q=-1,0,1$ indicates the type of transition. For ...


2

Yes, their outer product is defined as you said. Further, the product of operators is given by $$ (A \otimes 1_B)(F \otimes 1_B) = (AF) \otimes (1_B1_B) = (AF) \otimes 1_B $$ Therefore, $$ [A \otimes B, F \otimes 1_B] = [A,F] \otimes [B,I_B] = [A,F]\otimes {\bf 0} = 0 $$



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