New answers tagged

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

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 ...


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


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A CNOT gate can't be smaller than 4x4, because it operates on two qubits and $2^2=4$. The closest thing to an analogous gate, that applies to a single qubit, is the Z gate $\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$. You can think of Z, in the computational basis, as a negation of the global phase controlled by the target qubit. A ...


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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 ...


3

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 ...


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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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I'll try to add a rather intuitive take on the difference between pure and mixed states. Take first the simple example of a single spin-1/2 particle. Its pure states may always be written as superpositions of spin-up and spin-down states measured along some particular direction $z$. That is, we write $|\psi\rangle = a |\uparrow_z\rangle + ...


1

A pure state is a linear combination of basis states $|\psi\rangle = \sum_k c_k |b_k\rangle$. A pure state has unit 2-norm; pure states care about squared weight $\sum_k |c_k|^2=1$. Meaning the weights are amplitudes. A mixed state is a linear combination of adjoint-squared pure states $\rho = \sum_k p_k |\psi_k\rangle\langle\psi_k|$. A mixed state has ...


8

A Hilbert space $\cal H$ is complete which means that every Cauchy sequence of vectors admits a limit in the space itself. Under this hypothesis there exist Hilbert bases also known as complete orthonormal systems of vectors in $\cal H$. A set of vectors $\{\psi_i\}_{i\in I}\subset \cal H$ is called an orthonormal system if $\langle \psi_i |\psi_j \rangle ...


3

This completeness relation of the basis means that you can reach all possible directions in the Hilbert space. It means that any $|\psi \rangle$ can be made up from these basis vectors. If the sum of the projectors (the ket-bras) would not be the unit matrix, the vector $|\psi\rangle$ could have components which cannot be represented within your basis. ...


1

I think the question was already answered here, look at either my answer of the one by Adam. The punch line is that that the expectation value of a field (such as the field $\phi$ at the bottom of the mexican hat potential) is fixed by the way the source $j$ that couples to $\phi$ in the path-integral for the functional generator is sent to zero. As there ...


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and as we represent the wave function in general as $e^{ikx}$ This isn't true. In general, we represent the wave function as $$\psi(\mathbf{x}) = \langle\mathbf{x}|\psi\rangle$$ Only in the case that $|\psi\rangle = |\mathbf{p}\rangle$ (a momentum eigenstate) do we have $$\psi(\mathbf{x}) = \langle\mathbf{x}|\mathbf{p}\rangle = ...


2

A wave function is an abstract mathematical function which could completely describe about the system under consideration. We define a wave function such that we could derive whatever information from it, provided that will not affect the state of the system. The wave function is not any operator. It's simply a function of position and time. Any ...


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The simplest way to do this is to apply CNOT gate, so you're then looking at $\langle01|\rho|00\rangle$ instead of $\langle11|\rho|00\rangle$. Then you can post-select on the first qubit being $|0\rangle$ (i.e. keep redoing the experiment and throwing out runs where the first qubit measures as $|1\rangle$) and go about measuring the off-diagonal term of the ...


1

Many answers discuss the transition from the unstable state to the stable one. Let me thus discuss the issue of choosing the ground state itself. I will suppose a two dimensional Mexican hat potential. As Numrock realised, it has degeneracy. There is nothing which lift this degeneracy in principle. Then you can change the ground state without energy. This ...


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In relativistic QFT, this cannot be a process in time. The unstable initial state does not exist at all: An unstable ground state is impossible in relativistic QFT at temperature T=0 (i.e., the textbook theory in which scattering calculations are done) since it would be a tachyonic state with imaginary mass, while the Kallen-Lehmann formulas require $m^2\ge ...


0

A "wave function" is a mathematical model (or representation) of a given wave. A "function" is represented by the symbol $f$. It can be a function of distance (x), time (t), space (r), etc. and is usually represented by an equation. If the equation represents a wave, then the function is a wave function. For example, a simple wave with constant amplitude ...


1

By Wigner's general procedure of representing the little group, the $\theta(\Lambda,p)$ is the angle of rotation associated to the massless little group element $L(\Lambda,p)\in\mathrm{SE}(2) = \mathrm{SO}(2)\ltimes\mathbb{R}^2$ fulfilling $$ L(\Lambda,p) = l^{-1}(\Lambda k)\Lambda l(k)$$ where $l(k)$ is the Lorentz transformation carrying the null vector ...


2

The precise statement of "self-adjoint operators generate continuous unitary symmetries" is Stone's theorem. It guarantees that there is a bijection between self-adjoint operators $O$ on a Hilbert space and unitary strongly continuous one-parameter groups $U(t)$ that is given by $O\mapsto \mathrm{e}^{\mathrm{i}tO}$. The definition of the exponential for an ...


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Depends on conventions of the author, usually yes, though, $\langle A,B|=\langle A|\langle B|=\langle A|\otimes\langle B|$. If $|A\rangle$ is from $\mathcal{H}_1$ and $|B\rangle$ is from $\mathcal{H}_2$, then an operator that acts on $|A\rangle\otimes|B\rangle$ is a linear operator on $\mathcal{H}_1\otimes\mathcal{H}_2$, however if you have an operator, say ...


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The notion of tensor product is independent from the Hilbert space structure, it is defined for vector spaces on the field $\mathbb K$ (usually $\mathbb R$ or $\mathbb C$). A formal definition is given below (there are many equivalent approaches). First, if $V$ is a vector space, $V^*$ denotes its algebraic dual space, namely the vector space of the linear ...


15

$\lvert A\rangle \langle B \rvert$ is the tensor of a ket and a bra (well, duh). This means it is an element of the tensor product of a Hilbert space $H_1$ (that's where the kets live) and of a dual of a Hilbert space $H_2^\ast$, which is where the bras live. Although for Hilbert spaces their duals are isomorphic to the original space, this distinction ...


2

Generically, any square-integrable function is an admissible wave function, and the space of square-integrable complex functions indeed has uncountable dimension as a vector space over $\mathbb{C}$. And it is also true that the eigenstates of the Hamiltonian span the space of states, and that they are countably many. This is the content of the spectral ...


2

As a complement to another answer, I'll demonstrate moving from the coordinate representation (wave function) to the momentum representation. Recalling that $$\Psi(x,t) = \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$ and putting this expression into the (coordinate representation of the) TDSE, we have $$i\hbar\frac{\partial}{\partial ...


4

Why is the vector |S⟩ represented as Ψ for both bases when working out the components for the quantum mechanics case above? The first of the final two equations is simply an expression for the sifting property of the delta 'function'. $$f(x) = \int dx' f(x')\delta(x - x') $$ Let's back up just a bit and write the state (ket) as a weighted 'sum' of ...


1

Take the inverse Fourier transform of your last equation, $\psi(x,t)=\frac{1}{\sqrt{2\pi\hbar}}\int e^{i p x/\hbar}\phi(p,t) dp$, to see that the "coefficients" of $\psi(x,t)$ are not the same in the two representations: in the $x$ representation, the coefficients are $\psi$ and in the $p$ representation, the coefficients are $\phi$.


7

Let me rephrase those precise equations in the language of finite-dimensional linear algebra. You have a vector $A$ and two bases $\beta=\{e_i\}_i$ and $\beta'=\{e_i'\}_i$. This means you can write the components of $A$ with respect to $\beta$ as $$ A_i=e_i·A=\sum_j\delta_{ij}e_j·A $$ and the components with respect to $\beta'$ as $$ ...


1

The safest way to start with is the representation-free Schrodinger equation, $$ i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \hat{H}|\Psi(t)⟩. $$ Referring to your case, we take the separable Hamiltonian: $H=\frac{p^2}{2m}+V$ so that $$ i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \left(\frac{\hat{p}^2}{2m}+V\right)|\Psi(t)⟩. $$ Now is the time the ...


1

The state of $n$ qubits is described by a $2^n \times 2^n$ density matrix. A density matrix can contain complex values, but it is always Hermitian (equal to its own adjoint) and its Trace always equals 1. (Density matrices are flexible enough to describe both superpositions and classical uncertainty. If there is no uncertainty, i.e. you know the exact pure ...


1

Expectation values of constants or numbers are just those constants or numbers. The expectation value of the anti commutator of $\hat x$ and $\hat p$, that is, $\langle\{\hat x,\ \hat p\}\rangle$, for the Harmonic Oscillator, or coherent states of the Harmonic Oscillator, is equal to $0$. $$\langle\{\hat x,\ \hat p\}\rangle = \langle\hat x \hat p + \hat p ...


1

@Andrew's answer provides the big picture, but I'd like to give a few more specific pointers that hopefully may help. Questions 1-2: Does everything look correct so far? How to express these operators in terms of one-particle operators? So you want to set up single-particle analogues of the ladder operators using eigenstates of the 1st quantized ...


-1

A very simple example of a wave function is y = sin x This might describe the momentary shape of a wave made by wiggling a rope. This shape would move along the x-axis with time, so y= Sin(x-t) would be a very simplified example of a moving sin wave I'm sure I will be dissed for answering your question literally. I do have 4 years of tertiary maths and 3 ...


3

At the risk of revealing that I have completely misunderstood your question, a few thoughts... People sometimes talk about regular QM as being like "zero-dimensional QFT,"* and I think that correspondence is more or less what you are getting at here. I'm not sure to what extent this viewpoint has been or can be formalized. But here is my understanding of ...


10

A wave function is a complex-valued function $f$ defined on ${\mathbb R}^1$ (if your electron is confined to a line) or on ${\mathbb R}^2$ (if your electron is confined to a plane) or ${\mathbb R}^3$ (if your electron ranges over three-space), and satisfying $$\int |f|^2=1$$ (where the integral is defined over the entire line or plane or 3-space). Every ...


5

The wave function is the solution to the Shroedinger equation, given your experimental situation. With a classical system and Newton's equation, you would obtain a trajectory, showing the path something would follow: the equations of motion. For a quantum mechanical system you get a wave function, and the rules it obeys over time. With this you can determine ...


1

In fact in some sense the full state of the quantum system shouldn't break any symmetry. It's just that the overlap of the different vacuum states (in field theory) vanishes. So in other words the full quantum state is a superposition of all the different vacua, but when we make observations we "collapse the wave function" (feel free to insert your ...


1

The statement is correct. Let's be really specific. Is there a way to, in one measurement, determine whether you have: $$ | \psi_1 \rangle = \frac{1}{\sqrt{2}} \left( | 0 \rangle + | 1 \rangle \right) $$ or $$ | \psi_2 \rangle = | 1 \rangle $$ ? You should probably suspect that this is not possible. Whatever measurement you would get on the $| \psi_2 ...


-1

The quoted statement is, strictly speaking, false. But it's pretty clear what the author meant to say. If we interpret "state" to mean "pure state" then the state space for a single cubit is the projectivization of a two-dimensional complex vector space. Therefore it has one complex dimension, which is to say two real dimensions. The state space for a ...


2

Hint: The linear Hilbert space $H$ of $n$ qubits has $2^n$ complex dimensions. The set of density operators on $H$, which by definition are Hermitian (actually semipositive) and have trace $=1$ must then have real dimension $(2^n)^2-1$.


3

What does it mean for a particular mode of particles in an infinite square well to have a Fock state $|0\rangle$ with Gaussian wave function? As far as the many-particle setting is concerned, I am tempted to say that the short answer is "Nothing, really, because the $|0\rangle$ state is not Gaussian." :D Longer answer: The formal vacuum state of ...


2

I'm having a hard time totally understanding the question here but I think the resolution might be to think of different modes as different spatial dimensions. Recall that a three dimensional particle in box has three quantum numbers ($n_x$, $n_y$, and $n_z$). In terms of quantum information content (at a logical level), is there a difference between a 2D ...


0

Physically, you look for the ground state of your theory in order to make a correct predictive calculus around the ground state. The "random" choice of the ground state depends on the dimension of the theory, in fact, you can obtain a "hat" shape or a simple 2-dimensional shape with just 2 possibilities for the ground state (look the scalar quantum ...


1

Your question amounts to "explain slowly the first month of a quantum field theory course," so I am not going to be able to address everything you've brought up in your question. But, I will try to discuss some of the key points. For a free theory there is an equivalence (at least at physicist level of rigor) between the quantum field picture and the Fock ...


4

You need to use a more precise notion of the cloning process, in order to understand the general statement and its repercussions. I will give you some outline here (mainly following the explanations of B. Schumacher and M. Westmoreland given in the reference), with an emphasis on the most important aspects of it, but to fully appreciate the importance of the ...


0

Quantum Mechanical operations are all done in Hilbert Space. A Hilbert space is an infinite dimensional complex vector space. There are many reasons why we do so. For example, if you try to find the mechanics of body by using the Hamiltonian formulation in Classical Mechanics, you will study the object's flow in phase space. If you use use Lagrangian ...


3

The No-Cloning Theorem means that if you have an unknown state then it is not possible to make an identical copy. The original reference is to Wooters, A single quantum cannot be cloned. Of course, if you know the state, you can manufacture duplicates; or if you have many identical copies of the unknown state, provided by some quantum machine, you could ...



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