# Tag Info

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From a classical standpoint, it seems pretty clear that information can be easily lost. Nope, classical mechanics is reversible. It's so reversible that you can't even squeeze the state space: states that start out "very different" stay "very different". Those vague terms are formalized by Liouville's theorem. Because classical mechanics is reversible, ...

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First, observe that the Slater determinant you have written down is the linear combination of the singlet state and the z-spin-0 state of the triplet. Vice versa, you can produce the singlet and triplet states as linear combinations of Slater determinants. Whether you prefer the Slater determinant or the singlet/triplet formalism for writing down your ...

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Within the context of first quantization, spin itself is not characterized by the wave function: it exists as a separate (Hilbert) space. The total state of the particle is then a composite of its wave function (often projected onto the configuration or momentum basis) and spin state.

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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 $$\psi(\mathbf{x}) = \langle\mathbf{x}|\mathbf{p}\rangle = ... 2 A wave function is an abstract mathematical function which could completely describe about the system under consideration. We define a wave function such that we could derive whatever information from it, provided that will not affect the state of the system. The wave function is not any operator. It's simply a function of position and time. Any ... 0 ψ and ψ^* are not the same but are mutually connected.Taking a complex conjugate means finding the reflection of a point in the Argand plane about the real axis. So you cannot simply put ψ as a substitute for ψ^* and vice versa. So, if you need to replace ψ by ψ^*, you need to put a negative sign for where you see the operator i as an image ... 3 You can always change the symbol that stands for a dummy variable, but you can't change its interpretation. Your mistake is tantamount to starting from the equation$$x + 1 = 2$$which has solution x = 1, then declaring x is a dummy variable and replacing it with -x, for$$-x + 1 = 2$$which has solution x = -1. This is totally valid, but these two ... 3 Both are correct, actually. If you measure an observable for that wave function you'll either find the eigenvalue corresponding to state 1 with probability |c1|^2 (similarly for state 2), subject to the condition |c1|^2 + |c2|^2 = 1. Edit: What Griffiths is saying is that before you perform the measurement, the particle is neither in state 1 or 2, but in a ... 0 I figured it out. The problem was that I was simply adding the two phase shifts, ignoring the periodicity in the tangent and arctangent functions. In order to get the proper result/plot, one can resort to the addition formula for tangents$$\delta=\delta_1+\delta_2=\arctan ...

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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 ...

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You are free to choose which axis you make a measurement on and doing so will always yield and eigenvalue of the spin operator in that direction: you will always measure $\pm\hbar/2$ in whichever direction you choose to measure. The reason is that the state vector of the particle exists in a superposition of states with various probabilites $$|\psi\rangle ... 3 Given that the "state" of a single qubit cannot be independently factored and considered, how does the hadamard gate now affect the state of system? To apply a single-qubit operation M to an n-qubit system you hit the system with I \otimes M \otimes I \otimes ... \otimes I. The position of M within that tensor product determines which qubit you ... 3 How can I write a single Qubit operator O (given as a 2\times 2 matrix) as a multiQubit operator O' (given as a 2^k \times 2^k matrix) that acts as the identity on all inputs except the first where it acts as O traditionally would. This has a very direct answer:$$O' = O\otimes \underbrace{I_2 \otimes \cdots \otimes I_2}_{k-1\text{ ...

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As far as I know, and I'm only an undergraduate student, the boundary conditions in Schrodinger equations are there to hold some special subspace of the Hilbert space of the system or the Hilbert space as a whole. Bound states, for example, form a subspace on the Hilbert space. The boundary condition to that is that $\psi\sim e^{-r}$ at infinity, for every ...

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The choice of boundary conditions fixes the domain and thus a self-adjoint extension of your Schroedinger operator $$-\frac{\hbar^2}{2m} \Delta + V\:.$$ In turn, this choice determines the spectrum of that operator, i.e., the eigenvalues $E$. (There is no analogy with initial conditions here.) No circumstances! Boundary conditions are uncorrelated with ...

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In the examples you've given, the boundary conditions simply say "don't have infinite energy" and "don't be non-normalizable". These don't really have a physical interpretation. Moreover, no boundary conditions ever have an interpretation like "initial position and velocity" because the time-independent Schrodinger equation describes stationary states. ...

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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 ...

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$\lvert A\rangle \langle B \rvert$ 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 ...

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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 ...

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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 $\phi$.

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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 $$... 0 The wave function you've provided is a linear superposition of two distinct energy eigenfunctions, \psi_1(x) and \psi_2(x), that are assumed to have distinct energy eigenvalues, E_1 and E_2 respectively. It is not possible to predict with absolute certainty the outcome of a measurement of the energy observable. 1 Reflection, R and transmission, T, probabilities are determined in terms of the current flux. For the case you describe, yes the reflection coefficient will be unity and the transmission coefficient will be zero. However there will still be a non-zero probability of finding the particle in the classically forbidden region, however this decays ... 3 The way you combine quantum systems is not by summing their wavefunctions, but by taking the tensor product, see this question. In particular, if the systems you are composing are just single electrons described by square-integrable wavefunctions in L^2(\mathbb{R}^3,\mathrm{d}x), then a system of N electrons is described by a square-integrable function ... -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 ... 0 The short answer is No. Because there is no definite energy state of this particle. It is in a superposition of two energy states E_0 and E_1.So, all we know is that any measurement will result in the collapse of wavefunction in one of the energy eigenstates. But where will it collapse is completely random. All we know is that there is equal probability ... 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$$(where the integral is defined over the entire line or plane or 3-space). Every ... 5 The wave function is the solution to the Shroedinger equation, given your experimental situation. With a classical system and Newton's equation, you would obtain a trajectory, showing the path something would follow: the equations of motion. For a quantum mechanical system you get a wave function, and the rules it obeys over time. With this you can determine ... 3 What does it mean for a particular mode of particles in an infinite square well to have a Fock state |0\rangle with Gaussian wave function? As far as the many-particle setting is concerned, I am tempted to say that the short answer is "Nothing, really, because the |0\rangle state is not Gaussian." :D Longer answer: The formal vacuum state of ... 2 I'm having a hard time totally understanding the question here but I think the resolution might be to think of different modes as different spatial dimensions. Recall that a three dimensional particle in box has three quantum numbers (n_x, n_y, and n_z). In terms of quantum information content (at a logical level), is there a difference between a 2D ... 1 After revising the (very informative) responses to my previous queries, my revised understanding (which addresses my queries) is: If we measure the position then \Psi(x,t) collapses to \Psi(x,t) = \delta(x-z) (where z is the measured position), using the Fourier transform we get:$$\Phi(p,t) = \frac{1}{\sqrt{2 \pi \hbar}} ...

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The act of measuring cannot be represented by an operation in Linear Algebra, because postulate of QM says that upon measurement of observable $A$ with eigenvectors and eigenvalues given by $|a_i>$ and $a_i$ respectively on $|\psi>$ . $$|\psi>=\sum_i c_i |a_i>$$ Gives $|\psi_{after}>$. where $|\psi_{after}>$ can be any of $|a_i>$ with ...

2

I think you are getting something wrong in the 2nd and 3rd point, but i am going to try and give you an explanation. If i got something wrong in your questions, please do point it out. 1) If you measure the position, then Ψ(x)=δ(x-z) where z is the position that you measured(not a variable). If you plug this into the Fourier transform of Ψ(the first ...

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What we want us not just $$\int|\Psi (x)|^2\;dx$$ For example $$p=\int Ψ^*(x)(- i \hbar)\; \text{grad} Ψ(x)\;dx$$ If the wave function is not differeciable $\text{grad}$ will come into a terrible result

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Well, the trite answer is "no" because every bit of matter we see around us is in an entangled state. It's the normal classical world.

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