# Tag Info

3

What is meant by that expression is $\Delta\hat{x}=\hat{x}-\langle\hat{x}\rangle \text{Id}$, with $\text{Id}$ being the identity operator.

0

As noted in the comments already, in infinite dimensions the completeness relation reads: $$\int|x'\rangle\langle x'|dx'= \Bbb I, \tag{1}$$ where $\Bbb I$ is the identity operator. This is equation 1.10.11 on page 59 in Shankar. Addition discussion on this here.

0

$\int\langle x|D|x’\rangle \langle x’|f\rangle dx’= \langle x|D \int |x’\rangle \langle x’ | dx’ |f \rangle= \\ \: \: \: \: \: \: \langle x| D | f \rangle = | \frac{df}{dx}\rangle$ Since you integrate over orthonormal basis elements spanning the infinite-dimensional Hilbert space, $\int |x’\rangle \langle x’ | dx’ = \text{Id}$.

2

OP's calculation seems to match Zee's calculation; except for the final step. Here OP has made a mistake: $$\left< k | k \right> = (2 \pi)^4 \delta^{(4)}(0) \neq 1.$$ This is where the factor of $VT$ comes from: $$\left< k | k \right> = \left<k | 1 | k \right> = \int d^4 x \left< k | x \right> \left< x | k \right> = \int d^4 ... 0 The way to answer this question is to think in terms of a basis for the matrix, for convenience we can choose a basis that is hermitian, so for a 2-by-2 matrix it has basis:$$ e_{1} = \left(\begin{matrix} 1 & 0 \\ 0 & 0\end{matrix} \right), \qquad e_{2} = \left(\begin{matrix} 0 & 0 \\ 0 & 1\end{matrix} \right), \qquad e_{3} = \left(\begin{...

3

It doesn't matter, because $\langle j|V\rangle$ is just a number. It's written that way to reflect the fact that $\sum_i |i\rangle\langle i|$ is the identity operator. Acting with it on $|V\rangle$ from the left then gives an expansion of $|V\rangle$ in the basis $|i\rangle$.

1

There is a more slick, more mathematical way, of approaching the problem, by using the general theorems that eigenvalues of a Hermitian operator are real and that eigenvectors with distinct eigenvalues are orthogonal. In this case the result is a trivial corollary. However, I do not think the question would be asked if this is what was intended. I think in ...

2

The condition for two eigenfunctions to be orthogonal is that their inner product is zero. In Dirac notation this would mean: $$\langle \psi_n|\psi_m\rangle=0\quad m\neq n\tag{1},$$ and in wave function notation (as you have written $\psi_n(x)=\langle x|\psi_n\rangle$ in your question) this becomes: $$\langle \psi_n|\psi_m\rangle=\int_{-\infty}^{\infty}\... 7 No, this is not true. If A and B are anti-Hermitian, that is if:$$A^\dagger = -A, B^\dagger = -B$$then their tensor product is Hermitian:$$(A\otimes B)^\dagger = (-1)^2 A\otimes B$$6 No. Counterexample: A=0 and B is anything. 20 Suppose we don't know quantum mechanics yet and we want to calculate the expecatation value of an observable A. Could be momentum, spin whatever. It is given by$$\mathbb E(A)=\sum_i a_i\,p(a_i)$$Where a_i are the possible outcomes and p(a_i) are the probabilities of those outcomes. When the outcome is continuous this becomes an integral. To each of ... 0 Operators correspond to observables. Eigenvalues are the observables' possible values. And the why is: Well as I learnt , these are axioms of QM. No need for proof, the whole theory is built on these. 2 Look at the postulates of quantum mechanics, the second page in the link: With every physical observable q there is associated an operator Q, which when operating upon the wavefunction associated with a definite value of that observable will yield that value times the wavefunction. Postulates, laws, principles are an encapsulation of physical observations,... 11 I think the conceptual issue is that you've got it backwards. Given any set of eigenfunctions, you can have operators with those eigenfunctions, whose corresponding eigenvalues are anything you want. For example, if you're in a three-dimensional Hilbert space and the eigenvectors are the standard unit vectors, then$$A = \begin{pmatrix} 1 & 0 & 0 \\ ...

3

If a system is described by the eigenfunction $\psi$ of an operator $A$ then the value measured for the observable property corresponding to $A$ will always be the eigenvalue $a$ which can be calculated from the eigenvalue equation $\hat {A} | \Psi \rangle = a | \Psi \rangle \tag 1$ This is a mathematical description relating the eigenvalue and ...

0

Probably the clearest cases of superposition are entanglements. Here is my favorite abstract game: we have a team of three people who is trying to thwart me, their Inquisitor. Each round, I will split them up into three rooms so that no known physical processes can communicate between them. Each room has a screen, a timer, and two buttons labeled 0 and 1. ...

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The properties of a particle are described by its wavefunction which has all the info of the particle. When we interact with a quantum particle its wavefunction becomes extremely localised and we can read that info. When we don't interact with the particle the wavefunction of the particle is not localised and these properties get many values at the same ...

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I am now looking at the computer screen using polarizer glasses. If I turn my head 90 degrees, the screen darkens because the light coming from it is polarized, and the glasses doesn't allow the polarizing direction pass through my eyes. With my head up or tilted 90 degrees the states of the photons are defined. In the first case they pass through, and in ...

2

Your inequality can be read as $$f(\ell)\geq f(m)$$ where $f(x)=x(x+1)$. This can be reduced to $\ell\geq m$ by applying $f^{-1}$ to both sides. But there's a caveat. Our function $f$ is not an injection since multiple x-values can map to the same number. This means we can't define the inverse easily and we have to do more work. Our function is a parabola ...

2

You need to use the $\langle \psi|\chi\rangle=\int \psi^\dagger \chi\,d^3x$ to do Dirac single particle quantum mechanics. There is no conflict with Lorentz invariance because rewriting the Dirac equation as $i\partial_t \psi =H_{\rm Dirac} \psi$ has already broken explicit Lorentz invariance. Of course $\psi^\dagger \psi$ is not Lorentz invariant, but ...

0

In quantum field theory deriving operators with respect to parameters is routine. (Recall how the canonical momentum field is produced.) So, if you are careful, of course you may define such objects. For instance, $[a(\omega),a^\dagger (\omega')] = \delta(\omega-\omega')$ can yield $$[\partial_{\omega} a(\omega),a^\dagger (\omega')] = \partial_{\omega} \... 2 If you define the spatial probability density \rho(x,t) = \psi^*(x,t)\psi(x,t) and compute its derivative with respect to time, you obtain after some algebra$$\frac{\partial \rho}{\partial t} = -\frac{\partial}{\partial x} \left[\frac{i\hbar}{2m} \left(\frac{\partial \psi^*}{\partial x}\psi - \psi^* \frac{\partial \psi}{\partial x}\right)\right]\equiv -\...

0

Two things happen, depending on the magnitude and phase of the complex factor. Both can be seen by examining what happens when you insert the factor $f$ in $\mid\Psi\rangle = f\mid\psi\rangle$. Your primary observables are always bra-ket pairs, so calculate $\langle\Psi\mid\Psi\rangle=\langle\psi\mid f^*f\mid\psi\rangle = |f|^2\langle\psi\mid\psi\rangle$. ($... 2 An eigenfunction like the hydrogen ground state referred to in the question separates into time and space parts: $$\Psi(r,\theta,\phi,t) = \psi(r,\theta,\phi) ~ e^{-iEt/\hbar}$$ where$E$is the energy of the state. However the exponential part isn't a physical observable so we tend to ignore it and just write: $$\Psi(r,\theta,\phi,t) = \psi(r,\theta,\phi)... 4 Think about this like that,- when you multiply a vector by a scalar, you just re-scale original vector. Same idea applies here to vectors in Hilbert space, aka. ket. You just re-scale Hilbert vector by a complex number, an hence get another ket in same Hilbert space. 1 Hydrogen ground state identified by quantum numbers (n=1,\ell =0,m=0) has no dependancy on phase, so ground state has phase factor e^{i \phi} = 1. However for other Hydrogen energy levels situation can be different. For example Hydrogen state identified with quantum numbers (2,1,\pm 1) does depend on phase and as such state's wavefunction has full ... 9 The states of a quantum system, the kets, are elements of a complex Hilbert space (modulo a phase). A complex Hilbert space is nothing more than a fancy (in)finite dimensional vector space equipped with an inner product. So being the Hilbert space a vector space, all the rules which define a vector space apply. The field upon which the vector space is ... 2 It is complex, just with the imaginary component being zero. 1 The widely recommended approach for deriving completely asymmetric states of many particles is by using the Slater determinant. Thus, the asymmetric state of two particles can be expressed as$$ \Psi(x_1, x_2) = \frac{1}{\sqrt{2!}} \left|\begin{matrix} \phi_1(x_1) & \phi_2(x_1)\\ \phi_1(x_2) & \phi_2(x_2) \end{matrix}\right| = \frac{1}{\sqrt{2}}\... 1 You did a mistake in evaluating the derivative. $$\frac{d}{dt} \langle \psi \vert \psi \rangle = \left( \frac{d}{dt} \langle \psi \right)\vert\psi \rangle + \langle \psi \vert \left( \frac{d}{dt} \psi \rangle \right) \neq 2 \langle \psi \vert \left( \frac{d}{dt} \psi \rangle \right).$$ Recalling the Scrodinger equation$\vert \dot \psi \rangle= (- i \hat ...

5

The inner product in Quantum Mechanics is not symmetric if you interchange the terms, since $$\langle \phi | \psi \rangle = \langle \psi | \phi \rangle^*.$$ As a result, $$\frac{\text{d}}{\text{d}t}\langle \psi | \psi \rangle = \Big\langle \frac{\text{d}\psi}{\text{d}t}\Big|\psi\Big\rangle + \Big\langle \psi \Big| \frac{\text{d}\psi}{\text{d}t}\Big\rangle = ... 4 Ladder operators are not Hermitian, so they don't correspond to physical observables. Furthermore, Hermitian operators like the position and momentum operators don't correspond to some sort of physical manipulation of the system, at least in the way you are thinking. For example, taking a system described by some state vector |\psi\rangle and then looking ... 1 Hint : This is not precisely an answer but a suggestion to see the answer from a different point of view : that of the Pauli matrices. So, as @mike stone pointed out in his answer, the representation of your Hamiltonian \mathrm H with respect to the orthonormal basis \left(\psi_{1},\psi_{2}\right) is \mathrm H\boldsymbol{=}\alpha \begin{... 1 Consider the hermitian operator X = a^†b +b^†a. We can see that [X,a] = -b and [X,b]=-a, and [X,a^†] = b^† and [X,b^†]=a^†. Therefore U(\theta) = \exp(-iX\theta) is a unitary operator that satisfies$$U(\theta)aU^†(\theta) = a \cos\theta + ia\sin\theta bU(\theta)bU^†(\theta) = ia\sin(\theta) + b\cos\theta $$And so if you pick \theta = \... 1 Whenever you encounter quadratic equations, even without a linear term, there are always two answers. For example,$$x^2=4$$has two solutions, x=2 and x=-2. You have a similar situation in solving for your eigenvalues. Also, when finding the eigenkets, even if the eigenvalues are the same (which they aren't here), the eigenkets must be constructed to be ... 2 Your revised matrix$$ H=\alpha\left[\matrix{0&1\cr 1&0}\right] $$has eigenvalues \pm \alpha. You should be able to find the mutually orthonal eigenvectors now. 0 You can explicitly construct such a unitary map by mapping the modes generated by a to the ones generated by b and vice versa. Denote by |n,m\rangle\propto (a^\dagger)^n(b^\dagger)^m|0\rangle the eigenstates of the number operators, then the map$$U:|n,m\rangle\mapsto|m,n\rangle$$is unitary with a=U^\dagger bU as wished. 0 Since a Hermitian matrix can be diagonalized as$$ A = U^{\dagger} D U , $$where U is composed of the eigenvectors and D is a diagonal matrix with the eigenvalues on the diagonal, one can use it to compose A. For the two-dimensional case it would be represented by$$ A = |a\rangle \lambda_a \langle a| + |b\rangle \lambda_b \langle b| , $$where |a\... 0 The updated question is to show that there exists a state that satisfies CNOT\vert\psi\rangle = \vert\psi\rangle and U\otimes V\vert\psi\rangle = \vert\phi^+\rangle The eigenstates of CNOT with eigenvalue 1 are \vert 00\rangle, \vert 01\rangle and \vert 1+\rangle. Hence, our state is of the form$$\vert\psi\rangle = a\vert 00\rangle + b\vert 01\...

0

They don't hold particular meaning, no. As always, $|\langle\psi_j|\phi\rangle|^2$ is the probability of finding $|\psi_j\rangle$ having $|\phi\rangle$, so $\langle\psi_j|\phi\rangle$ is a quantity whose absolute value equals the square root of such probability. Similarly, $F_j|\phi\rangle$ is the state $|\psi_j\rangle$ multiplied by a number that is related ...

1

A fermionic theory is, by definition, one whose Hilbert space is a super-vector space, i.e., it admits a $\mathbb Z_2$ grading $$H=H_0\oplus H_1$$ Here $H_0$ denotes the "bosons" and $H_1$ the "fermions", i.e., the states that are even under the grading, and those that are odd, respectively. This grading is usually referred to as "...

1

I'll assume the question is, given the state $|\Psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$, how can we figure out whether this state is maximally entangled, as that is the question in the title of the post. A bipartite state $|\Psi\rangle$ is maximally entangled if and only if the corresponding reduced state $\rho_A\equiv\operatorname{Tr}_B(|... 1 What is true is that $$\langle p|\hat x|\psi\rangle = i\hbar \partial_p \langle p|\psi\rangle$$ so $$\langle p|\hat x|\bar p \rangle= i\hbar \partial_p \langle p| \bar p\rangle = i\hbar \partial_p \delta(p-\bar p).$$ Now $$\frac{\partial}{\partial p} \delta(p-\bar p)= - \frac{\partial}{\partial p'}\delta(p-\bar p)$$ so it all works out. 2 By setting up the integral the way you have,$X$does not end up in the integral. The implicit step you left out that is in the notes before the steps you give in your post is $$\langle p|X|\bar p\rangle=\iint\langle p|\bar x\rangle\langle\bar x|X|\hat x\rangle\langle\hat x|\bar p\rangle\,\text d\bar x\,\text d\hat x$$ And then$\langle\bar x|X|\hat x\rangle=...

1

I think the point at issue may be restated as: prove that, if $\rho_{tot}$ is the state of a composed system ($S+R$) and $\rho_S=\text{Tr}_R[\rho_{tot}]=|\varphi\rangle_S\langle\varphi|$ is pure, then $\rho_{tot}=\rho_S\otimes\rho_R=|\varphi\rangle_S\langle\varphi|\otimes\rho_R$. In other words, if the state of the system is pure, then the overall state must ...

1

Without going through all the calculations that you and other answers have done: In quantum mechanics, if we obtain degenerate eigenvalues for a given operator, we construct/choose the eigenvectors such that they are mutually orthogonal to each other and eigenvectors of other non-degenerate eigenvalues. Given your three choices for $\lambda=1$, $|\alpha\... 2 Yes, everything you have written is correct, although, maybe, it is better to clarify the meaning of some definitions. The "wavefunctions" of the Quantum Harmonic Oscillator are nothing but the representations in the position basis of the eigenstates of the Hamiltonian associated to the harmonic oscillator. Let us term the latter as$H_{HO}$. Then, ... 0 I think you are confusing there being infinite solutions with degeneracy. Let's label the vectors first to make things more clear. When I solved the eigensystem I got slightly different eigenvectors so I'm not sure what happened here. (I used Mathematica which is probably right) $$v_1=\pmatrix{0\\i\\1}\\v_2=\pmatrix{0\\-i\\1}\\v_3=\pmatrix{1\\0\\0}$$ This ... 4 Okay, let's just do this from the beginning. We have a Hermitian matrix: $$A = \left[ \begin{array}{c|c} I_{1\times 1} \\ \hline & \sigma_y \end{array} \right]$$ where$I_{1\times 1}$is the$1\times 1$identity matrix and$\sigma_y$is the second Pauli matrix. The upper block is$1\times 1$so is already diagonal: the eigenvalue is$1$and the ... 0 We postulate that different quantum states represent same physics state, when they only differ by multiplying a complex constant (i.e. all wavefunctions are physically same if they have same "direction"). Your 2nd and 3rd are same physics state, as they only differ by a$i\$.

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