New answers tagged

0

Note that $[H,A]$ is always the time derivative of $A$, modulo constants. If you want something physically different from a time derivative, then you should change the "generator" (of translations in time, i.e. time evolution) $H$. For example, the momentum operator $p$ generates translations, so $[p,A]$ gives you the spatial derivative of $A$ ...


1

There are at least two problems with the problem as stated: as @secavara mentioned, having a common basis of eigenvectors is not enough. We would certainly require the $|\psi_i\rangle$ to be common eigenvectors. Demanding $\langle\psi_i|B|\psi_j\rangle = \delta_{ij}$ is certainly too much, probably it was meant to be that $B$ is diagonal in this basis. As ...


1

A short answer to your question and to the confusion in the comments is that since $A$ and $B$ commute they are simultanously diagonalizable. Therefore your question is true up to some constants $\lambda_j$ in front of the $\delta_{ij}$. Your proof is essential correct and does not require hermiticity.


1

The equation $\langle\psi_i|B|\psi_j\rangle=\delta_{ij}$ does not seem right (unless $B$ is the identity, which would make this a strange question). To answer your questions in order: Not necessarily, although if this is a homework on quantum mechanics it would make sense that the operators in question correspond to observables and so must be Hermitian. ...


3

The "example" eq. (3) here is kind of a red herring. Let's just take the evolution equation eq. (1) and Ehrenfest's theorem eq. (2). What you're asking about is that taking the expectation value of eq. (1) and using eq. (2) results in $$ \frac{\mathrm{d} \langle A\rangle}{\mathrm{d}t} = \left \langle \frac{\mathrm{d}A}{\mathrm{d}t}\right\rangle,$$ ...


2

Here is a hint to assist your intuition, before you launch into abstractions and generalizations. A derivative is $(\hat f_\epsilon -\hat f_0)/\epsilon$ to lowest order in ε. Take $$ \hat f_\epsilon = \sigma_3 + 3\epsilon \sigma_1 ~. $$ Consider the state $$ |\psi_\epsilon\rangle= (\cos 𝜃−𝜖\sin 𝜃,\sin 𝜃+𝜖\cos 𝜃 )^T, $$ normalized to lowest order in ε. ...


0

A possible extension of operators will be operator-valued distributions. Functions contain scalars as the constant functions, distributions are a generalization of functions, but there is a wide class of distributions whose codomains are not a field, it is an algebra of operators instead. The context for this generalization is vector bundles over a manifold. ...


0

All operators met in textbook QM are linear. So, if one rephrases slightly your question Does there exist anything more general than linear operators in QM?, then the answer suggests itself: non-linear operators. As it happens, non-linear operators do play an important role in quantum field theory, where interactions can change the number of particles.


4

The $n=m$ contribution would give you the "diagonal" density distribution, whereas the $n\neq m$ gives you the "off-diagonal" one. Intuitively, if you only have a diagonal component, then the spatial profile at $x$ does not care in the slightest of its surroundings. If you were in a cubic lattice, it would correspond to a state localised ...


-1

The question asks for a constructive procedure: any vector $\mathbf{v}$ can be written in terms of two scalar functions multiplied by the two allowed vectors $\mathbf{x}$, and the gradient, respectively. These scalar functions depend on $\mathbf{x\cdot\nabla}$, $\mathbf{x}^2$ and $\nabla^2$. For simplicity let us forget about poles and cuts, so use for each ...


3

I gather you are interested in deriving all Hermite functions by the multiple application of the creation operator in coordinate space, or the ground state by laddering down from an excited state. I assume you are comfortable with Dirac's elegant ladder, in mainstream (not Griffiths) notation, $$ a|0\rangle=0\\ |n\rangle \equiv \frac{(a^\dagger)^n}{\sqrt{n!}}...


1

Since $f(x)$ must be either constant or balanced, there are only 4 possibilities for what $f(x)$ could be. For each of these four cases, you can explicitly write $U_f$ as a matrix, which you can easily check for unitarity.


1

You know that $\langle f|g\rangle$ is the inner product representing $\int f^*(x) g(x) dx$. $\langle f|X|g\rangle$ is just $\int f^*(x) xg(x) dx$. As to the problem (which looks to be out of Conquering the Physics GRE) $$\langle3|\hat a^2+\hat a\hat a^\dagger +\hat a^\dagger \hat a + (\hat a^\dagger)^2|3\rangle$$ All you have to do is distribute the bra-ket ...


0

I am assuming this is the harmonic oscillator, so we have the relations, $$a^\dagger | n \rangle = \sqrt{n+1}|n+1\rangle,$$ $$a|n\rangle = \sqrt{n}|n-1\rangle.$$ That is, the operators acting on a state produce a new state, along with a coefficient and $a |0\rangle = 0$ because it is the lowest allowed state. So let's go to your calculation: $$\langle 3 | (a+...


0

What you have in the case of $\langle\psi|\hat A|\psi\rangle$ is effectively a matrix sandwiched between a vector and a dual vector. If you like, it is a row vector on the left, a matrix in the middle, and a column vector on the right. From here you can use regular matrix multiplication rules. Where you have attempted to expand the raising/lowering operators ...


1

Your interpretation is correct. All you have to do is write the powers of $S_z$ in matrix form, and add a unit matrix factor for the terms without $S_z$.


1

First of all, $S$ is not a scalar, it is an operator! If $\{\chi_{s,m_s}\}$ is the simultaneous eigenbasis of $S_{z}$ and $S(S+1)$, then $$\tag{2} S(S+1) \chi_{s,m_s}= s(s+1) \hbar^{2} \chi_{s,m_s} $$ $$\tag{1} S_{z} \chi_{s,m_s} = m_{s}\hbar \chi_{s,m_s}$$ However, as $[S(S+1),S_{z}]=0$ and the quantum number $s$ is the same throughout this problem, the ...


3

In essence, the answer is yes. You do it by going to the "energy-space", i.e., you use the energy eigenbasis to span your Hilbert space. Let's say the kets forming the energy eigenbasis are $\vert E \rangle$ then the wavefunction of a state $\vert \psi\rangle$ in this basis would be given by $\psi(E) \equiv \langle E \vert \psi \rangle$. Just like $...


3

The basic idea of the Fourier transform is that of a transformation between two different bases in a vector space: In the position representation, the basis is that of states with well-defined positions $\phi_x$ (i.e. Dirac delta functions), and the state is written as $\psi = \int \psi(x)\phi_x \mathrm dx$, i.e. as a linear combination of the basis ...


1

Note: I suspect we might have a slightly different reading of the line from Weinberg's book: It can be shown that [the commutation relation you specify] is true of any vector $\mathbf{v}$ that is constructed from $\mathbf{x}$ or $\mathbf{\nabla}$. I feel that the stress is on the word "vector", not on the words "$\mathbf{x}$ or $\mathbf{\...


1

It seems OP is asking about the coefficient$^1$ $\frac{i}{\hbar}$ in front of the action $S$ in the Boltzmann factor $e^{\frac{i}{\hbar}S}$ of the path integral, cf. e.g. this related Phys.SE post. The coefficient is fixed by how the path/functional integral formulation is derived from the operator formalism in the first place (using an operator ordering and ...


4

In case you're planning to actually compute it, here's how: Derivatives commute with operator averages in the Heisenberg picture (since your state is constant; alternatively in the Schrodinger picture, you can take derivatives of the state), so you can do something like this: $$\frac{d}{dt}\sqrt{\langle x^2\rangle-\langle x\rangle^2}=\frac{1}{2\sqrt{\langle ...


4

Adding to @Vadim answer's: note that when a system is in an eigenstate of the Hamiltonian, it is stationary. If an operator does not have an explicit time dependence, its expectation value will be constant. This is true regardless of the details of the Hamiltonian or the operator. To see this, you can note that if the state is an eigenstate $H|\psi\rangle = ...


5

The standard deviation in question is not an operator. One could find the time evolution of $\langle\hat{x}\rangle$ and $\langle\hat{x}^2\rangle$ and then calculate the time-dependent standard deviation. How one calculates them depends on whether one uses the Schrödinger or the Heisenberg picture. In Schrödinger picture the wave function carries time ...


1

For clarification, we are working in the Heisenberg Picture, where the operators $\hat{O}$ follow the Heisenberg Equation of Motion: $$i\hbar \frac{\text{d}\hat{O}}{\text{d}t} = [\hat{O},\hat{H}].$$ Your problem is that you seem to assume that $[\hat{x}(t),\hat{p}(0)]=0$, which is not true. The fact that $\hat{p}(0)$ is a constant in time does not alter the ...


2

This is an interesting question because in general there is no unique answer, although there is a "natural" answer for the simplest observables, which include $xp$. The lack of uniqueness is linked to the ordering problem. The simplest quantization is due to Dirac; he tried to implement the idea that the commutator of two operators representing ...


0

Your expression $$H=\left(\frac{1}{6}\right)\left[S_{x}^{4}+S_{y}^{4}+S_{z}^{4}-\frac{1}{5} S(S+1)\left(3 S^{2}+3 S-1\right)\right]$$ is magnificently ambiguous. The $S_i$s are operators, but the expression $S(S+1)$ immediately telegraphs to all physicists that it represents the eigenvalue of the Casimir operator acting on an irrep, $$\hat S ^2= S(S+1) 1\!\!...


2

The identity is in fact an operator, but it's a trivial one. Every state is an eigenstate of the identity operator with eigenvalue one. The $S$ in your hamiltonian is not an operator, that is just a number, but the combination $$S(S+1)\mathbb{1}$$ is an operator and, again, is a trivial operator since it's just the identity rescaled by some quantity $S(S+1)$....


2

$$\langle p \rangle = -i\hbar\int\psi^*\frac{d \psi}{dx}dx$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\frac{d}{dt}\int\psi^*\frac{d\psi}{dx}dx$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\int\frac{d}{dt}\left(\psi^*\frac{d\psi}{dx}\right)dx \ \ \ \ \text{By Leibniz's Rule}$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\int \left(\frac{d\...


3

Yes, normally one uses a convenient normalization for creation and annihilation operators. In QM there is usually only one frequency, ω, in your problem, so you incorporate it into the normalization of $a$ and $a^\dagger\equiv a^*$, (11.10), and it virtually disappears from the problem, $$ D^2={\hbar \over m\omega}, \\ a=\frac{1}{\sqrt 2} \left ({X\over\sqrt{...


0

A preliminar point should be made clear, before discussing Uncertainty Relations. According to Quantum Mechanics a particle is never a wave or a superposition of waves. Classical waves have nothing to do with individual quantum particles. A relation with waves does exist, but it is much more subtle: waves can used to obtain the probability distribution of ...


0

You more or less answered your own question. The wave properties are described by a wave function that is governed by the dynamical equations of the system under investigation and the statistical properties appear when you make measurements of this wave function. Such measurements will sample the wave function in a random fashion so that the squared modulus ...


1

In both terms the indices $\mu$ and $\nu$ are free indices. That means that when you try to go from the first one to the second $k^2g^{\mu\nu}\rightarrow k^\mu k^\nu$ by breaking the square you can't write $k^2g^{\mu\nu}=k^\mu k_\mu g^{\mu\nu}=k^\mu k^\nu$ because You would be introducing a dummy index (one that is summed over) with the same letter as one ...


3

First, We have formula $$[AB,C]=A[B,C]+[A,C]B.$$ Since $L_i$ commutes with $H$, we can identify $[L_i^2,H]=0$ below $$ \begin{split} [L_i^2,H] &= L_i[L_i,H]+[L_i,H]L_i\\ &=0. \end{split} $$ Then we have $$ [L^2,H]=\sum[L_i^2,H]=0 $$ because of $L^2=\sum L_i^2$. I think the condition $[L_i,L^2]$ don't work for this question, because ...


1

$[\textbf{L}^2,H] = [\sum_i L_i^2,H] = \sum_i [L_i^2,H].$ If $[L_i,H]=0$, then so that $[f(L_i),H]=0$ where $f$ is a function of $L_i$, and hence $[L_i^2,H]=0$. Hence, $$ \sum_i [L_i^2,H] = 0 \quad \Rightarrow \quad [\textbf{L}^2,H] = 0.$$


3

The hamiltonian of a system has to be an hermitian operator since it's associated to a measurable quantity, the energy. By symply taking the classical hamiltonian$$H=\omega xp$$ and converting it directly to an operatore $$\hat H = \omega\hat x\hat p$$ you can easily see that this operator is not hermitian since, given $\hat x$ and $\hat p$ hermitian, $$\hat ...


1

There are multiple ways of proving this, as other answers show. Here is another one. There is a general identity that states that if $[x, y]=0$, then $[x,f(y)]=0$, where $f$ is some function. The result you are after follows immediately from this, with $f(y)=\sqrt y$. To prove this identity, let us first show that $[x,y]=0$ implies $[x, y^n]$, where $n$ is ...


5

The answers given by Philip and Alexander are correct, but let me add another snippet of information: one could calculate the commutation in the polar coordinates! Then: $$\hat{L}_i = \frac{\partial}{\partial \phi_i},$$ whereas $f(r)$ is obviously independent on angle $\phi_i$.


1

You could use the following rule that states that if you have two operators $A$ and $B$ such that $[B,[A,B]]=0$, then $$[A, f(B)] = f'(B)[A,B].$$


1

Hint 1: There is a basic identity (derived upon several restrictions) that in your case reads as - $$[L_i,f(r)]=[L_i,r]f'(r)$$ Applying the same relation you can show that if operator commutes with $r$, it commutes with any function of $r$ You can also use the following for $L_i$ to relax the expression - $$[AB,C]=A[B,C]+[A,C]B$$ @Philips answer also useful, ...


2

It follows from the fact that $\hat L_i$ are generators of the rotations. Generally, the vector operator $\hat V_i$ is not just "a list" of operators. It has to transform like a vector under rotations $\hat V_i' = R_{ij} \hat V_j$ (summation implied). This identity can be proven by considering an observable in a rotated frame of reference. The ...


1

Theorem: Let $\mathbf{A},\mathbf{B}$ be vector operators. then $\mathbf{A}\times\mathbf{B}$ is a vector operator. But e.g. $\mathbf{A}\mathbf{B}$ is a scalar. That is $[L_j,\mathbf{A}\mathbf{B}]=0$ The proof is done by straight forward algebra, using the definition of a vector operator $[L_j,A_i]=i\hbar\epsilon_{jik}A_k$. But I think your confusion isn't ...


1

@probably_someone gave you the answer, but your comment suggests you don't see how constant operators in the Schrodinger picture can give you time-dependent, indeed, oscillating, expectation values. Free quantum fields are just an infinity of elegantly packaged dumb oscillators. For simplicity, non-dimensionalize to $\hbar=1$, and pick up just one momentum ...


1

The classical electric field is the expectation value of the electric field operator applied to a particular quantum state of the electromagnetic field $|\psi\rangle$: $$\langle \mathbf{E}\rangle=\langle \psi|\mathbf{E}|\psi\rangle$$ If the state evolves in time, then the expectation value of the electric field operator can also evolve in time. For example, ...


0

Elaborating on @shehab 's answer.... Given $k^2g^{\mu\nu}-k^\mu k^\nu = k^2P^{\mu\nu}$, for an arbitrary co-vector $A_\nu$ we then have $$\begin{align*} (k^2g^{\mu\nu}-k^\mu k^\nu)A_\nu &=(k^2P^{\mu\nu})A_\nu\\ k^2 A^{\mu} - k^\mu (k^\nu A_\nu) &=k^2 (P^{\mu\nu}A_\nu). \end{align*}$$ Observe that, in general, $$k^2 A^{\mu} \neq k^\mu (k^\nu A_\nu)$$ ...


0

Your assumption that you can break the square term and raise index is incorrect: The indices will not be the same as the indices in the second term, it is always helpful to name your indices differently each time you "break" a term.


2

Integrating the RHS from your hint, using the divergence theorem (and spherical polar coordinates to better make the point): $$\int_V \mathrm{d}^3\mathbf{r}\, \nabla\cdot(f^*\nabla g-g\nabla f^*) = \oint_S r^2\mathrm{d}\Omega \, (f^*\nabla g-g\nabla f^*)\bigg\vert_{r=R}, $$ where the last integral is a surface integral evaluated at the edge of the volume $V$,...


1

In matrix form, $$\langle f|Ag\rangle = f^\dagger A g,$$ $$\langle Af|g\rangle= (Af)^\dagger g.$$ Using the matrix property of $(AB)^\dagger=B^\dagger A^\dagger$ on the latter expression, we get $$\langle Af|g\rangle = (Af)^\dagger g=f^\dagger A^\dagger g.$$ Hence if $A=A^\dagger$, then $$\langle Af|g\rangle= f^\dagger A^\dagger g= f^\dagger A g=\langle f|Ag\...


6

$\langle f|Ag\rangle=\langle f|A|g\rangle$. $\langle Af|g\rangle$: $(\langle Af|) = (|Af\rangle)^\dagger =(A|f\rangle)^\dagger = \langle f |A^\dagger$, so $\langle Af|g\rangle = \langle f |A^\dagger|g\rangle$ If $A = A^\dagger$, then $\langle f|Ag\rangle =\langle Af|g\rangle$.


1

I don't think is possible, because the derivative can not be inverted. For example, if you have a function $f(x)=x+1$ then its derivative is $f'(x)=1$. Now, if you try to invert the derivative (which means finding the primitive) you will see there you get the original function only up to an arbitrary constant ${\cal C}$ $$\int f'(x) dx = f(x) + {\cal C}.$$ ...


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