In physics, an operator is almost always either a square matrix or a linear mapping from one space of functions (often on $\mathbb{R}^N$ or $\mathbb{C}^N$) to the same or other like space of functions. Operators serve as *observables* and as *time evolution operators* in Quantum Mechanics. This tag ...

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Prove that $(e^{i\lambda A})^\dagger=e^{-i\lambda A^\dagger}$

Prove $$(e^{i\lambda A})^\dagger=e^{-i\lambda A^\dagger}$$ where $A$ is an operator. Can anyone explain how to go about this question? Writing it as a power series gets confusing.
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How to calculate the expectation value of position/momentum using path integrals?

We have the formula: \begin{equation} \langle F \rangle = \frac{\int Dx \times F[\phi] exp\{i/\hbar S[\phi]\}}{\int Dx \times exp\{i/\hbar S[\phi]\}} \end{equation} Now, I am wondering how a change ...
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2answers
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The equivalence between Heisenberg and Schroedinger pictures

In quantum mechanics, the two pictures of Schroedinger and Heisenberg are taken as equivalent, where in the former wavefunctions are time variants and operators are not, and in the latter it is the ...
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How does Dirac define the representative of $\{\langle\phi\frac{d}{dq}\}\psi\rangle = \langle\phi\{\frac{d}{dq}\psi\rangle\}$

On pate 89 of Dirac's book, The Principles of Quantum Mechanics, he writes: Let us treat the linear operator $\frac{d}{dq}$ according to the general theory of linear operators of section 7. We ...
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How to calculate the expectation value of position vector?

$$\psi (\vec{x})=Ae^{-(1/4a^2)|\vec{x}-\vec{x}_0|^2}e^{i\vec{p}_0\cdot \vec{x}/\hbar}$$ The wave function is like this, then how is the expectation value of position vector (not position) calculated? ...
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What is the Hermitian adjoint of operator $\hat{K}\psi = \psi^*$?

For many operators, their adjoint can be expressed as a function of other known operators, for example $$\hat{T}_a^\dagger = \hat{T}_{-a} \\ \hat{p}_x^\dagger = \hat{p}_x$$ where $\hat{T}_a \psi (x) = ...
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Adjoint of momentum operator

In position basis, we have, $$\langle x \mid \hat p \mid \Psi(t) \rangle = -\imath \hbar \frac{\partial{\langle x \mid \Psi(t) \rangle}}{\partial{x}} $$ Now I know $\hat{p}$ is a Hermitian operator ...
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$n^\text{th}$ operation of creation and annihilation operators on vacuum

My question is similar to the that posted in this link. In particular I would like to express the following expression in the most compact form: $(\hat{a}^\dagger(x)+\hat{a}(x))^n\vert0\rangle$, ...
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1answer
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How to pull out the momentum operator?

In the equation (1.7.17), how does operator $p$ get out of the bracket without any operation though $<a | $, $| x'>$ are function of $x'$? How to prove this?
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Representation of operator in “$a$-representation” [on hold]

I have an operator, $X$ say, which is given by a simple phase-shift of the annihilation operator $a.$ I've been asked to give the matrix of $X$ in the $a$-representation, but I was hoping someone ...
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Understanding the Quantum Vacuum State [duplicate]

In terms of the creation and annihilation operators $a_{j}$ and $a_{j}^{\dagger}$ (fermionic or bosonic, doesn't matter): Is the vacuum state $\mid\mathrm{vacuum}\rangle$ exactly the zero vector on ...
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Binomial expansion of non-commutative operators

I would like to determine the general expansion of $(A+B)^n$, where $[A,B]\neq0$, i.e. A and B are two generally no-commutative operators. How could I express this in terms of summations of the ...
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Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis

I am trying to calculate $\langle x\ |\ \hat{x}\ |\ p\rangle$. I can work in the $x$-basis like so: $$\langle x\ |\ \hat{x}\ |\ p\rangle=\int dx'\langle x\ |\ \hat{x}\ |\ x'\rangle\langle x'\ |\ ...
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How does the momentum operator act on state kets?

I have been going through some problems in Sakurai's Modern QM and at one point have to calculate $\langle \alpha|\hat{p}|\alpha\rangle$ where all we know about the state $|\alpha\rangle$ is that ...
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1answer
42 views

Total angular momentum operator

How do the eigenfunctions of the total angular momentum operator analytically look like? I mean the operator is given by $J = L+S$ so the eigenfunctions have to be tensor-product states, right? Can ...
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How to replace $T$-product with retarded commutator in LSZ formula?

I am reading Itzykson and Zuber's Quantum Field Theory book, and am unable to understand a step that is made on page 246: Here, they consider the elastic scattering of particle $A$ off particle $B$: ...
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Apply Hamiltonian to position eigenstates

Let $\hat{H}$ be the free Hamilton operator, is it then true that $$\langle {\bf r}| \hat{H} ~=~ - \frac{\hbar^2}{2m} \Delta \langle {\bf r}|~?$$ Where $\Delta\equiv \nabla^2$. I currently don't see ...
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Intuitive meaning of the exponential form of an unitary operator in Quantum Mechanics

I'm an undergraduate student in Chemistry currently studying quantum mechanics and I have a problem with unitary transformations. Here in my book, it is stated that Every unitary operator ...
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1answer
76 views

Prove that this operator is unitary

$\hat{O}\equiv(1/\sqrt{2\pi})\int e^{-iNz}dz$ $\hat{O}^\dagger\equiv(1/\sqrt{2\pi})\int e^{iN'x}dx$ We have the operator $\hat{O}$ and its Hermitian adjoint $\hat{O}^\dagger$, in the one dimensional ...
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Can someone clarify what should and should not be an operator in my verification of the 1D solution to the SE for a free particle?

I just worked out the 1D free particle solution to the Schrödinger equation. My wave function was \begin{equation} \psi(x,t) = Ae^{i(px-Et)/\hbar} \end{equation} So I plugged this into both sides ...
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Simultaneous eigenket

J. J. Sakurai states in his "Modern Quantum Mechanics", this fact as a theorem ($\pi$ is the parity operator): Suppose $$[H,\pi]=0$$ and $| n>$ is a nondegenerate eigenket of $H$ with ...
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Help understanding proof in simultaneous diagonalization

The proof is from Principles of Quantum Mechanics by Shankar. The theorem is: If $\Omega$ and $\Lambda$ are two commuting Hermitian operators, there exists (at least) a basis of common eigenvectors ...
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Unitary operator algebra and multiplying by identity

If $\hat{H}$ is Hermitian, with eigenvalues $a_k$, then $$\hat{H} = \sum_k a_k \left|\psi_k\right> \left<\psi_k\right|.$$ I read that it then follows that $$\begin{align*} e^{i\hat{H}} = ...
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Probability of getting a particular spin

I'm a beginner in quantum mechanics, and I'm a bit confused about states and the probability to measure certain values. I would like to understand at least the following simplified situation: ...
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Question about equation 2.27 from Pachos's Introduction to topological quantum computing

http://quince.leeds.ac.uk/~phyjkp/Files/IntroTQC.pdf above is the PDF that is hosted on his website. The equation is on page 22 (pg 30 in the pdf). In chapter 2. It is the second equation of the ...
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Fundamental Commutation Relations in Quantum Mechanics

I am trying to compile a list of fundamental commutation relations involving position, linear momentum, total angular momentum, orbital angular momentum, and spin angular momentum. Here is what I have ...
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1answer
48 views

Commutator between square position and square momentum [duplicate]

I need (as a part of one exercise) to find commutator between $\hat{x}^2$ and $\hat{p}^2$ and my derivation goes as follows: $$[\hat{x}^2,\hat{p}^2]\psi = [\hat{x}^2\hat{p}^2 - ...
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1answer
73 views

What is the difference between a parameter, a variable, and an operator in QM?

On the question why time isn't an operator, people will usually say that time is a parameter in QM (Time as a Hermitian operator in QM?) and not a variable. Can someone please distinguish between a ...
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How to tackle 'dot' product for spin matrices

I read a textbook today on quantum mechanics regarding the Pauli spin matrices for two particles, it gives the Hamiltonian as $$ H = \alpha[\sigma_z^1 + \sigma_z^2] + ...
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Computing $\langle0|T[Q(t_2)Q(t_1)]|0\rangle$

Given Hamiltonian $H=\frac{P^2}{2}+\frac{\omega^2}{2}Q^2$, compute $\langle0|T[Q(t_2)Q(t_1)]|0\rangle$, where $T$ is the time-ordering of the product, $|0\rangle$ is the ground state. Now set ...
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Why are the spin operators defined as they are?

$$\begin{align*}S_z &= \frac{\hbar}{2} \left(\left|+\right>\left<+\right| - \left|-\right>\left<-\right|\right)\\ S_y &= i\frac{\hbar}{2} \left(\left|-\right>\left<+\right| - ...
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Problem with momentum operator

Why is there no problem with the eigenfunction of the momentum operator being non-normalisable? How can it be a valid quantum state?
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RH side of the Uncertainty principle: when is it a number and when an expectation value?

The uncertainty principle between the position $x$ and the momentum $p$ is given by: $$ \sigma_x \sigma_p \geq \hbar/2,$$ whereas for the $x$ and $y$ components of the angular momentum is given by: ...
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How does an operator transform under time reversal?

We know that a time-reversal operator $T$ can be represented as $$T=UK$$ where $U$ is some unitary operator and $K$ is the complex conjugation operator. Then under time-reversal operation, a quantum ...
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Apparent spacetime dependence of creation and annihilation operators

I'm currently going through An Introduction to Quantum Field Theory by Hartmut Wittig I've stumbled upon. Having trouble with equation (2.29), I'm asking the question: Do creation and annihilation ...
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1answer
106 views

Is the Hilbert space spanned by both bound and continuous hydrogen atom eigenfunctions?

As e.g. Griffiths says (p. 103, Introduction to Quantum Mechanics, 2nd ed.), if a spectrum of a linear operator is continuous, the eigenfunctions are not normalizable, therefore it has no ...
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1answer
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Product on Tensor Products

I'm trying to understand how products on tensor products work. For instance, in quantum mechanics, you have ($x$ tensor $y$) times ($z$ tensor $a$), where $x$, $y$, $z$, $a$ are all operators acting ...
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Why we cannot describe operator for force $F$ in quantum mechanics?

In quantum mechanics we describe operators corresponding to momentum but we don't define operator for force what is the reason behind it?
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How much information does the Hamiltonian contain in quantum mechanics? [closed]

Given a Hamiltonian, let's say of a many-body system, through the Schrodinger equation,in principle we can find the eigenfunctions and their corresponding eigenvalues (spectrum). Now given an ...
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How are anyons possible?

If $|ψ\rangle$ is the state of a system of two indistinguishable particles, then we have an exchange operator $P$ which switches the states of the two particles. Since the two particles are ...
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How to find creation and annihilation operators? [duplicate]

I get confused when trying to find this. Please describe it as simply as possible, but keep in mind I have no budget whatsoever to pay for textbooks, so here goes: How do you find the creation and ...
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2answers
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Why are the charge operator $Q$ and the baryon number operator $B$ unbounded?

A friend recommended me to read PCT, Spin and Statistics, and All That written by R. F. Streater and A. S. Wightman. In page 5 to 6, here's what the authors of this book have to say: [...] In ...
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Shape of the state space under different tensor products

I am currently studying generalized probabilistic theories. Let me roughly recall how such a theory looks like (you can skip this and go to "My question" if you are familiar with this). Recall: In a ...
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Show that a function takes the following form using the definition for the function of an operator

If $f(z)$ is a function with a Taylor series expansion $$f(z)=\sum _{ n=0 }^{ \infty }{c_n z^n },$$ then we define $$f(M)=\sum _{ n=0 }^{ \infty }{c_n M^n }.$$ First consider ...
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The delta function as an eigenfunction of the position operator explanation

$\delta (\textbf{r})$ can be interpreted as a wavefunction. [...] It is non-vanishing only for $\textbf{r}=0$. [...] $\delta(\textbf{r})$ is an eigenfunction of the position operator with ...
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Kraus operator rank

All quantum operations $\mathcal{E}$ on a system of Hilbert space dimension $\mathcal{d}$ can be generated by an operator-sum representation containing at most $\mathcal{d^2}$ elements. Extending ...
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223 views

Ladder Operator killing state

So in my last question, @joshphysics showed me how to prove $K_\pm$ were ladder operators. Now I need to show that there is a lowest state, i.e $$\langle m_0|K_+=K_-|m_0\rangle=0$$ I am not ...
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2answers
61 views

When generalizing from discrete (but infinite) eigenstates to continuous eigenstates, Why do we change the definition?

The propagator function for discrete eigenstates is $$u(t)=\sum_{n=1}^{\infty}|E_n\rangle\langle E_n|e^{-iE_nt/ \hbar } \tag{1}\ .$$ But when we have continuous eigenstates, (like for the case of ...