What is Ladder operator: Definition and 23 Discussions

In linear algebra (and its application to quantum mechanics), a raising or lowering operator (collectively known as ladder operators) is an operator that increases or decreases the eigenvalue of another operator. In quantum mechanics, the raising operator is sometimes called the creation operator, and the lowering operator the annihilation operator. Well-known applications of ladder operators in quantum mechanics are in the formalisms of the quantum harmonic oscillator and angular momentum.

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  1. waadles

    I Displacement operation acting on individual quadrature components

    Hi all, I have a naive understanding of how operators work and wondered if someone could help me. I have tried to understand this myself, but alas, I think my knowledge is too premature to understand what I am reading online. Is someone able to explain? I want to perform the operation...
  2. Sciencemaster

    I Finding ##\partial^\mu\phi## for a squeezed state in QFT

    I'm trying to apply an operator to a massless and minimally coupled squeezed state. I have defined my state as $$\phi=\sum_k\left(a_kf_k+a^\dagger_kf^*_k\right)$$, where the ak operators are ladder operators and fk is the mode function $$f_k=\frac{1}{\sqrt{2L^3\omega}}e^{ik_\mu x^\mu}$$...
  3. binbagsss

    Exponential of ladder operator

    I'm just trying to follow the below And I understand all, I think, except what's happened to the term when A hits 1: [A,1] ? If I'm correct basically we're just hitting on the first operator so reducing the power by one each time of the operator in the right hand bracket thanks
  4. Garlic

    Landau levels: Hamiltonian with ladder operators

    Dear PF, I hope I've formulated my question understandable enough. Thank you for your time, Garli
  5. M

    A rather weird form of a coherent state

    As far as I know we can express the position and momentum operators in terms of ladder operators in the following way $${\begin{aligned}{ {x}}&={\sqrt {{\frac {\hbar }{2}}{\frac {1}{m\omega }}}}(a^{\dagger }+a)\\{{p}}&=i{\sqrt {{\frac {\hbar }{2}}m\omega }}(a^{\dagger }-a)~.\end{aligned}}.$$...
  6. M

    I Raising the ladder operators to a power

    Hi! I am working on homework and came across this problem: <n|X5|n> I know X = ((ħ/(2mω))1/2 (a + a+)) And if I raise X to the 5th, its becomes X5 = ((ħ/(2mω))5/2 (a + a+)5) What I'm wondering is, is there anyway to be able to solve this without going through all of the iterations the...
  7. P

    Normalization of the Angular Momentum Ladder Operator

    Homework Statement Obtain the matrix representation of the ladder operators ##J_{\pm}##. Homework Equations Remark that ##J_{\pm} | jm \rangle = N_{\pm}| jm \pm 1 \rangle## The Attempt at a Solution [/B] The textbook states ##|N_{\pm}|^2=\langle jm | J_{\pm}^\dagger J_{\pm} | jm \rangle##...
  8. Mayan Fung

    B Quantum Oscillator States: Is the Ladder Operator Enough?

    We learned that we can use the ladder operator to obtain the states of a quantum oscillator. However, I see no direct evidence to show that the solutions are complete. I mean, how can we know the energy state follows E is (E+hw). Why can't we have some more states in between? Does the derivation...
  9. J

    I QM - Ladder Operator QHO - factorization

    Hi, quick question with A being the lowering operator and A† the raising operator for a QHO (A A† - 1 + 1/2) ħω [Aψ] = A (A† A - 1 + 1/2) ħω ψ By taking out a factor of A. Why has the ordering of A A† swapped around? I would have thought taking out a factor of A would leave it as A (A† - 1 +...
  10. AwesomeTrains

    Ladder operator commutator with arbitary function

    Hey there! 1. Homework Statement I've been given the operators a=\sqrt\frac{mw}{2\hbar}x+i\frac{p}{\sqrt{2m\hbar w}} and a^\dagger=\sqrt\frac{mw}{2\hbar}x-i\frac{p}{\sqrt{2m\hbar w}} without the constants and definition of the momentum operator: a=x+\partial_x and a^\dagger=x-\partial_x with...
  11. S

    I Angular momentum ladder operator derivation

    In the Griffiths textbook for Quantum Mechanics, It just gives the ladder operator to be L±≡Lx±iLy With reference to it being similar to QHO ladder operator. The book shows how that ladder operator is obtained, but it doesn't show how angular momentum operator is derived. Ive searched the...
  12. Raptor112

    Matrix Representation for Combined Ladder Operators

    Due to the definition of spin-up (in my project ), \begin{eqnarray} \sigma_+ = \begin{bmatrix} 0 & 2 \\ 0 & 0 \\ \end{bmatrix} \end{eqnarray} as opposed to \begin{eqnarray} \sigma_+ = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} \end{eqnarray} and the annihilation operator is...
  13. S

    Proof of ladder operator identity

    Homework Statement Define n=(x + iy)/(2)½L and ñ=(x - iy)/(2)½L. Also, ∂n = L(∂x - i ∂y)/(2)½ and ∂ñ = L(∂x + i ∂y)/(2)½. with ∂n=∂/∂n, ∂x=∂/∂x, ∂y=∂/∂y, and L being the magnetic length. Show that a=(1/2)ñ+∂n and a†=(1/2)n -∂ñ a and a† are the lowering and raising operators of quantum...
  14. I

    Quantum Mechanics - Lowering Operator

    Homework Statement let A be a lowering operator. Homework Equations Show that A is a derivative respects to raising operator, A†, A=d/dA† The Attempt at a Solution I start by defining a function in term of A†, which is f(A†) and solve it using [A , f(A†)] but i get stuck after that. Can...
  15. K

    Quantum States and ladder operator

    In any textbooks I have seen, vacuum states are defined as: a |0>= 0 What is the difference between |0> and 0? Again, what happens when a+ act on |0> and 0? and Number Operator a+a act on |0> and 0?
  16. fluidistic

    Ladder operator for harmonic oscillator, I don't get a mathematical

    If the ladder operator ##a=\sqrt {\frac{m\omega}{2\hbar}}x+\frac{ip}{\sqrt{2m\hbar \omega}}## and ##a^\dagger=\sqrt {\frac{m\omega}{2\hbar}}x-\frac{ip}{\sqrt{2m\hbar \omega}}## then I get that the number operator N, defined as ##a^\dagger a## is worth ##\frac{m \omega...
  17. B

    Quantum Harmonic Oscillator ladder operator

    Homework Statement What is the effect of the sequence of ladder operators acting on the ground eigenfunction \psi_0 Homework Equations \hat{A}^\dagger\hat{A}\hat{A}\hat{A}^\dagger\psi_0The Attempt at a Solution I'm not sure if I'm right but wouldn't this sequence of opperators on the ground...
  18. H

    How to show only multiples of unit energy are allowed in QHO by ladder operator

    I'd like to know how to do it without solving via Hermite polynomials, so that I can check by both methods when I solve other problems. I have tried to figure out myself but I need some help. Let's say H:=x^2+p^2 , and a:= x-ip . So that [a,a^+]=2\hbar , H a \varphi_i = (E_i -...
  19. U

    Does the existence of a ladder operator imply that the eigenvalues are discrete?

    Hi! I don't know much about QM. I'm reading lecture notes at the moment. Angular momentum is discussed. The ladder operators for the angular-momentum z-component are defined, it is shown that <L_z>^2 <= <L^2>, so the z component of angular momentum is bounded by the absolute value of angular...
  20. K

    Ladder Operator for Harmonic Oscillator: a|0> = |0>

    For harmonic oscillator, let |0> be the ground state, so which statement is correct? a|0> =|0> or a|0> = 0 (number) where a is the destroy operator
  21. B

    Quantum Mechanics Ladder Operator and Dirac Notation

    Homework Statement I'm given the eigenvalue equations L^{2}|\ell,m> = h^2\ell(\ell + 1)|\ell,m> L_z|\ell,m> = m|\ell L_{\stackrel{+}{-}}|\ell,m> = h\sqrt{(\ell \stackrel{-}{+} m)(\ell \stackrel{+}{-} m + 1)}|\ell, m \stackrel{+}{-} 1> Compute <L_{x}>. Homework Equations Know...
  22. B

    Quantum SHO Ladder Operator in Mathcad

    A Quantum I problem set asks me to graph the first 15 states of the simple harmonic oscillator. Our department uses mathcad heavily, so I think I should write a function that applies the ladder operator repeatedly to generate the wave function. I'm having trouble getting it to actually return a...
  23. P

    Ladder Operator Theorem: Uniting Operators Through Unitary Transformations

    Is there a theorem that says if [a, a^\dagger] = [b, b^\dagger] = 1 then there is a unitary operator U such that b = UaU^\dagger ?
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