Physics, asked by Alexxender, 10 months ago

is the hermitian of ladder operator possible??if yes then
how it will operate.Ans plzzzzzzzzzzzzzzzzzzzzzzzz

Answers

Answered by diksha254134
0

Explanation:

Operators in quantum mechanics aren’t merely a convenient way to keep track of

eigenvalues (measurement outcomes) and eigenvectors (definite-value states). We

can also use them to streamline calculations, stripping away unneeded calculus (or

explicit matrix manipulations) and focusing on the essential algebra.

As an example, let’s now go back to the one-dimensional simple harmonic oscilla-

tor, and use operator algebra to find the energy levels and associated eigenfunctions.

Recall that the harmonic oscillator Hamiltonian is

H =

1

2m

p

2 +

1

2

mω2

cx

2

, (1)

where p is the momentum operator, p = −i¯hd/dx (in this lesson I’ll use the symbol

p only for the operator, never for a momentum value). As in Lesson 8, it’s easiest

to use natural units in which we set the particle mass m, the classical oscillation

frequency ωc, and ¯h all equal to 1. Then the Hamiltonian is simply

H =

1

2

p

2 +

1

2

x

2

, (2)

with p = −id/dx.

The basic trick, which I have no idea how to motivate, is to define two new

operators that are linear combinations of x and p:

a− =

1

2

(x + ip), a+ =

1

2

(x − ip). (3)

These are called the lowering and raising operators, respectively, for reasons that

will soon become apparent. Unlike x and p and all the other operators we’ve worked

with so far, the lowering and raising operators are not Hermitian and do not repre-

sent any observable quantities. We’re not especially concerned with their eigenvalues

or eigenfunctions; instead we’ll focus on how they mathematically convert one en-

ergy eigenfunction into another. (Note: Most authors use the notation a and a

instead of a− and a+, but Griffiths uses the −/+ notation, and I too think it’s

more appropriate. In conventional units, by the way, we would insert factors of m,

ωc, and ¯h into these definitions as needed to make a− and a+ dimensionless. It’s a

good exercise to figure out exactly how to do this, especially if you’re even a little

uncomfortable with my use of natural units.)

The first thing to notice about a− and a+ is that if you multiply them together,

you almost get the Hamiltonian

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