# Eigenvector of raising operator

• Pouyan
In summary: Psi|\Psi\rangle=1##.In summary, in order to show that the raising operator at has no right eigenvectors, we must first define a vector |Ψ> = ∑cn|n> (for n=0 to ∞) and use the given equation at|n> = √(n+1)|n+1> to find the relationship between the coefficients cn. By examining the eigenvector equation a^t|Ψ> = λ|Ψ>, it can be shown that the only possible solution for all n is cn = 0, indicating that there are no right eigenvectors for the raising operator at.

## Homework Statement

show that the raising operator at has no right eigenvectors

## Homework Equations

We know at|n> = √(n+1)|n+1>

## The Attempt at a Solution

we define a vector |Ψ> = ∑cn|n> (for n=0 to ∞)

at|Ψ>=at∑cn|n>=∑cn(√n+1)|n+1>

But further I give up!

Pouyan said:

## Homework Statement

show that the raising operator at has no right eigenvectors

## Homework Equations

We know at|n> = √(n+1)|n+1>

## The Attempt at a Solution

we define a vector |Ψ> = ∑cn|n> (for n=0 to ∞)

at|Ψ>=at∑cn|n>=∑cn(√n+1)|n+1>

But further I give up!

You have to try. It's not that hard.

Pouyan
PeroK said:
You have to try. It's not that hard.
Well I do see this :

at|Ψ>=at∑cn|n>=∑cn(√n+1)|n+1>
In a solution i do see:

Latex code : \sum_{n=0}^{\infty}c_{n}|n>=\sum_{n=1}^{\infty}c_{n-1}\sqrt{n}|n>
I know how to iterate a sum serie but why shall we say so ?!

and the only possible value is c_n=0

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Pouyan said:
Well I do see this :

at|Ψ>=at∑cn|n>=∑cn(√n+1)|n+1>
In a solution i do see:

View attachment 230995

Latex code : \sum_{n=0}^{\infty}c_{n}|n>=\sum_{n=1}^{\infty}c_{n-1}\sqrt{n}|n>
I know how to iterate a sum serie but why shall we say so ?!

and the only possible value is c_n=0

Two things.

1) You should be writing down what it means for ##a^t## to have an eigenvector:

##a^t|\Psi \rangle = \lambda |\Psi \rangle##

2) Sometimes you can see things better using the ##\sum## notation. And sometimes you can see things better with the terms written out. If you are stuck, you should always try both ways. In this case, for example, let:

##|\Psi \rangle = c_0 |0 \rangle + c_1 |1 \rangle + c_2 |2 \rangle \dots ##

You are right that you can show that ##c_n = 0##. Can you show this fully? Also, what about the case where ##\lambda = 0## as a possible eigenvalue?

Pouyan
PeroK said:
Two things.

1) You should be writing down what it means for ##a^t## to have an eigenvector:

##a^t|\Psi \rangle = \lambda |\Psi \rangle##

2) Sometimes you can see things better using the ##\sum## notation. And sometimes you can see things better with the terms written out. If you are stuck, you should always try both ways. In this case, for example, let:

##|\Psi \rangle = c_0 |0 \rangle + c_1 |1 \rangle + c_2 |2 \rangle \dots ##

You are right that you can show that ##c_n = 0##. Can you show this fully? Also, what about the case where ##\lambda = 0## as a possible eigenvalue?
this is my problem, I don't get it
How can I show

is right if and only if c_n = 0...?
I know the only solution for all n must be c_n=0

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Pouyan said:
this is my problem, I don't get it
How can I show View attachment 231005

is right if and only if c_n = 0...?
I know the only solution for all n must be c_n=0

You haven't followed either of my suggestions. You must at least write down what it means to have an eigenvector. Maths doesn't solve itself. You need to start working with the equations.

Anyway, the rules of PF are clear: you must start to show some of your work.

Pouyan said:
In a solution i do see:
$$\sum_{n=0}^{\infty}c_{n}|n\rangle=\sum_{n=1}^{\infty}c_{n-1}\sqrt{n}|n\rangle$$
You might be getting confused because that's wrong. What it should say is
$$\hat a^\dagger \sum_{n=0}^{\infty}c_{n}|n\rangle=\sum_{n=0}^{\infty}c_{n}\sqrt{n+1}|n+1\rangle = \sum_{n=1}^{\infty}c_{n-1}\sqrt{n}|n\rangle.$$ From this alone, you can't conclude anything about the coefficients. You need to use the eigenvector equation @PeroK mentioned above.

PeroK
The equations given by @PeroK and @vela imply a certain recurrance relationship on your coefficients, i.e.

if ##c_1=1## then ##c_2=\dots## then etc..

Try to write it out, and see what it implies, bearing in mind that one usually wants eigenvectors that can be normalized

## What is an eigenvector of a raising operator?

An eigenvector of a raising operator is a vector that, when operated on by the raising operator, results in a scalar multiple of itself. In other words, it is a special vector that is unchanged up to a constant factor when acted upon by the raising operator.

## What is the significance of the eigenvectors of a raising operator?

The eigenvectors of a raising operator are important because they correspond to the states of a physical system that have well-defined values for the observable quantity associated with the raising operator. These states are also known as eigenstates.

## How are eigenvectors of a raising operator related to eigenvalues?

The eigenvalues of a raising operator correspond to the possible values that can be measured for the observable quantity associated with the operator. Each eigenvector has a corresponding eigenvalue, and the eigenvalue represents the amount by which the eigenvector is scaled when operated on by the raising operator.

## Are the eigenvectors of a raising operator unique?

No, the eigenvectors of a raising operator are not unique. In fact, there are usually an infinite number of eigenvectors for a given raising operator, each with a different eigenvalue. However, the set of eigenvectors for a particular operator is always orthogonal, meaning they are perpendicular to each other.

## Can any vector be an eigenvector of a raising operator?

No, not all vectors can be eigenvectors of a raising operator. The vector must satisfy certain conditions, such as being in the vector space of the operator and being linearly independent from other eigenvectors. In addition, the vector must also satisfy the eigenvalue equation for the operator.