Linear Operators: Identifying and Solving

[Axiom17]

Homework Statement

I have some operators, and need to figure out which ones are Linear (or not).

For example:

1. $$\hat{A} \psi(x) \equiv \psi(x+1)$$

Homework Equations

I have defined the Linear Operator:

$$\hat{A}[p\psi_{1}+q\psi_{2}]=p\hat{A}\psi_{1}+q\hat{A}\psi_{2}$$

The Attempt at a Solution

I just don't know where to start with this

 

[arkajad]

Homework Statement

I have some operators, and need to figure out which ones are Linear (or not).

For example:

1. $$\hat{A} \psi(x) \equiv \psi(x+1)$$

Homework Equations

I have defined the Linear Operator:

$$\hat{A}[p\psi_{1}+q\psi_{2}]=p\hat{A}\psi_{1}+q\hat{A}\psi_{2}$$

The Attempt at a Solution

I just don't know where to start with this

Your confusion comes from popular but dangerous notation. It should read:

$$(A\psi )(x)=\psi (x+1)$$

which you read: the effect of acting with A on $$psi$$ is a new function, denoted $$A\psi$$ whose value at every point x is equal to the value of the original function at the point x displaced by 1.

Now you look at your

$$\hat{A}[p\psi_{1}+q\psi_{2}]$$

Here you have A acting on the sum of functions. This sum of functions is a function. So from the definition of A:

$$(\hat{A}[p\psi_{1}+q\psi_{2}])(x)=(p\psi_{1}+q\psi_{2})(x+1)$$

Now, how are sums of functions defined? Isn't it true that they are defined like:

$$(f+g)(x)=f(x)+g(x)$$

$$(cf)(x)=cf(x)$$.

Paying attention to such details you will get your answer.

 

[Axiom17]

I understand a bit more but I still don't quite get it. This is what I have so far:

$$\left(A\left[P\psi_{1}+Q\psi_{2}\right]\right)(x) = \left(AP\psi_{1}+AQ\psi_{2}\right)(x) = AP\psi_{1}(x)+AQ\psi_{2}(x)$$

.. but then I don't see how to relate that to $\psi(x+1)$

 

[arkajad]

I understand a bit more but I still don't quite get it. This is what I have so far:

$$\left(A\left[P\psi_{1}+Q\psi_{2}\right]\right)(x) = \left(AP\psi_{1}+AQ\psi_{2}\right)(x) = AP\psi_{1}(x)+AQ\psi_{2}(x)$$

.. but then I don't see how to relate that to $\psi(x+1)$

You have missed an important step. You want to calculate

$$\left(A\left[P\psi_{1}+Q\psi_{2}\right]\right)(x)$$

Here A acts on a certain function(which is a linear combination of two other functions) and you want to evaluate the result at x. For this you use the definition of A, this is definition that you wrote at the very beginning. Whatever is the function A acts on, it shifts its argument by 1. Here A acts on what is in the [] bracket. It will shifts its argument by 1. Thus:

$$\left(A\left[P\psi_{1}+Q\psi_{2}\right]\right)(x)=(P\psi_{1}+Q\psi_{2})(x+1)$$

You used just the definition of A, what A does to whatever it acts on. Now you continue using the definition of the sum of two functions and the product of a number and a function.

So, continue, and see where you will end. But first, let me know if you really-really understand what I wrote above.

 

[Axiom17]

I think I understand what you've wrote. I see that basically have:

$$(A\left[function])(x)=(function)(x+1)$$

which is in this case:

$$(A\left[P\psi_{1}+Q\psi_{2}\right])(x)=(P\psi_{1}+Q\psi_{2})(x+1)$$

Then using the relation $(f+g)(x)=f(x)+g(x)$ the RHS of the equation becomes:

$$(P\psi_{1})(x+1)+(Q\psi_{2})(x+1)$$

(which hopefully is correct).

Then using the other relation [itex](cf)(x)=cf(x)[/tex] this becomes:<br /> <br /> $$P\psi_{1}(x+1)+Q\psi_{2}(x+1)$$<br /> <br /> .. but that's what I had 2 steps before! <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f641.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":frown:" title="Frown :frown:" data-smilie="3"data-shortname=":frown:" />[/itex]

 

[arkajad]

Now, look at your right hand side:

$$\hat{A}[p\psi_{1}+q\psi_{2}]=p\hat{A}\psi_{1}+q\hat{A}\psi_{2}$$

Which you want to calculate at x

$$(p\hat{A}\psi_{1}+q\hat{A}\psi_{2})(x)$$

First you notice that you have here a sum of two functions evaluated at x. Therefore the first step is to us general formula (f+g)(x)=f(x)+g(x). That is

$$(p\hat{A}\psi_{1}+q\hat{A}\psi_{2})(x)=(p\hat{A}\psi_{1})(x)+(q\hat{A}\psi_{2})(x).$$

Then you need to use (cf)(x)=cf(x) and at the end the definition of A (in two terms). See if you get the same as from the LHS (left hand side).

All this exercise is nothing but careful application of definitions! It's pretty formal. Nothing really deep. But it teaches you to be careful with putting and moving parenthesis right.

BTW: P and p, as well as Q and q should be the same. You better use p,q everywhere.

 

[Axiom17]

I've got a bit confused now but perhaps I'm not too far away

So this is the operator:

$$\hat{A}\psi(x)=(A\psi)(x)=\psi(x+1)$$

This is the definition of a linear operator:

$$\hat{A}[p\psi_{1}+q\psi_{2}]=p\hat{A}\psi_{1}+q\hat{A}\psi_{2}$$

Thus obtain this:

$$\left(A\left[p\psi_{1}+q\psi_{2}\right]\right)(x) = \left(Ap\psi_{1}+Aq\psi_{2}\right)(x) = Ap\psi_{1}(x)+Aq\psi_{2}(x)$$

Then this step:

$$(p\hat{A}\psi_{1}+q\hat{A}\psi_{2})(x)=(p\hat{A}\psi_{1})(x)+(q\hat{A}\psi_{2})(x)$$

Now using $(cf)(x)=cf(x)$ can change the RHS of the above to:

$$p\hat{A}\psi_{1}(x)+q\hat{A}\psi_{2}(x)$$

Now using $(f+g)(x)=f(x)+g(x)$ can change the above (the RHS) to:

$$(p\hat{A}\psi_{1}+q\hat{A}\psi_{2})(x)$$

.. which gives the LHS!

SO that means that the operator $\hat{A} \psi(x) \equiv \psi(x+1)$ is a linear operator.. right?

*note1: not sure what should be the correct notation as far as to use $\hat{A}$ or A throughout.*

*note2: I see I've changed around the order of terms, does this matter? I.e. is Ap the same as pA?*

 

[arkajad]

All right. See how it should look like:

Definition of A: for any $$\psi$$ and any x:
$$(A\psi)(x)=\psi(x+1)$$

Theorem: A is linear

Proof: What it means A is linear? It means I need to prove that for any two functions $$\psi_1,\psi_2$$ and any numbers p,q we have
$$A(p\psi_1+q\psi_2)=pA\psi_1+qA\psi_2.$$
The above equations is eqaulity of two functions. To check that two functions are equal I need to check that there values at every point are equal. So I calculate the LHS:

$$(A(p\psi_1+q\psi_2))(x)=(p\psi_1+q\psi_2)(x+1)$$

(from the definition of A).

$$(p\psi_1+q\psi_2)(x+1)=p\psi(x+1)+q\psi_2(x+1)$$

(from the definition of sum and products of functions.

Now I calculate the RHS at x:

$$(pA\psi_1+qA\psi_2)(x)$$

First I use the definition of sum and products
$$(pA\psi_1+qA\psi_2)(x)=p(A\psi_1)(x)+q(A\psi_2)(x)$$
And now I use definition of A in each of these terms:
$$p(A\psi_1)(x)+q(A\psi_2)(x)=p\psi_1(x+1)+q\psi_2(x+1)$$

I see that LHS=RHS.

Calculating LHS I first used definition of A (once) and then sum and products of functions.
Calculating RHS I first used sum and product definitions, and the definition of A (twice).

Think of it. Contemplate until you get it. You will use similar reasonings many times in the future.