Linear operators, quantum mechanics

[cookiemnstr510510]

Homework Statement

Show that two of the operators are linear and one is not

Relevant Equations

A(f+g)=Af+Ag

Hello,
I am struggling with what each piece of these equations are. I generally know the two rules that need to hold for an operator to be linear, but I am struggling with what each piece of each equation is/means.
Lets look at one of the three operators in question.
A(f(x))=(∂f/∂x)+3f(x)
I need to get an understanding of what each piece of the above equation says. Correct me when I am wrong.
A is the operator, but this also says that f(x) is the operator? does the partial derivative represent a partial derivative of our operator? If that is true our operator A is now equal to the partial derivative of our operator + 3 times our operator?

To show linearity we need to show A(f(x)+g(x)) is true and the scalar property, but let's first stick with this first one.

So I am already confused because we have an f(x) in the problem and our linearity rules have another f(x).

Here is my attempt at a solution, but it is practically worthless because I do not get what each piece means:

A(f(x)+g(x))=[(∂f/∂x)(f(x)+g(x))]+[3f(x)(f(x)+g(x))]
"= (∂f/∂x)*f(x) + (∂f/∂x)*g(x) + 3f(x)*f(x)+ 3f(x)*g(x)
"=A*f(x) + A*g(x) Not sure if this is right, not sure what A represents...
Any help would be appreciated, I think I am close, but I am not able to make sense of each variable.
I have searched far and wide online, but can't make sense of the info out there, too many elementary steps are skipped.
Any help would be appreciated!

Thank you.

 

[nrqed]

Homework Statement:: Show that two of the operators are linear and one is not
Homework Equations:: A(f+g)=Af+Ag

Hello,
I am struggling with what each piece of these equations are. I generally know the two rules that need to hold for an operator to be linear, but I am struggling with what each piece of each equation is/means.
Lets look at one of the three operators in question.
A(f(x))=(∂f/∂x)+3f(x)
I need to get an understanding of what each piece of the above equation says. Correct me when I am wrong.
A is the operator, but this also says that f(x) is the operator? does the partial derivative represent a partial derivative of our operator? If that is true our operator A is now equal to the partial derivative of our operator + 3 times our operator?

To show linearity we need to show A(f(x)+g(x)) is true and the scalar property, but let's first stick with this first one.

So I am already confused because we have an f(x) in the problem and our linearity rules have another f(x).

Here is my attempt at a solution, but it is practically worthless because I do not get what each piece means:

A(f(x)+g(x))=[(∂f/∂x)(f(x)+g(x))]+[3f(x)(f(x)+g(x))]

A is the operator that acts on a function. The result of applying the operator A on a function is to take the derivative of the function and add to the result three times the function.
Now, what you wrote as the action on f(x) + g(x) is incorrect. It should be

A(f(x)+g(x))=(∂f/∂x+∂g/∂x)+3(f(x)+g(x))

 

[cookiemnstr510510]

A is the operator that acts on a function. The result of applying the operator A on a function is to take the derivative of the function and add to the result three times the function.
Now, what you wrote as the action on f(x) + g(x) is incorrect. It should be

A(f(x)+g(x))=(∂f/∂x+∂g/∂x)+3(f(x)+g(x))

Do you mean the result of A on a function is to take the partial derivative of the function?

 

[cookiemnstr510510]

A is the operator that acts on a function. The result of applying the operator A on a function is to take the derivative of the function and add to the result three times the function.
Now, what you wrote as the action on f(x) + g(x) is incorrect. It should be

A(f(x)+g(x))=(∂f/∂x+∂g/∂x)+3(f(x)+g(x))

I think I get it, so our initial equation: A(f(x))=(∂f/∂x)+3f(x) is defining what the operator A does. And what it is telling us is that A, the operator, takes a function and does two things to it:
1. it takes the partial derivative of the function, and
2. adds to it 3 times the initial function.
If that is it that makes sense to me.

 

[cookiemnstr510510]

Okay so here is my new attempt at the whole problem:
Complete question attached.
The two properties I need to fulfill are:
A(f(x)+g(x))=A(f(x))+A(g(x))
A(cf(x))=cA(f(x))

first example:
A(f(x))=(∂f/∂x)+3f(x)
A(f(x)+g(x))=(∂f/∂x)+(∂g/∂x)+3f(x)+3g(x)
A(f(x))+A(g(x))=(∂f/∂x)+3f(x)+(∂g/∂x)+3g(x) to me the RHS of this equation looks like LHS (just in different form)

A(f(x))=(∂f/∂x)+3f(x)
A(cf(x))=c*(∂f/∂x)+c*3f(x)
cA(f(x))=c[(∂f/∂x)+3f(x)] LHS=RHS, Linear

Next example:
B(f(x))=(1/2)f(x)(∂f/∂x)
B(f(x)+g(x))=(1/2)f(x)(∂f/∂x)+(1/2)g(x)(∂g/∂x)
B(f(x))+B(g(x))=(1/2)f(x)(∂f/∂x)+(1/2)g(x)(∂g/∂x) LHS=RHS, Linear

B(f(x))=(1/2)f(x)(∂f/∂x)
B(cf(x))=c(1/2)f(x)(∂f/∂x)
c*B(f(x))=c[(1/2)f(x)(∂f/∂x)] LHS=RHS, Linear

Last example attached.
I immediately see a problem with it. What I say this operator saying is this: The operator C takes a function f(x) in and outputs the definite integral from 0-->5 ∫(x-y)^3 f(y)dy. I don't understand how we could even set up the properties to see if linear? what is the f(y) doing in the integral, what is f(y) and why are integrating W.R.T y?
I am assuming this one is not linear, but could someone show me why? I don't see how I could go about showing this.
Thanks!

 

[nrqed]

I think I get it, so our initial equation: A(f(x))=(∂f/∂x)+3f(x) is defining what the operator A does. And what it is telling us is that A, the operator, takes a function and does two things to it:
1. it takes the partial derivative of the function, and
2. adds to it 3 times the initial function.
If that is it that makes sense to me.

Yes, exactly.

Note that I skipped a step when I wrote the effect of A on g+f. To be really detailed, one shoudl write

A[f(x)+g(x)]=(∂(f(x)+g(x))/∂x)+3(f(x)+g(x))

Using the fact that taking the derivative of a sum is the sum of the derivatives, one gets what I wrote earlier:

∂f(x)/∂x+∂g(x)/∂x +3(f(x)+g(x))

which is of course

∂f(x)/∂x+∂g(x)/∂x +3f(x)+3g(x)

Which is A(f(x))+ A(g(x))

 

[cookiemnstr510510]

Yes, exactly.

Note that I skipped a step when I wrote the effect of A on g+f. To be really detailed, one shoudl write

A[f(x)+g(x)]=(∂(f(x)+g(x))/∂x)+3(f(x)+g(x))

Using the fact that taking the derivative of a sum is the sum of the derivatives, one gets what I wrote earlier:

∂f(x)/∂x+∂g(x)/∂x +3(f(x)+g(x))

which is of course

∂f(x)/∂x+∂g(x)/∂x +3f(x)+3g(x)

Which is A(f(x))+ A(g(x))

Oh wow, okay so I was wrong again! thank you.
So since our initial operator, A, takes a function and takes the partial derivative of it first. So for sum of two functions (f(x)+g(x)) we are taking the partial derivative of this entire quantity, not each individual part. so it would then look like your answer: (∂(f(x)+g(x))/∂x)

 

[nrqed]

Okay so here is my new attempt at the whole problem:
Complete question attached.
The two properties I need to fulfill are:
A(f(x)+g(x))=A(f(x))+A(g(x))
A(cf(x))=cA(f(x))

first example:
A(f(x))=(∂f/∂x)+3f(x)
A(f(x)+g(x))=(∂f/∂x)+(∂g/∂x)+3f(x)+3g(x)
A(f(x))+A(g(x))=(∂f/∂x)+3f(x)+(∂g/∂x)+3g(x) to me the RHS of this equation looks like LHS (just in different form)

A(f(x))=(∂f/∂x)+3f(x)
A(cf(x))=c*(∂f/∂x)+c*3f(x)
cA(f(x))=c[(∂f/∂x)+3f(x)] LHS=RHS, Linear

Your conclusion is correct. Although it is confusing when you write LHS=RHS, but I guess you mean the LHS and RHS of the equations near the top, in which case you are 100% correct.

Next example:
B(f(x))=(1/2)f(x)(∂f/∂x)
B(f(x)+g(x))=(1/2)f(x)(∂f/∂x)+(1/2)g(x)(∂g/∂x)

No, to get B(f(x)+g(x)), you must simply take the expression for B(f(x)) and replace every instance of f(x) by f(x)+g(x), which is not what you did.

 

[PeroK]

Oh wow, okay so I was wrong again! thank you.
So since our initial operator, A, takes a function and takes the partial derivative of it first. So for sum of two functions (f(x)+g(x)) we are taking the partial derivative of this entire quantity, not each individual part. so it would then look like your answer: (∂(f(x)+g(x))/∂x)

You have some serious misconceptions that you need to iron out. To take your example:

$A(f(x)) = f(x)\frac{df}{dx}$

Now, if you want to operate on $f(x) + g(x)$, a trick to see how to do this is first to let $h(x) = f(x) + g(x)$ and now:

$A(f(x) + g(x)) = A(h(x)) = h(x)\frac{dh}{dx} = (f(x) + g(x))\frac{d}{dx}(f(x) + g(x)$

In addition to clearing up why the operation works like it does, it also highlights the point that a function is a function and you cannot treat it any differently just because you write it as a sum. For example, we have:

$\cos (2x) = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$

If an operator acts on the function $\cos 2x$, it cannot operate differently on these other variations. You can't come out with a different answer for $A(\cos(2x))$ just because you write it as $A(\cos^2 x - \sin^2 x)$

Last example attached.
I immediately see a problem with it. What I say this operator saying is this: The operator C takes a function f(x) in and outputs the definite integral from 0-->5 ∫(x-y)^3 f(y)dy. I don't understand how we could even set up the properties to see if linear? what is the f(y) doing in the integral, what is f(y) and why are integrating W.R.T y?
I am assuming this one is not linear, but could someone show me why? I don't see how I could go about showing this.
Thanks!

Note that a definite integral turns a function into a number. For example:

$\int_0^{\pi} \sin x \ dx = 2$

Simlarly, if you take a function of two variables, $f(x, y)$ and integrate with respect to one variable, then you get a function of the other. E.g.

$\int_0^1 xy \ dx = [\frac 1 2 yx^2]_0^1 = \frac y 2$

This allows you to define an operator by taking a function and integrating it in some way. To take your example$$C(f(x)) = \int_0^5 (x-y)^3f(y) \ dy$$
The RHS is a function of $x$, as the $y$ variable gets integrated out, as it were. So, $C$ takes a function of $x$ and spits out a function of $x$. $C$, therefore, is an operator.

To see whether $C$ is linear or not, you need to think about the properties of the integral. Hint: a definite integral is generally a linear "thing"!

 

[S.G. Janssens]

Note that a definite integral turns a function into a number. For example:

$\int_0^{\pi} \sin x \ dx = 2$

Simlarly, if you take a function of two variables, $f(x, y)$ and integrate with respect to one variable, then you get a function of the other. E.g.

$\int_0^1 xy \ dx = [\frac 1 2 yx^2]_0^1 = \frac y 2$

This allows you to define an operator by taking a function and integrating it in some way. To take your example$$C(f(x)) = \int_0^5 (x-y)^3f(y) \ dy$$
The RHS is a function of $x$, as the $y$ variable gets integrated out, as it were. So, $C$ takes a function of $x$ and spits out a function of $x$. $C$, therefore, is an operator.

To see whether $C$ is linear or not, you need to think about the properties of the integral. Hint: a definite integral is generally a linear "thing"!

Physicists may find the following pedantic, but it can be very helpful to distinguish between the function $f$ and the value $f(x)$ of that function at the point $x$ in its domain. So, suppose you consider functions as vectors in some otherwise unspecified vector space $V$. Then the operator $C$ defined in the quote above is defined as
$$
C : V \to V, \qquad (Cf)(x) = \int_0^5 (x - y)^3 f(y)\,dy.
$$
That is, $C$ takes the function $f \in V$ and maps it to $Cf$, and we tell what $Cf$ is by specifying the value $(Cf)(x)$ at any point of the domain of $Cf$.

To see if $C$ is linear, check whether $C(f + g) = Cf + Cg$ and $C(a \cdot f) = a \cdot Cf$.

 

[PeroK]

Yes, I nearly added that myself, but I thought I'd said enough already. Not pedantic at all, IMO.

 

[cookiemnstr510510]

You have some serious misconceptions that you need to iron out. To take your example:

$A(f(x)) = f(x)\frac{df}{dx}$

Now, if you want to operate on $f(x) + g(x)$, a trick to see how to do this is first to let $h(x) = f(x) + g(x)$ and now:

$A(f(x) + g(x)) = A(h(x)) = h(x)\frac{dh}{dx} = (f(x) + g(x))\frac{d}{dx}(f(x) + g(x)$

First, thank you for your response. Although I am still a bit confused, I am making some sort of progress.
Let me stick with the example quoted and go a bit further with it.
So my initial operator A is acting on some function f(x) and when it acts on that function it gives us $f(x)\frac{df}{dx}$.

So our operator is acting on a function f. If I wanted to show that this operator was either linear or not I would use my two linearity properties (addition and scalar multiplication). To show addition I would need my operator to act on not only one function f(x) but on f(x)+g(x)--> $A(f(x) + g(x))$
I like your idea about letting h(x)=f(x)+g(x).
So initially my operator turns a function f(x) into itself times its derivative, $f(x)\frac{df}{dx}$.
So whatever my operator is acting on it will do those two things.
In the case of $A(f(x) + g(x))$ A has to spit out the thing inside of it, or f(x)+g(x) and the derivative of the thing inside of it which is $\frac{d}{dx}(f(x) + g(x)$

For linearity we want A(f(x)+g(x))=A(f(x))+A(g(x))
In our case this operator would not be linear, because we are multiplying two sums like (a+b)(c+d)?

 

[PeroK]

First, thank you for your response. Although I am still a bit confused, I am making some sort of progress.
Let me stick with the example quoted and go a bit further with it.
So my initial operator A is acting on some function f(x) and when it acts on that function it gives us $f(x)\frac{df}{dx}$.

So our operator is acting on a function f. If I wanted to show that this operator was either linear or not I would use my two linearity properties (addition and scalar multiplication). To show addition I would need my operator to act on not only one function f(x) but on f(x)+g(x)--> $A(f(x) + g(x))$
I like your idea about letting h(x)=f(x)+g(x).
So initially my operator turns a function f(x) into itself times its derivative, $f(x)\frac{df}{dx}$.
So whatever my operator is acting on it will do those two things.
In the case of $A(f(x) + g(x))$ A has to spit out the thing inside of it, or f(x)+g(x) and the derivative of the thing inside of it which is $\frac{d}{dx}(f(x) + g(x)$

For linearity we want A(f(x)+g(x))=A(f(x))+A(g(x))
In our case this operator would not be linear, because we are multiplying two sums like (a+b)(c+d)?

Yes, that the operator that is not linear. Technically, to show something is not linear you need a counterexample. In this case:

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

$\ \ \ \ \ \ \ \ \ \ \ \ = f(x)f'(x) + g(x)g'(x) + [f(x)g'(x) + g(x)f'(x)]$

$\ \ \ \ \ \ \ \ \ \ \ \ = A(f(x)) + A(g(x)) + [f(x)g'(x) + g(x)f'(x)]$

And, because that last expression in the square bracket is not always zero, then we see that $A$ is not linear.