Understanding inner product space and matrix representations of Operat

[cookiemnstr510510]

Homework Statement

Image attached

Relevant Equations

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(scroll to bottom for problem statement)
Hello,
I am wondering if someone could break down the problem statement in simpler terms (not so math-y).
I am struggling with understanding what is being asked.
I will try to break it down to the best of my ability

Problem statement:Consider the inner product space consisting of all linear combinations of sin(x) and cos(x).
My Understanding: To me this part is referring to the length that sin(x) and cos(x) can make with all possible linear combinations of each other. This doesn't make sense to me because inner product space seems to be talking about a literal space. After re-reading a chapter on inner product in my linear algebra book what it seemed to say was that the inner product space allows us to define things like length, angle etc. between vectors. So maybe that's what they are referring to in the problem statement.

Problem statement: what is the matrix representation of the operator i(d/dx) if we take sin(x) and cos(x) to be our basis functions?
My understanding: if sin and cos(x) are our basis then this is like saying i(hat) and j(hat) are our basis for R-2. I don't understand anything more than that.

I think Ill start with that rather than going through all parts. I want to have a good understanding of this and I am open for any suggestions. I have tried to brush up on my linear algebra, but I struggle with making connections between this and Lin Alg.

Thanks

 

[Orodruin]

Homework Statement:: Image attached
Relevant Equations:: ...

To me this part is referring to the length that sin(x) and cos(x) can make with all possible linear combinations of each other.

It is unclear what you mean by ”the length”. A linear combination of sin(x) an cos(x) is a function on the form f(x) = A sin(x) + B cos(x).

Homework Statement:: Image attached
Relevant Equations:: ...

This doesn't make sense to me because inner product space seems to be talking about a literal space.

What do you mean by ”literal” space. The space being referred to is the function space consisting of all possible linear combinations of sin(x) and cos(x).

Homework Statement:: Image attached
Relevant Equations:: ...

Problem statement: what is the matrix representation of the operator i(d/dx) if we take sin(x) and cos(x) to be our basis functions?
My understanding: if sin and cos(x) are our basis then this is like saying i(hat) and j(hat) are our basis for R-2. I don't understand anything more than that.

i(d/dx) is a (linear) operator on your function space. What do you get of you act with it on a general function in your function space?

 

[BvU]

Bit hard to help here (guidelines require an attempt, not just 'dunno')
And you are (understandably) not very specific in where precisely you are stuck.

So let me just ask:
Can you write $|p\rangle$ in terms of the basis functions ? $|q\rangle$ ? $\langle p|q \rangle$ ?

Is this enough to start on b) ?

 

[Orodruin]

Bit hard to help here (guidelines require an attempt, not just 'dunno')
And you are (understandably) not very specific in where precisely you are stuck.

So let me just ask:
Can you write $|p\rangle$ in terms of the basis functions ? $|q\rangle$ ? $\langle p|q \rangle$ ?

Is this enough to start on b) ?

The OP has made an attempt at interpreting the problem statement inside the problem statement. The problem is with this interpretation. He has also asked to start with (a) only.

 

[BvU]

He has also asked to start with (a) only.

I missed that .

 

[Orodruin]

I missed that .

It happens. I just pointed it out to highlight it.

 

[BvU]

It happens. I just pointed it out to highlight it.

What happens ?
You obliging someone to start with a) ?
Or me missing that requirement in the original post ?
Even if you pointed it out there, I still don't see the necessity of starting with a).
I am hopeful @cookiemnstr510510 can do this on his/her own once the initial step is taken.

 

[BvU]

Ah, misunderstanding. He/she wasn't asked, he/she has asked. My bad.

I feel free to point to b) to get started.

 

[cookiemnstr510510]

i(d/dx) is a (linear) operator on your function space. What do you get of you act with it on a general function in your function space?

I believe I understand what you are saying, which is the inner product space is made of up all linear combinations of sin(x) and cos(x) which, like you said, is a function of the form f(x)=Asin(x)+Bcos(x). I am assuming A and B can be anything? complex or real? That makes sense to me, I went on Desmos and graphed about ten cosine and sine functions with different coefficients out front so I have a visual representation of what we mean.

Now to answer the quote above if we have an operator O=i(d/dx), this operator takes in a function, takes the partial derivative of it and multiplies it by i. So it might look something like this:
O(f(x))=O(sin(x))+O(cos(x))=i$\frac{d}{dx}(sin(x))$+i$\frac{d}{dx}(cos(x)$

or does our operator act on sin(x) and cos(x) separately, like this:
f(x)=sin(x), g(x)=cos(x)
O(f(x))=(i$\frac{d}{dx}(sin(x)))$+cos(x)
O(g(x))=(i$\frac{d}{dx}(cos(x)))$+sin(x)

Thanks for your help! It's a struggle learning what is being asked and how to approach.

 

[Orodruin]

I am assuming A and B can be anything? complex or real? That makes sense to me, I went on Desmos and graphed about ten cosine and sine functions with different coefficients out front so I have a visual representation of what we mean.

In general, you can make this into a real or complex vector space depending on what you intend. However, based on the rest of the problem, the intended vector space is over the complex numbers.

The operator acts on any function f(x) such that Of(x) = i f’(x)

 

[PeroK]

or does our operator act on sin(x) and cos(x) separately, like this:
f(x)=sin(x), g(x)=cos(x)
O(f(x))=(i$\frac{d}{dx}(sin(x)))$+cos(x)
O(g(x))=(i$\frac{d}{dx}(cos(x)))$+sin(x)

Thanks for your help! It's a struggle learning what is being asked and how to approach.

Note that $\frac{d}{dx}$ is just the derivative. It's not any different just because we are now calling it an operator. If you take a function and differentiate it you get another function. That's what an operator does. So $\frac{d}{dx}$ has always been an operator, mapping differentiable functions into their derivatives.

If you want to know how $O = i \frac{d}{dx}$ acts on any function, just ask yourself how do I differentiate a function (and then multiply it by $i$)?

In your example:

$i\frac{d}{dx} (\sin x) = i \cos x$

That's all there is to it.

 

[cookiemnstr510510]

Note that $\frac{d}{dx}$ is just the derivative. It's not any different just because we are now calling it an operator. If you take a function and differentiate it you get another function. That's what an operator does. So $\frac{d}{dx}$ has always been an operator, mapping differentiable functions into their derivatives.

If you want to know how $O = i \frac{d}{dx}$ acts on any function, just ask yourself how do I differentiate a function (and then multiply it be $i$)?

In your example:

$i\frac{d}{dx} (\sin x) = i \cos x$

That's all there is to it.

Ahh, okay. This is starting to make more sense.

So since we are using sin(x) and cos(x) as our basis functions, and we are operating on them, we just differentiate each piece and multiply by i.

I understand that the inner product space defined is made from the linear combination of sin(x) and cos(x), which @Orodruin helped me understand is represented by f(x)=Asin(x)+Bcos(x). The problem then shows me the form of an inner product for functions (the bra ket = integral...).
So when the problem statement says "the inner product space consisting of all linear combinations of sin(x) and cos(x)" and gives me the "bra" "ket" integral, are these saying the same thing? Are we saying that the linear combination of sin(x) and cos(x) is representing the exact same thing as the bra-ket integral?

And let's say I wanted to solve this integral, would my |f>=sin(x) and |g>=cos(x)?

Thanks

 

[PeroK]

Ahh, okay. This is starting to make more sense.

So since we are using sin(x) and cos(x) as our basis functions, and we are operating on them, we just differentiate each piece and multiply by i.

I understand that the inner product space defined is made from the linear combination of sin(x) and cos(x), which @Orodruin helped me understand is represented by f(x)=Asin(x)+Bcos(x). The problem then shows me the form of an inner product for functions (the bra ket = integral...).
So when the problem statement says "the inner product space consisting of all linear combinations of sin(x) and cos(x)" and gives me the "bra" "ket" integral, are these saying the same thing? Are we saying that the linear combination of sin(x) and cos(x) is representing the exact same thing as the bra-ket integral?

And let's say I wanted to solve this integral, would my |f>=sin(x) and |g>=cos(x)?

Thanks

Yes. Once you have established that functions have the properties of vectors (function spaces are an important example of vector spaces) and once you have defined an inner product on your function space:$$\langle f|g\rangle = \int f(x)^*g(x)dx$$ then you are free to use whatever notation you use to do your linear algebra, including the Dirac notation.

If you were studying pure maths then you would have to prove that this integral has the properties of an inner product. In QM, you might want to take time out to prove things like that or just get on with it!

One of the advantages of pure mathematics is that once you have proved that the integral behaves like an inner product, then you can do a lot of things generically without going back to an integral each time.

There is another major advantage of using the bra-ket notation that you'll come to soon. For now, yes it does represent the integral.

What you have here is:

1) A function space (with an inner product) consisting of all linear combinations of $\cos x$ and $\sin x$. You could write this as$$V = \{a\sin x + b\cos x: \ a, b \in \mathbb{C}\}$$

2) A vector space (with an inner product) consisting of all vectors of the form $$V' = \{a|f\rangle + b|g \rangle: \ a, b \in \mathbb{C}\}$$ Where $|f\rangle, g \rangle$ represent $\sin x, \cos x$.

What you are doing mathematically is setting up a correspondence between $V$ and $V'$ so you can switch back and forwards between them.

 

[Orodruin]

Are we saying that the linear combination of sin(x) and cos(x) is representing the exact same thing as the bra-ket integral?

No, the linear combination represents a vector in the function space. The braket integral represents the inner product between two vectors.

 

[FactChecker]

With basis vectors $\sin(x)$ and $\cos(x)$, any linear combination $f(x) = A\sin(x)+B\cos(x)$ is represented by the vector $[A,B]^T$. Can you give the matrix that, operated on $[A,B]^T$ would give the derivative of $f(x)$ in that representation?

 

[cookiemnstr510510]

With basis vectors $\sin(x)$ and $\cos(x)$, any linear combination $f(x) = A\sin(x)+B\cos(x)$ is represented by the vector $[A,B]^T$. Can you give the matrix that, operated on $[A,B]^T$ would give the derivative of $f(x)$ in that representation?

If I understand what you are saying correctly, I believe it would be:
[A,B]^T=[A;B]
[what matrix]⋅[A;B]=Acos(x)-Bsin(x)

[cos(x),0; 0 -sin(x)]⋅[A;B]

 

[FactChecker]

If I understand what you are saying correctly, I believe it would be:
[A,B]^T=[A;B]
[what matrix]⋅[A;B]=Acos(x)-Bsin(x)

[cos(x),0; 0 -sin(x)]⋅[A;B]

Close, but not exactly. $d/dx [A\sin(x)+B\cos(x)] = A\cos(x)-B\sin(x) = -B\sin(x)+A\cos(x)$ So there you want a matrix, $M$, such that $M[A;B] = [-B;A]$. You don't need to mention $sin(x)$ and $cos(x)$ in the definition of $M$, since they are the basis vectors.