Shankar Exercise 1.1.3 (Vector Spaces)

In summary: I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.I do understand how this proves anything? What am I misunderstanding? It said f(0)= 4, not f(x)=4. In my head 8 would be an acceptable answer.Your working shows that you perhaps lack a good enough grasp of linear algebra and formal mathematics. You seem to be confused between an individual function (such as ##f(x)##) and a vector space, which is an infinite set of functions. Normally,
  • #1
Stephen Wright
2
0
<Mentor's note: moved from a technical forum, therefore no template.>

I'm long out of college and trying to teach myself QM out of Shankar's.
I'm trying to understand the reasoning here because I think that I am missing something...

1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L) ?
3) How about functions that obey f(0)=4 ?

If the functions do not qualify, list what go wrong.

The first 2 questions seem straight forward. X and L go to zero, and outside those boundaries you would get the null vector. If the function is periodic, you are certain to return a vector in the f(x).

#3 is tripping me up. I thought of an example function... let's say f(x)=x+4, so f(0)=4.
How is this not a vector space?

I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

I do understand how this proves anything? What am I misunderstanding?
 
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  • #2

Homework Statement


I'm long out of college and trying to teach myself QM out of Shankar's.
I'm trying to understand the reasoning here because I think that I am missing something...

1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L) ?
3) How about functions that obey f(0)=4 ?

If the functions do not qualify, list what go wrong.

The first 2 questions seem straight forward. X and L go to zero, and outside those boundaries you would get the null vector. If the function is periodic, you are certain to return a vector in the f(x).

#3 is tripping me up. I thought of an example function... let's say f(x)=x+4, so f(0)=4.
How is this not a vector space?

I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

I do understand how this proves anything? What am I misunderstanding?

Homework Equations


The first 2 questions seem straight forward. X and L go to zero, and outside those boundaries you would get the null vector. If the function is periodic, you are certain to return a vector in the f(x).

The Attempt at a Solution



#3 is tripping me up. I thought of an example function... let's say f(x)=x+4, so f(0)=4.
How is this not a vector space?

I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

I do understand how this proves anything? What am I misunderstanding? It said f(0)= 4, not f(x)=4. In my head 8 would be an acceptable answer.
 
  • #3
Stephen Wright said:
#3 is tripping me up. I thought of an example function... let's say f(x)=x+4, so f(0)=4.
How is this not a vector space?

I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

I do understand how this proves anything? What am I misunderstanding? It said f(0)= 4, not f(x)=4. In my head 8 would be an acceptable answer.

Your working shows that you perhaps lack a good enough grasp of linear algebra and formal mathematics. You seem to be confused between an individual function (such as ##f(x)##) and a vector space, which is an infinite set of functions. Normally, you would denote a vector space by ##V## and have something like ##V## is the set of functions that vanish on the endpoints.
 
Last edited:
  • #4
Well for 3) it's not closed under addition, take two functions with f(0)=g(0)=4, does h(x)=f(x)+g(x) satisfy this condition?
 
  • #5
3 is a collection of functions all satisfying ##f(0)=4 ##. These functions can't form a vector space over ##\mathbb R ## (or ##\mathbb C## for that matter), because ##2f(0)=8\neq 4##. There is no homogeneity.
 
  • #6
Stephen Wright said:
<Mentor's note: moved from a technical forum, therefore no template.>

I'm long out of college and trying to teach myself QM out of Shankar's.
I'm trying to understand the reasoning here because I think that I am missing something...

1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L)?
3) How about functions that obey f(0)=4?

If the functions do not qualify, list what go wrong.
The way the question is worded might be confusing you. In each case, you have a set of functions, and the question is asking you if this set along with the usual definitions of adding functions and multiplying by a scalar satisfy the requirements of being a vector space. So first, you have to know all of the conditions required of a vector space. There's probably a list in the book, or if not, you can check Wikipedia.

The first 2 questions seem straight forward. X and L go to zero, and outside those boundaries you would get the null vector. If the function is periodic, you are certain to return a vector in the f(x).
What you wrote here doesn't make sense. For the first question, let ##V = \{f\ |\ f(0)=0\text{ and }f(L)=0\}##, and let ##f, g \in V##. The fact that ##f## is an element of ##V## means you know that ##f(0)=0## and ##f(L)=0##. Likewise for ##g##. One of the conditions of a vector space is that of closure under addition. In other words, the function ##f+g## should also be in ##V##. So you can check to see if ##(f+g)(0)=0## and if ##(f+g)(L)=0##. If this is the case, then the function ##(f+g) \in V##. It turns out that's true because ##(f+g)(0) = f(0)+g(0) = 0+0 = 0## and ##(f+g)(L) = f(L)+g(L) = 0+0 = 0##. So ##V## is closed under addition.

You have to verify all of the other requirements before concluding ##V## is a vector space.

#3 is tripping me up. I thought of an example function... let's say f(x)=x+4, so f(0)=4.
How is this not a vector space?

I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

I do understand how this proves anything? What am I misunderstanding?
Given the explanation above, do you now understand why set 3 isn't closed under addition?
 
  • #7
For each of the cases 1), 2) and 3) you will have a set of functions (probably infinite number, but it doesn't effect the problem). To show that they form a vector space ##V##, you have to show that for all members of the set, the conditions in Definition 1 are satisfied.
 

1. What is Shankar Exercise 1.1.3?

Shankar Exercise 1.1.3 is a problem set found in the book "Principles of Quantum Mechanics" by Ramamurti Shankar. It is a set of exercises designed to help students understand the concept of vector spaces in the context of quantum mechanics.

2. What is a vector space?

A vector space is a mathematical structure that consists of a set of objects (vectors) that can be added together and multiplied by scalars (numbers). In the context of quantum mechanics, vector spaces are used to represent physical states and operators.

3. What is the significance of Shankar Exercise 1.1.3?

Shankar Exercise 1.1.3 is significant because it helps students develop a deep understanding of vector spaces, which is crucial for understanding the mathematical framework of quantum mechanics. It also prepares students for more advanced concepts in quantum mechanics.

4. What are some key concepts covered in Shankar Exercise 1.1.3?

Shankar Exercise 1.1.3 covers concepts such as vector addition, scalar multiplication, linear independence, and spanning sets. These concepts are fundamental to understanding vector spaces and their applications in quantum mechanics.

5. How can I approach solving Shankar Exercise 1.1.3?

To solve Shankar Exercise 1.1.3, it is important to carefully read and understand the problem statement. Then, use the concepts covered in the exercise and any relevant equations to arrive at a solution. It may also be helpful to refer to other resources or ask for assistance from a teacher or classmate if needed.

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