Mastering Linear Vector Spaces: Exploring Bases, Dependence, and Operators

[fahd]

hi there..i am stuck wid these 2 problems from the subject mathematical methods for physicists and the topic is "linear vector spaces"

Q1) If S={|1>,|2>,...|n>} is a basis for a vector space V, show that every set with more than n vectors is linearly dependent? (where |> is a dirac bracket)

Q2)Show that the differential operator
p=h/i (d/dx)
is linear and hermitian in the space of all deifferentiable wave functions
[phi(x)] that, say, vanish at both ends of an interval (a,b)?


i am totally confused with these two questions..we were not taught this topic that well and they expect us to know these questions because similar ones like these wud be in the test tomrrow..please help me ..i dun want to loose marks.I ALSO KNOW THAT according to the rules..i need to show u what iv dun so far..but please understand..what do i show you..im totally confused! please revert!

 

[Allday]

to get started

hey Fahd,

if you're going to continue on in physics, particularly quantum mechanics, this is important stuff to know. to get you started think about the definitions of the terms involved. linearly dependent means that at least one of the vectors in the 'greater than n' set can be written in terms of others in the set. also recall that the definition of a basis is that any vector in the space can be written in terms of these basis vectors. start out by thinking about how you can write every vecor in the 'greater than n' set in terms of the n vectors given as the basis.


for the momentum operator recall that linearity just means

O[a f(x) + b g(x)] = a O f(x) + b O g(x)

and hermitian just means
$$\int_a^b f(x)^*p g(x) \,dx = \int_a^b (p f(x))^* g(x) \,dx$$

do a little integration by parts and you should be set


gabe

 

[fahd]

hey Fahd,
if you're going to continue on in physics, particularly quantum mechanics, this is important stuff to know. to get you started think about the definitions of the terms involved. linearly dependent means that at least one of the vectors in the 'greater than n' set can be written in terms of others in the set. also recall that the definition of a basis is that any vector in the space can be written in terms of these basis vectors. start out by thinking about how you can write every vecor in the 'greater than n' set in terms of the n vectors given as the basis.
for the momentum operator recall that linearity just means
O[a f(x) + b g(x)] = a O f(x) + b O g(x)
and hermitian just means
$$\int_a^b f(x)^*p g(x) \,dx = \int_a^b (p f(x))^* g(x) \,dx$$
do a little integration by parts and you should be set
gabe

thanks allday
just wonering
what do i take as f(x) and g(x) in the second question as stated by you
thanks

 

[Allday]

you can't use any particular function because the relation has to hold for the entire vector space ie (every differentiable wave function). the only objects you can use are those functions and their derivatives. The important thing to know (and this comes up all the time in derivations) is that integration by parts allows you to move a derivative under an integral from one function to the other at the cost of a boundry term and a minus sign.
ill show you some of the steps in the last part.
$$\int_a^b f^*(x)\frac{h}{i}\frac{dg}{dx} \,dx$$
$$= \frac{h}{i}[f^*(x=b)g(x=b)-f^*(x=a)g(x=a)] - \int_a^b \frac{h}{i}\frac{df^*}{dx}g(x)$$
how is this related to the rhs? the tricky part about these problems is dealing only with the abstract label of the function which represents all the functions in a certain family. here you'll have to make some assumptions about how the functions behave at the boundries of the region that they're defined (x=a and x=b)

 

[fahd]

thanks

hey..allday
thanks a lot for ur help
i finally understood it well...got both the questions..i was initially wondering where did g(x) and f(x) come from when they don't belong to the question..now i know!
thanks again!

 

[Allday]

glad to hear that it makes sense. i know it feels great to finally understand a particularly abstract concept.

gabe