Understanding Dirac Notation: A Simplified Explanation for Scientists

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Dirac notation simplifies the representation of quantum states and operators, where a wave function is denoted as a ket |psi> and its complex conjugate as a bra <psi|. Operators, such as the Hamiltonian H, act on kets similarly to traditional wave functions, allowing expressions like H|psi> = E|psi> to represent eigenvalue equations. The inner product of bra and ket, <i|j>, signifies the integral of the product of two wave functions over all space, while expressions like <i|Q|j> indicate operator action between states. Resources like Sakurai and Dirac's books are recommended for understanding this notation, with some users finding Sakurai more accessible than Ballentine. Overall, grasping Dirac notation requires familiarity with concepts from linear algebra and quantum mechanics, which can be challenging for beginners.
  • #31
Storm Butler said:
I do have a few books on linear algebra which i gave up after i realized i didn't have a good enough understanding of matrices
Hm...I don't think there are any books on linear algebra that don't explain matrices. I mean, linear algebra is the mathematics of linear operators between finite vector spaces, and a matrix is just a specific way to represent a linear operator. So you definitely shouldn't feel that you need to understand matrices before you start studying linear algebra.

(I'm feeling a bit nostalgic here. The very first post I wrote at Physics Forums was about the relationship between linear operators and matrices. One of the reasons I wrote it was that I wanted to remind myself how to use LaTeX).
 
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  • #32
lol ok thanks, and don't worry about sounding nostalgic the more you guys can help me the better i understand 99% of the people on this site probably have a better understanding of physics and math than i do. Also yes, the books i have to go over matrices but i found that they seemed a little abstract and decided to settle with something that had a lot more examples and problems in them in order to go over matrices.
 
  • #33
also what counts as vector space? is it just space in which vectors are present or something entirely different?
 
  • #35
Storm Butler said:
also what counts as vector space? is it just space in which vectors are present or something entirely different?
A vector is by definition a member of a vector space, so you define the concept "vector space" first.

You worry about abstractions, but I think the best way to define a vector space is to get very abstract. A vector space (over the real numbers) is a triple (X,\ S:\mathbb R\times X\rightarrow X,\ A:X\times X\rightarrow X) that satisfies eight specific properties. (The × is a Cartesian product and the notation f:U→V means "f is a function from X into Y"). The real numbers are called "scalars" in this context. The function S is called "multiplication by a scalar" (or "scalar multiplication", but do not confuse this with "scalar product" which is something else entirely), and A is called "addition". The conventional notation is to write kx instead of S(k,x) and x+y instead of A(x,y)=x+y. The eight specific conditions that must be satisfied for the triple (X,S,A) to be a vector space are listed on the Wikipedia page that Count Iblis linked to.

Once we're done with the definition, we can allow ourselves to be a bit sloppy with the terminology and refer to X as a vector space. This is a convention that you should be aware of. For example, the set \mathbb R^2 of ordered pairs of real numbers isn't really a vector space all by itself, but we still call it a vector space, because we know that if we define (a,b)+(c,d)=(a+c,b+d) and k(a,b)=(ka,kb), the triple (\mathbb R^2,S,A) is a vector space. We often say that S and A define a vector space structure on X. I think it would be a good exercise for you to verify that they do, i.e. that all of the eight conditions are satisfied when X,S and A are defined that way.

A complex vector space is defined by replacing the real numbers in the definition above with complex numbers. They can be replaced with other types of "numbers" as well to get a different type of vector space, but don't worry about that. You only need to understand complex vector spaces.

Storm Butler said:
Also yes, the books i have to go over matrices but i found that they seemed a little abstract and decided to settle with something that had a lot more examples and problems in them in order to go over matrices.
That sounds like a good idea for now, but you will eventually have to get used to abstractions. If you keep studying mathematics, you will see that it gets much, much more abstract than you would expect after only studying linear algebra.
 
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  • #36
ok so I've been re-reading my calculus, linear algebra, and matrix/vector books and I've come across two things that are tripping me up, vector calculus and differential equations. for vector calc the one thing that i wasnt sure about is how do you integrate/ differentiate the vectors, do you simply use the magnitude and then the new differentiated for is the direction or is there something else to do. for the differential equations i have been reading differential equations for dummies and it said all you really needed for a backround was the calculus one for dummies which i read (as well as calculus two fr dummies) however i got very confused very early while reading the book. the thing that i found the most strange was when the author was explaining how to find the integrating factor for the equation dy/dt +2y=4. basically he says to multiply by an unknown factor or equation so it would look something like, U(t)dy/dt+ 2U(t)y=4U(t). Then he says this "now you have to choose U(t) so that you can recognize the left side of the equation as the derivative of some expression. This way it can easily be integrated. Here's the key the left side of the previous equation looks very much like differentiation the product of U(t)y. so try to choose U(t) so that the left side of the equation is indeed the derivative of U(t)y. Doing so makes he integration easy.
Th derivative of U(t)y by t is: d[U(t)y]/dt=U(t)dy/dt+ dU(t)y/dt", this is the part that confused me number one maybe I am just not reading it correctly or i just completely missed something in calc but how is that the derivative of U(t)y. secondly the equation was 2U(t)y not U(t)y. then the next part that tripped me up was this "comparing the previous two equations term by term gives you: dU(t)/dt=2U(t). how did the two get back into the equation. These were just two parts that i got lost at and any help would be greatly appreciated.
 
  • #37
Storm Butler said:
Th derivative of U(t)y by t is: d[U(t)y]/dt=U(t)dy/dt+ dU(t)y/dt", this is the part that confused me number one maybe I am just not reading it correctly or i just completely missed something in calc but how is that the derivative of U(t)y.
I bet you misread
\frac{dU(t)}{dt} y​
or something similar.

secondly the equation was 2U(t)y not U(t)y.
You were computing the derivative of U(t)y. The differential equation you're trying to solve has nothing to do with that.

then the next part that tripped me up was this "comparing the previous two equations term by term gives you: dU(t)/dt=2U(t). how did the two get back into the equation.
He's simply stating the thing he wants to be true: "the left hand side is equal to the derivative of U(t)y". (And then simplifying)
 
  • #38
ok for the first one i know that he specifically meant d[U(t)y]/dt only because the brakets were written in, but even if it was dU(t)/dt*y wouldn't it just be the derivative of U(t) times y and therefore not be the equation that he gave? (idk i am not too familiar with the d/dx notation I am more used to the F'(x) or y'). Then for the second thing, jwhat do you mean by the differential equation you're trying to solve has nothing to do with that. i guess i really didn't get what the purpose or goals of the differential equations is, like what is this problem trying to do and how is it helping me. Then for the last part, what's the signifigance of the left hand side being equal to the derivative of U(t)y?
 
  • #39
Storm Butler said:
i know that he specifically meant d[U(t)y]/dt only because the brakets were written in,
I was referring to the bit that you wrote as dU(t)y/dt -- since you didn't specify, I assumed that was the part that was bothering you. What did you think the derivative of [U(t)y] -- the product of two functions in t -- should have been?


Then for the second thing, jwhat do you mean by the differential equation you're trying to solve has nothing to do with that.
I mean you need to pay attention to what you're doing. Yes, the overall goal is to solve the differential equation. But in his search, the thing he is doing right now is trying to find the derivative of [U(t)y] -- and the differential equation has absolutely nothing to do with that task.

Then for the last part, what's the signifigance of the left hand side being equal to the derivative of U(t)y?
That should be made clear by the next step he does after all of this.

Think back to when you learned how to solve quadratic equations -- in particular, the method of "completing the square". The difficulty in solving ax²+bx+c=0 a quadratic equation is that you have two terms involving x. One method of solution is to find a way to rearrange the equation to combine those terms into one, into the form "the square of something involving x equals something not involving x", which you know how to solve.

Your professor is about to pull a very similar trick here. He's looking for a way to combine both of the terms involving y into a single term, putting the equation into the form "the derivative of something involving y equals something that doesn't involve y".
 
  • #40
well for the derivative part i just thought about it as it being the derivative of U(t) and then the y would just be treated as a constant since it has no "t" terms. So i guess it was/is the derivative part that i found screwy. Also, i get what you are saying about the other two things now (i think :D), and its not a professor its the book called "Differential Equations for Dummies" and its by Steve holtzer ( i think).
 

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