# Intro Quantum Mechanics - Dirac notations

• Graham87
In summary, Dirac notation can be used to represent quantum mechanical operators and their actions on kets. In this conversation, the topic of spin operators S+ and S- are discussed, and it is explained that they flip the arrow ket, but also map the other arrow ket to zero. The concept of eigenvalues and eigenstates is also mentioned, and it is clarified that these operators do not have a basis of orthogonal eigenstates. The use of Dirac notation is shown to be more efficient and superior, and it is recommended to practice using it to solve problems.

#### Graham87

Homework Statement
Familiarizing Dirac notations in intro to quantum mechanics
Relevant Equations
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I am learning Dirac notations in intro to quantum mechanics. I don’t understand why the up arrow changes to down arrow inside the equation in c).

My own calculation looks like this:

Your calculation makes no sense as you have replaced an operator with a ket.

To see why the first calculation is correct, think of what the raising and lowering operators do - by definition.

malawi_glenn, Graham87 and vanhees71
PeroK said:
Your calculation makes no sense as you have replaced an operator with a ket.

To see why the first calculation is correct, think of what the raising and lowering operators do - by definition.
You mean like this?
How does that change the arrow ket?

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Graham87 said:
You mean like this?
How does that change the arrow ket?
View attachment 305682
What is the effect of S+ on |up> vs. on |down>?

Graham87
malawi_glenn said:
What is the effect of S+ on |up> vs. on |down>?
It raises the eigenvalue by h?

Graham87 said:
It raises the eigenvalue by h?
On both? Can |up> be raised?
This should be covered in your coursebook.
Which one do you have?

vanhees71, PeroK and Graham87
malawi_glenn said:
On both? Can |up> be raised?
This should be covered in your coursebook.
Which one do you have?
Oh, sry, I misread the question. Well, it looks like S+ effect on the up arrow is that it flips it. Still trying to figure it out.
We have Griffith.
Is it because of this?

You can think of |up> as a column vector (1,0)T and similarly for |down>

S+ on |up> has eigenvalue 0
S- on |down> also has eigenvalue 0

You can confirm this with the matrix representation too

vanhees71 and Graham87
malawi_glenn said:
You van think of |up> as a column vector (1,0)T and similarly for |down>

S+ on |up> has eigenvalue 0
S- on |down> also has eigenvalue 0

You can confirm this with the matrix representation too
Ahaa! Got it. I did it by matrix calculation.
So with the Dirac form you just have to memorise that S+- flips the arrows?

vanhees71 and malawi_glenn
Graham87 said:
Ahaa! Got it. I did it by matrix calculation
Great but that is the slow way of doing it. Once you get comfortable, Dirac notation is superior and efficient. Try to confirm this with the formula you posted and redo the problem with just dirac notation

Graham87
Graham87 said:
Ahaa! Got it. I did it by matrix calculation.
So with the Dirac form you just have to memorise that S+- flips the arrows?
For spin 1/2 yes.
But you also have to consider the eigenvalues.

For instance spin 1, S+ on |+1> gives 0 as eigenvalue. And S- on |-1> also has 0 as eigenvalue

Graham87
Graham87 said:
Ahaa! Got it. I did it by matrix calculation.
So with the Dirac form you just have to memorise that S+- flips the arrows?
No. ##S_+## turns ##|\downarrow\rangle## into ##|\uparrow\rangle## but maps ##|\uparrow\rangle## to zero and similarly for ##S_-## but with the states reversed. This means that ##S_+ + S_-## flips the arrows.

Graham87
Graham87 said:
Ahaa! Got it. I did it by matrix calculation.
So with the Dirac form you just have to memorise that S+- flips the arrows?
##S_+## is the raising operator. In general, subject to the appropriate scalar factor, the raising operator raises each eigenstate to the next level up (and annihilates the highest eigenstate). And, ##S_-## is the lowering operator. In general, subject to the appropriate scalar factor, the lowering operator lowers each eigenstate to the next level down (and annihilates the lowest eigenstate).

Graham87
malawi_glenn said:
For instance spin 1, S+ on |+1> gives 0 as eigenvalue. And S- on |-1> also has 0 as eigenvalue
I don't understand this sentence. These states are not eigenstates of S+ and S- (they are eigenstates of Sz

hutchphd said:
I don't understand this sentence. These states are not eigenstates of S+ and S- (they are eigenstates of Sz
Yeah I should not have written eigenvalue there.

I meant ## {\hat S}_+ |s= 1,m = 1 \rangle = 0##

## \sqrt{2} \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}##
Perhaps its not ok to say that ##\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}## is an eigenvector of ## \sqrt{2} \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} ## with eigenvalue 0? I've seen this in Linear Algebra books at least.

@Graham87 this is always useful to know for a ket where ##m = m_{\text{max}}##
## {\hat S}_+ |s,m = m_{\text{max}} \rangle = 0##
and similar for ket with ##m = m_{\text{min}}##
## {\hat S}_- |s,m = m_{\text{min}} \rangle = 0##

This is also taken care of in the formula you gave. But it is very handy to know.

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Graham87 and hutchphd
hutchphd said:
I don't understand this sentence. These states are not eigenstates of S+ and S- (they are eigenstates of Sz
Technically, ##|\uparrow\rangle## is however an eigenstate of ##S_+## with eigenvalue zero. However, ##|\downarrow\rangle## is not. As ##S_\pm## are not Hermitian they do not have a basis of orthogonal eigenstates.

vanhees71 and malawi_glenn
Orodruin said:
Technically, |↑⟩ is however an eigenstate of S+ with eigenvalue zero. However, |↓⟩ is not. As S± are not Hermitian they do not have a basis of orthogonal eigenstates.

Does that technicality ever have a practical import?

hutchphd said:
Does that technicality ever have a practical import?
Maybe not, but you wrote that did not understand that ##|\uparrow \rangle## had eigenvalue 0 with the operator ##\hat S_+##. In linear algebra it is ok for a vector to be an eigenvector to a matrix with eigenvalue 0 (I am pretty, but not 100%, sure of that).

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hutchphd said:
Does that technicality ever have a practical import?
Yes. Because they have zero eigenvalue they are mapped to zero by the respective operators.

vanhees71
My apologies for to @malawi_glenn. I guess there was nothing improper about the way it was stated. (The physics was never in doubt !)

vanhees71, berkeman and malawi_glenn
hutchphd said:
My apologies for to @malawi_glenn. I guess there was nothing improper about the way it was stated. (The physics was never in doubt !)
ah cool I got a bit worried, I do teach linear algebra for high school students but we do not cover eigenvectors there so I thought I had forgotten something important :)

hutchphd, vanhees71 and berkeman
malawi_glenn said:
Maybe not, but you wrote that did not understand that ##|\uparrow \rangle## had eigenvalue 0 with the operator ##\hat S_+##. In linear algebra it is ok for a vector to be an eigenvector to a matrix with eigenvalue 0 (I am pretty, but not 100%, sure of that).
By definition, an eigenvector ##v## of an operator ##A## is a vector such that ##Av = \lambda v## for some scalar ##\lambda##. There is nothing preventing the eigenvalue to be zero.

hutchphd and malawi_glenn
Note that ##v = 0## can never be an eigenvector (by definition). This is because ##v = 0## would be an eigenvector of every matrix or linear transformation (with eigenvalue ##0##). And that would be a nuisance.

PS in fact, with any eigenvalue. Which is even worse!

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Orodruin and vanhees71

## 1. What is Dirac notation in quantum mechanics?

Dirac notation, also known as bra-ket notation, is a mathematical notation used to describe the state of a quantum system. It was developed by British physicist Paul Dirac and is widely used in quantum mechanics to represent quantum states, operators, and measurements.

## 2. How is Dirac notation used in quantum mechanics?

Dirac notation is used to represent the state of a quantum system, which can be described as a vector in a mathematical space known as a Hilbert space. In this notation, the state of a system is represented by a ket vector |ψ⟩, while its dual vector, known as the bra vector, is represented by ⟨ψ|. Operators, such as measurements or transformations, are represented by matrices acting on these vectors.

## 3. What are the advantages of using Dirac notation in quantum mechanics?

Dirac notation provides a concise and elegant way to represent complex mathematical concepts in quantum mechanics. It allows for easy manipulation of quantum states and operators, making calculations and equations more manageable. It also provides a visual representation of quantum states and their transformations, making it easier to understand and interpret the results of experiments.

## 4. Is Dirac notation the only way to represent quantum states?

No, Dirac notation is not the only way to represent quantum states. Other notations, such as matrix notation, are also used in quantum mechanics. However, Dirac notation is widely preferred due to its simplicity and ease of use.

## 5. Can Dirac notation be used for all types of quantum systems?

Yes, Dirac notation can be used for all types of quantum systems, including single particles, multiple particles, and even continuous systems. It is a universal notation that is applicable to all quantum systems and is not limited by the number of particles or the type of system being studied.