Intro Quantum Mechanics - Dirac notations

[Graham87]

Homework Statement

Familiarizing Dirac notations in intro to quantum mechanics

Relevant Equations

See photo

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:

 

[PeroK]

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.

 

[Graham87]

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?

 

[malawi_glenn]

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]

What is the effect of S+ on |up> vs. on |down>?

It raises the eigenvalue by h?

 

[malawi_glenn]

It raises the eigenvalue by h?

On both? Can |up> be raised?
This should be covered in your coursebook.
Which one do you have?

 

[Graham87]

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?

 

[malawi_glenn]

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

 

[Graham87]

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?

 

[malawi_glenn]

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

 

[malawi_glenn]

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

 

[Orodruin]

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.

 

[PeroK]

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).

 

[hutchphd]

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

 

[malawi_glenn]

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.

 

[Orodruin]

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.

 

[hutchphd]

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?

 

[malawi_glenn]

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).

 

[Orodruin]

Does that technicality ever have a practical import?

Yes. Because they have zero eigenvalue they are mapped to zero by the respective operators.

 

[hutchphd]

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

 

[malawi_glenn]

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 :)

 

[Orodruin]

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.

 

[PeroK]

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!