Solv Trivial Spin Problem Involving E^iπS_y

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In summary, the conversation revolved around proving the equation e^{i\pi S_y}|S\ 0\rangle = (-1)^S |S\ 0\rangle for a general case. Different approaches were discussed, including using Wigner's formula and expanding in the basis of S_y's eigenvectors. Ultimately, it was found that the Wigner formula led to a coefficient of (-1)^s, proving the equation for the general case.
  • #1
kakarukeys
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A trivial problem, but I am stuck.

Prove that
[tex]e^{i\pi S_y}|S\ 0\rangle = (-1)^S |S\ 0\rangle[/tex]

I proved the S = 1 case, by expanding [tex]|S\ 0\rangle[/tex] in the basis of [tex]S_y[/tex]'s eigenvectors. How to do for general case?
 
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  • #2
i think its something like this
you have Sy|S k> = -S+k for k=0,1,...,2S+1
exp(i*pi*Sy)|S k> = exp(i*pi*(-S+k))|S k>

now

exp(i*pi*(-S+k)) = exp(-i*pi*S)*exp(i*pi*k) if k = 0 then

exp(i*pi*Sy)|S 0> = exp(i*pi*(-S))|S 0> = cos(Pi*S)|S 0> and cos(Pi*S) = (-1)^S
 
  • #3
What did you mean?

[tex]|S\ k\rangle[/tex] is eigenstate of [tex]S_z[/tex] not [tex]S_y[/tex]
 
  • #4
U can write

[tex]\hat{S}_{y}=\frac{1}{2i}\left(\hat{S}_{+}-\hat{S}_{-}\right) [/tex]

and u know the action of the lowering & rising ladder operators on the standard basis [itex] |s,m_{s}\rangle [/itex]



Daniel.
 
  • #5
your correct, didnt think about that :)
 
  • #6
Sure, I used that trick and proved the S=1 case (3 dimensional matrices)

For general case. The problem is, we are dealing with the exponentiation of the operator, and the dimension of the matrix can be very large.
 
  • #7
The dimension of the matrices will always be 2...

Daniel.
 
  • #8
dextercioby said:
The dimension of the matrices will always be 2...

Daniel.

:confused: :confused: :confused:
d = 2 is spin half case.
 
  • #9
I'm sorry.It can't be 2.The dimension of the matrix is [itex] \left(2s+1,2s+1\right) [/itex],where "s" is an integer (because of the "0" value of [itex] m_{s} [/itex]).

I'm getting 0.I don't know why.Here's what i do.

[tex] e^{i\pi\hat{S}_{y}}=\mathcal{D}(0,-\pi,0) [/tex]

[tex] \mathcal{D}(0,-\pi,0)|s,0\rangle =d^{(j)}_{0,0}(-\pi) |s,0\rangle [/tex]

Now,using Wigner's formula

[tex] d^{(j)}_{m,m'}(\beta)=\left[\frac{(j+m)!(j-m)!}{(j+m')!(j-m')!}\right]^{1/2} \sum_{t}\left(\begin{array}{c} j+m'\\j-m-t\end{array}\right)\left(\begin{array}{c} j-m'\\t\end{array}\right) (-1)^{j-m'-t}\left(\cos\frac{\beta}{2}\right)^{2l+m+m'}\left(\sin\frac{\beta}{2}\right)^{2j-2t-m-m'} [/tex]

for the case [itex] j=s,m=m'=0,\beta=-\pi [/itex]

,i get 0... :


Tell me what u did...And how.


Daniel.
 
  • #10
I found a solution and it looks as something rather fishy happens here...I thought i applied the Wigner formula correctly,but there are books which prove me wrong...Galindo & Pascual,first volume,page ~210 make a discussion on this issue...

I can post it,if u don't have the book.

Something is really weird.I saw that the Wigner functions in Sakurai are different than the ones taken by Constantinescu & Magyari,but they both would lead to 0 instead of [itex] (-1)^{j} [/itex]...


Daniel.
 
  • #11
OMG, it's an awesome formula, I had never seen it before

I have exam tomorrow, will give you a reply after that.
 
  • #12
It's the form i found in Constantinescu & Magyari book.They're both Romanian & I'm a patriot.:approve:


Daniel.
 
  • #13
I checked the formula in the book by Constantine you said
okay, cosine 90 is zero, so the only surviving term in the binomial series is when the power of the cosine is zero.

I have
[tex]\langle s\ m | D(0, -\pi, 0) | s\ 0 \rangle = \frac{\sqrt{(s+m)!(s-m)!}}{s!}\ ^sC_{s-m/2}\ ^sC_{-m/2}(-1)^{s+m/2}[/tex]

[tex]\ ^sC_{s-m/2}[/tex] makes sense only if [tex]0 \leq m \leq 2s[/tex]
[tex]\ ^sC_{-m/2}[/tex] makes sense only if [tex]-2s \leq m \leq 0[/tex]

so the only non-zero term is when [tex]m = 0[/tex].

and the coefficient is [tex](-1)^s[/tex]

Is that the correct way to use Wigner's formula?
 
  • #14
How did u get that [itex] m/2 [/itex] in the exponent and the binomial coefficients...?

Daniel.
 
Last edited:
  • #15
I used Wigner's formula to compute
[tex]\langle s\ m | D(0, -\pi, 0) | s\ 0 \rangle[/tex]

there is a non-zero term in the series (the term without any cosine)
 
  • #16
Alright

[tex] \langle s,m|\mathcal{D}\left(0,-\pi,0\right)|s,0\rangle=d^{(s)}_{m,0}(-\pi)\delta_{m,0} [/tex]

[tex] d^{(s)}_{m,0}(-\pi)=\frac{\sqrt{(s+m)!(s-m)!}}{s!}\sum_{t}\left(\begin{array}{c}s\\s-m-t\end{array}\right)\left(\begin{array}{c}s\\t\end{array}\right)(-1)^{s-t}\left[\cos\left(-\frac{\pi}{2}\right)\right]^{2s+m}\left[\sin\left(-\frac{\pi}{2}\right)\right]^{2s-2t-m} [/tex]

And you impose that the power of the 0 (of the cosine) be 0.

[tex] 2s+m=0\Rightarrow m=-2s [/tex]

But on the other hand,[itex]m=0 [/itex] from the states orthonormalization...

Hmmm...I'm getting that [itex] (-1)^{s-t} [/itex] with a summation after "t" and an "s=0"...Which gives "+1".

What do you think?

Daniel.
 
  • #17
no
something is wrong with your formula
your power of the cosine should contain t, the dummy index
 
  • #18
Yes,you're right,there was no typo in my book,just in my formulas...:redface:

So here goes.To prevent a [itex] 0^{0}[/itex],i'll write it again,for an arbitrary [itex] \beta [/itex] and then take it [itex] -\pi [/itex].

[tex] \langle s,0|\mathcal{D}\left(0,\beta,0\right)|s,0\rangle=d^{(s)}_{0,0}(\beta) [/tex] (1)

,where

[tex] d^{(s)}_{0,0}(\beta)=\sum_{t}\left(\begin{array}{c}s\\s-t\end{array}\right)\left(\begin{array}{c}s\\t\end{array}\right)\left(-1\right)^{s-t}\left(\cos\frac{\beta}{2}\right)^{2t}\left(\sin\frac{\beta}{2}\right)^{2s-2t} [/tex] (2)

Okay.Now,i'm interested in canceling the exponent of cosine,since that "cosine",when evaluated on a rotation of [itex] -\pi [/itex],would yield 0,and so i need no cosine term in the sum after "t".

Therefore

[tex] 2t=0\Leftrightarrow t=0 [/tex] (3)

and the sum is reduced to only one term,the one corresponding to (3).

[tex] d^{(s)}_{0,0}(-\pi)=\left(\begin{array}{c}s\\s\end{array}\right)\left(\begin{array}{c}s\\0\end{array}\right)(-1)^{s}(-1)^{2s}=(-1)^{s} [/tex] (4)

Q.e.d.

Daniel.
 

Related to Solv Trivial Spin Problem Involving E^iπS_y

1. What is the "Solv Trivial Spin Problem Involving E^iπS_y"?

The "Solv Trivial Spin Problem Involving E^iπS_y" is a scientific problem related to the behavior of particles with spin in a magnetic field. It involves finding the eigenvalues and eigenvectors of the operator E^iπS_y, which represents the spin in the y-direction.

2. Why is this problem important in science?

This problem is important in science because it helps us understand the behavior of particles with spin in magnetic fields, which is essential in fields such as quantum mechanics and solid state physics. It also has practical applications in technologies such as magnetic resonance imaging (MRI).

3. How is this problem solved?

This problem is solved using mathematical techniques such as matrix diagonalization and eigenvalue analysis. These methods allow us to find the eigenvalues and eigenvectors of the operator E^iπS_y, which provide information about the spin states of particles in a magnetic field.

4. What are the implications of solving this problem?

Solving this problem can help us better understand the behavior of particles with spin in magnetic fields, which has implications in various fields of science and technology. It can also lead to further developments in quantum mechanics and our understanding of fundamental particles.

5. Are there any real-world applications of this problem?

Yes, there are several real-world applications of this problem. As mentioned earlier, it has practical applications in MRI technology, which uses magnetic fields to image the inside of the human body. It also has implications in the development of quantum computers, which rely on the behavior of particles with spin in magnetic fields for their functioning.

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