A query regarding Rotational Invariance

[Student149]

We know that Bell States follow the Rotational Invariance property i.e. the probability of results on measurement of two entangled particles do not change if the initial measurement basis (say $u$) is rotated by an angle θ to a new basis (to say $v$).

Lets take the Bell State $\psi = \frac 1 {\sqrt 2}(\uparrow \uparrow + \downarrow \downarrow)$. Here both the terms have the probability $50$% in basis $u$ or rotated basis $v$.

How does the Rotational Invariance property get affected if the probability of two terms is unequal i.e. if the entangled state is of the form $\psi = (α\uparrow \uparrow + β\downarrow \downarrow)$, $α^2+β^2=1$, $α≠β$?

My guess is they remain unchanged as it is a generic case of Bell States.

 

[Nugatory]

How well does your guess work for the extreme case $\alpha=0$ and $\beta=1$?

 

[Student149]

How well does your guess work for the extreme case $\alpha=0$ and $\beta=1$?

haha. That is why it was a guess. I stated that assuming non zero real values. I am not sure how to compute the generic case hence the Q?

 

[PeroK]

We know that Bell States follow the Rotational Invariance property i.e. the probability of results on measurement of two entangled particles do not change if the initial measurement basis (say $u$) is rotated by an angle θ to a new basis (to say $v$).

Lets take the Bell State $\psi = \frac 1 {\sqrt 2}(\uparrow \uparrow + \downarrow \downarrow)$. Here both the terms have the probability $50$% in basis $u$ or rotated basis $v$.

How does the Rotational Invariance property get affected if the probability of two terms is unequal i.e. if the entangled state is of the form $\psi = (α\uparrow \uparrow + β\downarrow \downarrow)$, $α^2+β^2=1$, $α≠β$?

My guess is they remain unchanged as it is a generic case of Bell States.

You shouldn't be guessing, you should be calculating! I make it that the probability of getting spin up in a direction making angle $\theta$ with the z-axis is:
$$\beta^2 + (1 - 2\beta^2)\cos^2(\frac \theta 2)$$
You might want to double check that. It works out for $\beta = \frac 1 {\sqrt 2}$.

 

[PeroK]

haha. That is why it was a guess. I stated that assuming non zero real values. I am not sure how to compute the generic case hence the Q?

I can post the details if you are interested.

 

[Student149]

I can post the details if you are interested.

Thank you. I would be thankful..

 

[PeroK]

Thank you. I would be thankful..

The key is to work out the $\uparrow_z$ and $\downarrow_z$ states in an arbitrary basis for a measurement direction defined by spherical angles $\theta, \phi$ and denoted by $n$. These are:
$$\uparrow_z = \cos(\frac \theta 2)\uparrow_n - \sin(\frac \theta 2)\downarrow_n$$ and
$$\downarrow_z = e^{-i\phi}(\sin(\frac \theta 2)\uparrow_n + \cos(\frac \theta 2)\downarrow_n)$$
You can plug these into the required Bell state and expand. For the state you used in the OP, I get:
$$\psi = [\alpha \cos^2(\frac \theta 2) + \beta e^{-2i\phi}\sin^2(\frac \theta 2)]\uparrow_n\uparrow_n + [\alpha \sin^2(\frac \theta 2) + \beta e^{-2i\phi}\cos^2(\frac \theta 2)]\downarrow_n\downarrow_n + \\ \frac 1 2\sin\theta(\beta e^{-2i\phi} - \alpha)(\uparrow_n\downarrow_n + \downarrow_n\uparrow_n)$$
It would be good if you could double check that, as well as the probability calculations from that

 

[Student149]

You shouldn't be guessing, you should be calculating! I make it that the probability of getting spin up in a direction making angle θ with the z-axis is:
β2+(1−2β2)cos2⁡(θ2)
You might want to double check that. It works out for β=12.

The key is to work out the $\uparrow_z$ and $\downarrow_z$ states in an arbitrary basis for a measurement direction defined by spherical angles $\theta, \phi$ and denoted by $n$. These are:
$$\uparrow_z = \cos(\frac \theta 2)\uparrow_n - \sin(\frac \theta 2)\downarrow_n$$ and
$$\downarrow_z = e^{-i\phi}(\sin(\frac \theta 2)\uparrow_n + \cos(\frac \theta 2)\downarrow_n)$$
You can plug these into the required Bell state and expand. For the state you used in the OP, I get:
$$\psi = [\alpha \cos^2(\frac \theta 2) + \beta e^{-2i\phi}\sin^2(\frac \theta 2)]\uparrow_n\uparrow_n + [\alpha \sin^2(\frac \theta 2) + \beta e^{-2i\phi}\cos^2(\frac \theta 2)]\downarrow_n\downarrow_n + \\ \frac 1 2\sin\theta(\beta e^{-2i\phi} - \alpha)(\uparrow_n\downarrow_n + \downarrow_n\uparrow_n)$$
It would be good if you could double check that, as well as the probability calculations from that

I would. much appreciated.

 

[PeroK]

I would. much appreciated.

It all seems to check out. I can simplify the equations a bit:
$$p(\uparrow_n | \psi) = \frac 1 2 + (\frac 1 2 - |\beta|^2)\cos \theta = \frac 1 2 - (\frac 1 2 - |\alpha|^2)\cos \theta$$ $$p(\downarrow_n | \psi) = \frac 1 2 + (\frac 1 2 - |\alpha|^2)\cos \theta = \frac 1 2 - (\frac 1 2 - |\beta|^2)\cos \theta$$
Note that unless $|\alpha^2| = |\beta^2| = \frac 1 2$ these probabilities vary with $\theta$. Note also that if $\theta = \frac \pi 2$ (i.e. we are measuring in the x-y plane), then the probabilities are $1/2$ independent of $\alpha, \beta$.

 

[PeroK]

PS physically we should have expected the probabilities to depend on $\theta$ but not on $\phi$. I could have set $\phi = 0$ and made things easier.

 

[vanhees71]

Why don't the probs. add to 1? Also the very idea in #1 seems to be flawed to me. This is a state in the triplet, i.e., for $S=1$ and thus not rotationally invariant.

 

[Student149]

Why don't the probs. add to 1? Also the very idea in #1 seems to be flawed to me. This is a state in the triplet, i.e., for $S=1$ and thus not rotationally invariant.

I am confused. I think they do add to 1 in #9. Please elaborate what is flawed in #1 (I am a novice in QM, but I can't find a flaw)

 

[vanhees71]

The two-spin state in #1 is
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|1/2,1/2 \rangle + |-1/2,-1/2 \rangle).$$
This is a state belonging to total spin $S=1$, i.e., the spin-triplet state and thus is not rotationally invariant.

I also do not understand the notation in #9. The probabilities for the measurement of which observable are stated there? Also the numbers don't add to 1 but to $1/2(1-\cos \theta)+(|\alpha|^2+|\beta|^2)\cos \theta=1/2(1+\cos \theta)$.

 

[Student149]

1. I picked my example from here:

2. The $- (\frac 1 2 )\cos \theta$ is twice. Hence, it cancels $+\cos \theta$

3. If the direction of measurement is at an angle $\theta$ w.r.t. the original direction vector the probability of spin up or down in that new direction (it can be alternatively considered as the basis rotation w.r.t the original basis if i understand correctly)..

 

[PeroK]

The two-spin state in #1 is
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|1/2,1/2 \rangle + |-1/2,-1/2 \rangle).$$
This is a state belonging to total spin $S=1$, i.e., the spin-triplet state and thus is not rotationally invariant.

I also do not understand the notation in #9. The probabilities for the measurement of which observable are stated there? Also the numbers don't add to 1 but to $1/2(1-\cos \theta)+(|\alpha|^2+|\beta|^2)\cos \theta=1/2(1+\cos \theta)$.

The probabilities in post #9 fairly obviously add to $1$.

 

[PeroK]

I also do not understand the notation in #9.

$p(\uparrow_n | \psi)$ is the probablity of getting spin up in direction $\vec n$, at an angle of $\theta$ to the z-axis, given the state $\psi$ in the OP.

 

[vanhees71]

The probabilities in post #9 fairly obviously add to $1$.

Ok, then I don't understand the notation.

 

[vanhees71]

$p(\uparrow_n | \psi)$ is the probablity of getting spin up in direction $\vec n$, at an angle of $\theta$ to the z-axis, given the state $\psi$ in the OP.

There are two particles. Do you mean the spin of one of the particles or the total spin? It's hard to discuss, if the expressions are not well defined.

Let's do a well-defined example. I suppose that $|\pm 1/2 \rangle$ are the usual eigenstates of $\hat{\sigma}_z$ with eigenvalues $\pm 1/2$ ($\hbar=1$).

The two-spin state is $|\Psi \rangle \langle \Psi|$ with
$$|\Psi \rangle=\alpha |1/2,1/2 \rangle + \beta |-1/2,-1/2 \rangle, \quad |\alpha|^2+|\beta|^2$$
Now let's measure a spin component in direction taking an angle $\theta$ with the 3-direction. To be concrete, consider $\phi=0$ in the usual spherical coordinates. Then
$$\hat{\sigma}_{\theta}=\sin \theta \hat{\sigma}_x + \cos \theta \hat{\sigma}_z.$$
The eigenvectors are
$$|\chi_+ \rangle=\cos(\theta/2) |1/2 \rangle + \sin(\theta/2) |-1/2 \rangle$$
and
$$|\chi_- \rangle = -\sin(\theta/2) |1/2 \rangle + \cos(theta/2)|-1/2 \rangle.$$
The probability to measure $+1/2$ for the 1st spin thus is
$$P_+=\sum_{j =\pm 1/2} |\langle \chi_+,j|\Psi \rangle|^2.$$
Now
$$\langle \chi_+,1/2|\Psi \rangle=\alpha \cos(\theta/2), \quad \langle \chi_+,-1/2|\Psi \rangle=\beta \sin(\theta/2)$$
and thus
$$P_+=|\alpha|^2 \cos^2(\theta/2)+|\beta|^2 \sin^2 (\theta/2).$$
For $P_-$ we need
$$\langle \chi_-,1/2|\Psi \rangle=-\alpha \sin(\theta/2), \quad \langle \chi_-,-1/2 \rangle=\beta \cos(\theta/2)$$
and thus
$$P_-=|\alpha^2| \sin^2(\theta/2)+|\beta|^2 \cos^2(\theta/2).$$
I hope, I've not made a mistake, but at least now the probabilities clearly add to 1 :-)).

 

[PeroK]

There are two particles. Do you mean the spin of one of the particles or the total spin? It's hard to discuss, if the expressions are not well defined.

Yes, sorry, the spin of one particle.

 

[Student149]

There are two particles. Do you mean the spin of one of the particles or the total spin? It's hard to discuss, if the expressions are not well defined.

Let's do a well-defined example. I suppose that $|\pm 1/2 \rangle$ are the usual eigenstates of $\hat{\sigma}_z$ with eigenvalues $\pm 1/2$ ($\hbar=1$).

The two-spin state is $|\Psi \rangle \langle \Psi|$ with
$$|\Psi \rangle=\alpha |1/2,1/2 \rangle + \beta |-1/2,-1/2 \rangle, \quad |\alpha|^2+|\beta|^2$$
Now let's measure a spin component in direction taking an angle $\theta$ with the 3-direction. To be concrete, consider $\phi=0$ in the usual spherical coordinates. Then
$$\hat{\sigma}_{\theta}=\sin \theta \hat{\sigma}_x + \cos \theta \hat{\sigma}_z.$$
The eigenvectors are
$$|\chi_+ \rangle=\cos(\theta/2) |1/2 \rangle + \sin(\theta/2) |-1/2 \rangle$$
and
$$|\chi_- \rangle = -\sin(\theta/2) |1/2 \rangle + \cos(theta/2)|-1/2 \rangle.$$
The probability to measure $+1/2$ for the 1st spin thus is
$$P_+=\sum_{j =\pm 1/2} |\langle \chi_+,j|\Psi \rangle|^2.$$
Now
$$\langle \chi_+,1/2|\Psi \rangle=\alpha \cos(\theta/2), \quad \langle \chi_+,-1/2|\Psi \rangle=\beta \sin(\theta/2)$$
and thus
$$P_+=|\alpha|^2 \cos^2(\theta/2)+|\beta|^2 \sin^2 (\theta/2).$$
For $P_-$ we need
$$\langle \chi_-,1/2|\Psi \rangle=-\alpha \sin(\theta/2), \quad \langle \chi_-,-1/2 \rangle=\beta \cos(\theta/2)$$
and thus
$$P_-=|\alpha^2| \sin^2(\theta/2)+|\beta|^2 \cos^2(\theta/2).$$
I hope, I've not made a mistake, but at least now the probabilities clearly add to 1 :-)).

So, just to be clear/reconfirm:

if Alice and Bob have 1 particle each and Alice decides to measure its particle (in a direction that has an angle $\theta$ with the original direction $z$) the probability of both Alice and Bob getting a $\uparrow \uparrow$ and getting a $\downarrow \downarrow$ in new direction is given by the eqns in #9 correctly. right?

 

[PeroK]

So, just to be clear/reconfirm:

if Alice and Bob have 1 particle each and Alice decides to measure its particle (in a direction that has an angle $\theta$ with the original direction $z$) the probability of both Alice and Bob getting a $\uparrow \uparrow$ and getting a $\downarrow \downarrow$ in new direction is given by the eqns in #9 correctly. right?

No. Post #9 was for either Alice's particle or Bob's particle.

The probability that the spins are opposite in this direction for the Bell state $\frac 1 {\sqrt 2}(\uparrow \uparrow + \downarrow \downarrow)$ is:
$$\frac 1 2 \sin^2 \theta (1 - \cos (2\phi))$$
This shows that the state is not as rotationally invariant as you may have believed. In fact, in the y-direction, with $\theta = \phi = \frac \pi 2$, we have total anticorrelation:
$$\frac 1 {\sqrt 2}(\uparrow_z \uparrow_z + \downarrow_z \downarrow_z) = -\frac 1 {\sqrt 2}(\uparrow_y \downarrow_y + \downarrow_y \uparrow_y)$$

 

[Student149]

No. Post #9 was for either Alice's particle or Bob's particle.

The probability that the spins are opposite in this direction for the Bell state $\frac 1 {\sqrt 2}(\uparrow \uparrow + \downarrow \downarrow)$ is:
$$\frac 1 2 \sin^2 \theta (1 - \cos (2\phi))$$
This shows that the state is not as rotationally invariant as you may have believed. In fact, in the y-direction, with $\theta = \phi = \frac \pi 2$, we have total anticorrelation:
$$\frac 1 {\sqrt 2}(\uparrow_z \uparrow_z + \downarrow_z \downarrow_z) = -\frac 1 {\sqrt 2}(\uparrow_y \downarrow_y + \downarrow_y \uparrow_y)$$

Now I am terribly lost to the point of giving up any hope of understanding even the basics unless I have a complete formal background in physics. The original Q was regarding arbitrary $α, β$ and I assumed there is only one angle $θ$ (as described in the video). I have no clue about the other details (my background is CS), namely $\phi$ etc. Thank you for the help though..

 

[PeroK]

Now I am terribly lost to the point of giving up any hope of understanding even the basics unless I have a complete formal background in physics. The original Q was regarding arbitrary $α, β$ and I assumed there is only one angle $θ$ (as described in the video). I have no clue about the other details (my background is CS), namely $\phi$ etc. Thank you for the help though..

The video considers rotation in a plane (defined by a single angle $\theta$). The state has rotational invariance in a plane. But, nature is three dimensional (hence rotations in 3D are defined by two angles $\theta$ and $\phi$). That Bell state does not have 3D rotational invariance.

And, in answer to your question, you lose all rotational invariance unless you have $|\alpha| = |\beta| = \frac 1 {\sqrt 2}$.

 

[PeroK]

PS the Bell state $\frac 1 {\sqrt 2}(\uparrow \downarrow - \downarrow \uparrow)$, which represents the spin-0 singlet state, has full 3D rotational invariance - which the other Bell states do not.

 

[Student149]

PS the Bell state $\frac 1 {\sqrt 2}(\uparrow \downarrow - \downarrow \uparrow)$, which represents the spin-0 singlet state, has full 3D rotational invariance - which the other Bell states do not.

Got it. But if we limit ourselves to a plane (which I was assuming as its in the video example hence the confusion) we can set $\phi$=0 in #21 then from it we can get the probability of opposite spin. But, it doesn't answer about the probability of $\uparrow \uparrow$ and $\downarrow \downarrow$ each given arbitrary α, β.

PS Sorry for testing your patience.

 

[PeroK]

Got it. But if we limit ourselves to a plane (which I was assuming as its in the video example hence the confusion) we can set $\phi$=0 in #21 then from it we can get the probability of opposite spin. But, it doesn't answer about the probability of $\uparrow \uparrow$ and $\downarrow \downarrow$ each given arbitrary α, β.

PS Sorry for testing your patience.

With $\phi = 0$, the probability of getting opposite spins is:
$$\frac 1 2 |\alpha - \beta|^2 \sin^2 \theta $$

 

[Student149]

With $\phi = 0$, the probability of getting opposite spins is:
$$\frac 1 2 |\alpha - \beta|^2 \sin^2 \theta $$

I totally understand.

But how do we know about the same spin probability of each case ↑↑ and ↓↓ given arbitrary α, β?

 

[PeroK]

I totally understand.

But how do we know about the same spin probability of each case ↑↑ and ↓↓ given arbitrary α, β?

You use the coefficients of the state expressed in the relevant basis. From a few posts ago:
$$\psi = [\alpha \cos^2(\frac \theta 2) + \beta e^{-2i\phi}\sin^2(\frac \theta 2)]\uparrow_n\uparrow_n + [\alpha \sin^2(\frac \theta 2) + \beta e^{-2i\phi}\cos^2(\frac \theta 2)]\downarrow_n\downarrow_n$$ $$ + \frac 1 2\sin\theta(\beta e^{-2i\phi} - \alpha)(\uparrow_n\downarrow_n + \downarrow_n\uparrow_n)$$

 

[Student149]

You use the coefficients of the state expressed in the relevant basis. From a few posts ago:
$$\psi = [\alpha \cos^2(\frac \theta 2) + \beta e^{-2i\phi}\sin^2(\frac \theta 2)]\uparrow_n\uparrow_n + [\alpha \sin^2(\frac \theta 2) + \beta e^{-2i\phi}\cos^2(\frac \theta 2)]\downarrow_n\downarrow_n$$ $$ + \frac 1 2\sin\theta(\beta e^{-2i\phi} - \alpha)(\uparrow_n\downarrow_n + \downarrow_n\uparrow_n)$$

This has been very helpful. Thanks a ton again.

 

[vanhees71]

So, just to be clear/reconfirm:

if Alice and Bob have 1 particle each and Alice decides to measure its particle (in a direction that has an angle $\theta$ with the original direction $z$) the probability of both Alice and Bob getting a $\uparrow \uparrow$ and getting a $\downarrow \downarrow$ in new direction is given by the eqns in #9 correctly. right?

Ok, that's a different question.

To answer the question about joint measurements when both are measuring the spin component in direction $\theta$ (using the same states as in my previous posting) it's most simple to first express the old basis by the new, i.e.,
$$|1/2 \rangle=\cos(\theta/2) |\chi_+ \rangle - \sin[\theta/2]|\chi_- \rangle, \quad |-1/2 \rangle = \sin(\theta/2) |1/2 \rangle + \cos(\theta/2) |\chi_- \rangle.$$
The state ket the particle is prepared in is thus given in this basis by
$$|\Psi \rangle = [\alpha \cos^2(\theta/2) + \beta \sin^2(\theta/2)] |\chi_+,\chi_+ \rangle + (\beta-\alpha) \cos(\theta/2) \sin(\theta/2) (|1/2,-1/2 \rangle+|-1/2,1/2 \rangle) + (\alpha \sin^2 \theta/2 + \beta \cos^2 \theta/2) |-1/2,-1/2 \rangle.$$
From this you read off immediately for the probabilities for the 4 possible outcomes of joint measurements of the spin components in $\theta$ direction
$$P_{++}=|\alpha \cos^2 \theta/2 + \beta \sin^2 \theta/2|^2,$$
$$P_{+-}=P_{-+}=|\alpha-\beta|^2 \sin^2 \theta/2 \cos^2 \theta/2,$$
$$P_{--}=|\alpha \sin^2 \theta/2+\beta \cos^2 \theta/2)|^2.$$
You can simplify a bit using some trigonometry:
$$P_{++}=\frac{1}{4} [|\alpha+\beta|^2 + 2 (|\alpha|^2-|\beta|^2)\cos \theta + |\alpha-\beta|^2 \cos^2 \theta],$$
$$P_{+-}=P_{-+}=\frac{1}{4} |\alpha-\beta|^2 \sin^2 \theta,$$
$$P_{--}=\frac{1}{4}[|\alpha+\beta|^2 - 2 (|\alpha|^2-|\beta|^2) \cos \theta + |\alpha-\beta|^2 \cos^2 \theta].$$
As expected, the probabilities in general depend on $\theta$. That's because any spin-triplet state is not rotation invariant, because the total spin is $S=+1$. The spin-singlet state, i.e., $S=0$ is the only rotationally invariant state, i.e.,
$$|\Psi ' \rangle=\frac{1}{\sqrt{2}}(|1/2,-1/2 \rangle-|-1/2,1/2 \rangle),$$
as already mentioned in some postings above.