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My professor when talking about the EPR paradox said that the singlet spin state,

$$ |00\rangle=\frac{1}{\sqrt 2}(|+-\rangle-|-+\rangle) $$

is symmetric under rotation because its density matrix is

$$\rho=\frac{1}{4}(1-\vec{\sigma_1}\cdot\vec{\sigma_2}),$$

where $\vec{\sigma_1}$ and $\vec{\sigma_2}$ are the Pauli matrices for the first and second particle.

This is because for example a rotation around the $y$ axis that sends $\vec{e_z}$ to $\vec{e_x}$ and $\vec{e_x}$ to $-\vec{e_z}$, sends ${\sigma_i^z}$ to ${\sigma_i^x}$ and ${\sigma_i^x}$ to $-{\sigma_i^z}$ and so leaves $\rho$ unchanged.

But my professor said that $\rho$ for the state

$$ |10\rangle=\frac{1}{\sqrt 2}(|+-\rangle+|-+\rangle) $$

is

$$\rho=\frac{1}{2}(1+\vec{\sigma_1}\cdot\vec{\sigma_2}).$$

So also this state should be symmetric under rotations because there is a scalar product.
But this state isn't symmetric because, while the uncertainty of total spin third component $S^z$ is $0$ (the states is an eigenstate), the uncertainties of $S_x$ and $S_y$ are not $0$. This for example follows from the fact that $\langle S_x \rangle=\langle S_y \rangle=0$ and $\langle S_x^2 \rangle+\langle S_y^2 \rangle=\langle S^2 \rangle = 2\hbar$. How to solve this contraddiction?

EDIT:

I think I figured out what happened: the $\rho$ matrix for $|10\rangle$ state can't be the one in the OP as shown in this calculation:

$$\vec{\sigma_1}\cdot\vec{\sigma_2}= \frac{1}{2}[(\vec{\sigma_1}^2+\vec{\sigma_1})^2-\vec{\sigma_1}^2-\vec{\sigma_2}^2]= \frac{1}{2}(\frac{2}{\hbar})^2 S_{tot}^2-3 = \frac{2}{\hbar^2}S_{tot}^2-3$$

$$\rho=\frac{1}{\hbar^2}S_{tot}^2-1$$

$$\rho|10\rangle=|10\rangle$$ $$\rho|11\rangle=|11\rangle$$ $$\rho|1-1\rangle=|1-1\rangle$$

So $\rho$ can't be the projector onto the $|10\rangle$ state. Can someone give me a confirmation about this?

Mattia
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    Sorry, but your question is, IMHO, very unclear. Could you rephrase it and explicitly state what you are asking for? – Tobias Fünke Feb 13 '23 at 19:28
  • I'm asking how to solve the contraddiction that from the rho matrix the state seems symmetric but the indeterminations of the three components are not the same so the state can't be symmetric. – Mattia Feb 13 '23 at 20:19
  • There is no contradiction. The single state transforms trivially under rotations. The $S_z=0$ component of the triplet state transforms like a vector component. There is no contradiction. The forms of the density matrices you wrote down don't matter and don't tell you much at all about the transformations of the states. There is no contradiction. – hft Feb 13 '23 at 21:29
  • But your basis has rotated: eigenstates of $S_z$ have rotated to eigenstates of $S_x$, and the uncertainties have shifted the 0 from the z to the x-axis. The dot product of the σ matrices is part of the Casimirs, all rotationally invariant. – Cosmas Zachos Feb 13 '23 at 22:03
  • But to see if the density operator is symmetric I must apply the rotation operator only to the density operaror itself, not also to the states. Because all operators are invariant under basis change. And when you apply a transformation to the states and to the operators you are in fact performing a change of basis. – Mattia Feb 13 '23 at 23:02
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    No; the two density matrices you wrote are obscure ways to write $|00\rangle \langle 00|$ and $|10\rangle \langle 10|$, respectively. The composition from doublets and the σ.σ are red herrings! Just rotate these two operators. The first is invariant, the second not. You really slid down a rabbit hole of irrelevancies... – Cosmas Zachos Feb 13 '23 at 23:17
  • @CosmasZachos yes, but the second operator as written in the OP seems invariant too. Why? – Mattia Feb 13 '23 at 23:20
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    That's your rabbit hole. Repeat your uncertainty observation for the simple operators I wrote... – Cosmas Zachos Feb 13 '23 at 23:27
  • The formulas in your edit are correct. Recall you are looking at 4x4 matrices. But in they are misnormalized. They are not trace 1 idempotent matrices... The second ρ you wrote is the traditional exchange operator $P_{12}$. – Cosmas Zachos Feb 14 '23 at 16:01
  • This answer. might be helpful. – Cosmas Zachos Feb 14 '23 at 16:11
  • @Mattia You are right that the expression for the second density matrix was incorrect. – hft Feb 14 '23 at 16:12
  • The Second ρ you wrote is the standard exchange operator P. The antisymmetrizing projector is $(1-P)/2$, and the symmetrizing projector is $(1+P)/2$; you may plug in for the σ.σ expression, but you might confuse yourself... – Cosmas Zachos Feb 14 '23 at 18:01
  • @CosmasZachos I think that now my doubt is resolved. The density operator i wrote was symmetric but wasn't the density operator of that state. – Mattia Feb 14 '23 at 18:08
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    Absolutely! The idempotent trace 1 operator is the symmetrizer, and rotations connect the 3 symmetric states among themselves. – Cosmas Zachos Feb 14 '23 at 19:27

1 Answers1

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My professor when talking about the EPR paradox said that the singlet spin state:

$$ |00\rangle=|+-\rangle-|-+\rangle $$

is symmetric under rotation

Yes. This is a total-spin zero ($S^2=0$) state and also has $S_z = 0$. This means it transforms trivially under rotations.

because its density matrix is

$$\rho=\frac{1}{4}(1-\vec{\sigma_1}\cdot\vec{\sigma_2})$$

...But my professor said that $\rho$ for the state: $$ |10\rangle=|+-\rangle+|-+\rangle $$

$$\rho=\frac{1}{2}(1+\vec{\sigma_1}\cdot\vec{\sigma_2})$$

So also this state should be symmetric under rotations because there is a scalar product.

No. No it should not be symmetric. The state you wrote above is the $S_z=0$ part of the triplet. It transforms like a vector component into the other components of the triplet.

But this state isn't symmetric...

Yes. The components of the triplet do transform under rotations.

How to solve this contraddiction? [sic]

There is no contradiction. Also, as discussed below, one of the expressions for one of the density matrices you wrote above is incorrect.

Update:

I think I figured out what happened: the $\rho$ matrix for $|10\rangle$ state can't be the one in the OP as shown in this calculation:

Yes. I agree. The density matrix for $|10\rangle$ is: $$ |10\rangle\langle 10| \propto \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right) $$ whereas the expression $1 + \vec{\sigma_1}\cdot\vec{\sigma_2}$ is: $$ 1 + \vec{\sigma_1}\cdot\vec{\sigma_2} \propto \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right)\;, $$ so they are not the same.

However, the expression for $|00\rangle\langle 00| = \frac{1}{4}\left(1 - \sigma_1\cdot\sigma_2\right)$ is fine.

So $\rho$ can't be the projector onto the $|10\rangle$ state. Can someone give me a confirmation about this?

I can confirm in the sense provided in the update above.

hft
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  • But the density matrix is the projector to the state so I would think that the state is rotationally invariant if and only if the density matrix is rotationally invariant – Mattia Feb 13 '23 at 22:57
  • You might think that, but you would be wrong--as evidenced by this very question you are asking and the solution to the apparent "contradiction." – hft Feb 13 '23 at 23:02
  • To give you some more examples. A "singlet" is like a $Y_00$ spherical harmonic. It transforms trivially. A "triplet" is like the three $Y_{1m}$ spherical harmonics. They transform like components of an ordinary vector. – hft Feb 13 '23 at 23:03
  • It is also quite helpful to remember how to do quantum mechanical "addition of angular momenta." It's unfortunately a lot of group theory and representation theory that most trained physicists have internalized. But, you might recall that the direct product of two spin-1/2 representations can be decomposed into the direct sum of a spin-1 (triplet) representation and a spin-0 (singlet) representation. – hft Feb 13 '23 at 23:05
  • But why is that wrong? Also my professor said that many times. – Mattia Feb 13 '23 at 23:11
  • Also in 3d euclidean space the projection operator onto a subspace is invariant iff the subspace is invariant under rotations. For example the projection onto a plane is invariant under rotation around the axis the plane is perpendicular to, and the plane is invariant under that rotation. – Mattia Feb 13 '23 at 23:14
  • I can likely explain in more detail later today, but ultimately I think it would be good for you to understand how the states transform without trying to impose some incorrect understanding of the "transformation" of pauli matrices. Remember, Pauli matrices are just matrices, they don't "transform." You need to look at the states. – hft Feb 14 '23 at 00:38
  • What's the basis that matrices are written in? – Mattia Feb 14 '23 at 17:29
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    I think I'm using the convention $|\uparrow\rangle \to \left(\begin{matrix}1 \ 0\end{matrix}\right)$ and $|\downarrow\rangle \to \left(\begin{matrix}0 \ 1\end{matrix}\right)$ and ordering conventions such that, for example, $|\uparrow\downarrow\rangle = |\uparrow\rangle\otimes|\downarrow\rangle \to \left(\begin{matrix}1 \ 0\end{matrix}\right)\otimes \left(\begin{matrix}0 \ 1\end{matrix}\right) = \left(\begin{matrix}0 \ 1 \ 0 \ 0\end{matrix}\right)$ – hft Feb 14 '23 at 17:45
  • And, I'm ordering such that the "first" spin is "on the right" and the "second" spin is "on the left." So, for example, $\sigma^{(1)}_z = 1\otimes \sigma_z = \left(\begin{matrix}1 & 0 & 0 & 0\ 0 & -1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & -1\end{matrix}\right)$ – hft Feb 14 '23 at 17:48
  • Unfortunately, there area a lot of conventions that you have to check/confirm when you write down an explicit matrix form... and the explicit form will be different for different conventions – hft Feb 14 '23 at 17:48