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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?

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    $\begingroup$ Sorry, but your question is, IMHO, very unclear. Could you rephrase it and explicitly state what you are asking for? $\endgroup$ Commented Feb 13, 2023 at 19:28
  • $\begingroup$ 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. $\endgroup$
    – Mattia
    Commented Feb 13, 2023 at 20:19
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    $\begingroup$ 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... $\endgroup$ Commented Feb 13, 2023 at 23:17
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    $\begingroup$ That's your rabbit hole. Repeat your uncertainty observation for the simple operators I wrote... $\endgroup$ Commented Feb 13, 2023 at 23:27
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    $\begingroup$ Absolutely! The idempotent trace 1 operator is the symmetrizer, and rotations connect the 3 symmetric states among themselves. $\endgroup$ Commented Feb 14, 2023 at 19:27

1 Answer 1

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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.

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  • $\begingroup$ 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 $\endgroup$
    – Mattia
    Commented Feb 13, 2023 at 22:57
  • $\begingroup$ 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." $\endgroup$
    – hft
    Commented Feb 13, 2023 at 23:02
  • $\begingroup$ 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. $\endgroup$
    – hft
    Commented Feb 13, 2023 at 23:03
  • $\begingroup$ 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. $\endgroup$
    – hft
    Commented Feb 13, 2023 at 23:05
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    $\begingroup$ 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, $|\uparrow\downarrow\rangle = |\uparrow\rangle\otimes|\downarrow\rangle \to \left(\begin{matrix}1 \\ 0\end{matrix}\right)\otimes \left(\begin{matrix}0 \\ 1\end{matrix}\right) = \left(\begin{matrix}0 \\ 1 \\ 0 \\ 0\end{matrix}\right)$ $\endgroup$
    – hft
    Commented Feb 14, 2023 at 17:45

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