I have a Bell state ${\Psi}^{-}= \frac{1}{\sqrt2} (|01\rangle - |10\rangle).$
How can I prove that this state is invariant (up to a global phase), when doing the same unitary $U$ on each qubit?
That is, how can I show that, for all $2\times 2$ unitaries $U$, we have:
$$(U\otimes{U})|{\Psi}^{-}\rangle = e^{i\theta}|{\Psi}^{-}\rangle?$$
I know $U$ acts only on one qubit but the last expression acts on two qubits.
Let $W$ be the unitary matrix representing the swap operation on two qubits, which we can write as follows. $$ W = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ It is evident that $W$ commutes with $U\otimes U$ for every $2\times 2$ unitary matrix $U$ — the actions of $W(U\otimes U)$ and $(U\otimes U) W$ are clearly the same on the standard basis, for instance, so they must be equal as matrices by linearity. This implies that $$ (U\otimes U) W (U \otimes U)^{\dagger} = W \tag{1} $$ for any $2\times 2$ unitary matrix $U$. Of course, we also have
$$ (U\otimes U) (\mathbb{I}\otimes\mathbb{I}) (U \otimes U)^{\dagger} = \mathbb{I}\otimes\mathbb{I}, \tag{2} $$
for $\mathbb{I}$ denoting the $2\times 2$ identity matrix, because $U$ is unitary.
Finally, observing that
$$ \vert {\Psi^-}\rangle\langle{\Psi^{-}}\vert = \frac{1}{2} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = \frac{\mathbb{I}\otimes\mathbb{I} - W}{2}, $$
we conclude by linearity from $(1)$ and $(2)$ that $$ (U\otimes U)\vert {\Psi^-}\rangle\langle{\Psi^{-}}\vert(U \otimes U)^{\dagger} = \vert {\Psi^-}\rangle\langle{\Psi^{-}}\vert, $$ which is equivalent to $(U\otimes U)\vert {\Psi^-}\rangle = e^{i\theta}\vert {\Psi^-}\rangle.$
This is getting silly. I can't resist joining, though, as all other answers rely on $U$ being unitary, which is not necessary.
First observe that $A \otimes I |\phi^+\rangle = I \otimes A^T |\phi^+\rangle$ for any matrix $A$, and that $|\psi^-\rangle = iY \otimes I |\phi^+\rangle$. Using these two facts we get that $$U \otimes U |\psi^-\rangle = UYU^TY \otimes I |\psi^-\rangle.$$ Now we parametrize $U$ as $$U = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$$ and compute $UYU^TY$ explicitly to see that $$UYU^TY = (\alpha \delta - \beta\gamma)I = \det(U)I,$$ and therefore $$U \otimes U |\psi^-\rangle = \det(U)|\psi^-\rangle.$$
And another...
Let $|u_0\rangle, |u_1\rangle$ be normalized eigenvectors of $U$ for the eigenvalues $\lambda_0, \lambda_1$ respectively. Then $U \otimes U$ has the following eigendecomposition:
But notice $$ (\langle v_i|\otimes \langle v_i|) |\Psi^-\rangle = \frac{\langle v_i|0\rangle\langle v_i|1\rangle - \langle v_i|1\rangle\langle v_i|0\rangle}{\sqrt{2}} = 0 $$ and so $|\Psi^{-}\rangle \in \mathrm{span}\{|u_0\rangle \otimes |u_1\rangle, |u_1\rangle \otimes |u_0\rangle\}$ is solely contained in the second eigenspace. Overall this implies $$ (U \otimes U) |\Psi^-\rangle = \lambda_0 \lambda_1 |\Psi^-\rangle\,. $$
NB: this proof shows that $|\Psi^-\rangle$ is an eigenvector of $M \otimes M$ for any normal matrix $M \in \mathbb{C}^{2\times 2}$.
You can find $$U|0\rangle \otimes U|1\rangle$$ and $$U|1\rangle \otimes U|0\rangle$$ in the standard basis, where $$|0\rangle \equiv \begin{bmatrix} 1\\0 \end{bmatrix},\\ |1\rangle \equiv \begin{bmatrix} 0\\1 \end{bmatrix},$$ $$|01\rangle = |0\rangle \otimes |1\rangle,\\ |01\rangle = |1\rangle \otimes |0\rangle.$$ You know that a 2x2 unitary $U$ should take this form: $$ U = e^{i\phi/2}\begin{bmatrix}e^{i\phi_1}\cos{\theta} & e^{i\phi_2}\sin{\theta}\\ -e^{-i\phi_2}\sin{\theta} & e^{-i\phi_1}\cos{\theta} \end{bmatrix} $$ (see https://en.wikipedia.org/wiki/Unitary_matrix). Then,
Using this,
$U|0\rangle \otimes U|1\rangle =\\ = e^{i\phi/2}\begin{bmatrix}e^{i\phi_1}\cos{\theta}\\-e^{-i\phi_2}\sin{\theta}\end{bmatrix} \otimes e^{i\phi/2}\begin{bmatrix} e^{i\phi_2}\sin{\theta}\\ e^{-i\phi_1}\cos{\theta} \end{bmatrix}\\ = e^{i\phi}\begin{bmatrix} e^{i\phi_1}\cos{\theta}e^{i\phi_2}\sin{\theta}\\ e^{i\phi_1}\cos{\theta}e^{-i\phi_1}\cos{\theta}\\ -e^{-i\phi_2}\sin{\theta}e^{i\phi_2}\sin{\theta}\\ -e^{-i\phi_2}\sin{\theta}e^{-i\phi_1}\cos{\theta} \end{bmatrix}\\ = e^{i\phi}\begin{bmatrix} e^{i(\phi_1+\phi_2)}\cos{\theta}\sin{\theta}\\ \cos^2{\theta}\\ -\sin^2{\theta}\\ -e^{-i(\phi_2+\phi_1)}\sin{\theta}\cos{\theta} \end{bmatrix},$
$U|1\rangle \otimes U|0\rangle =\\ e^{i\phi/2}\begin{bmatrix} e^{i\phi_2}\sin{\theta}\\ e^{-i\phi_1}\cos{\theta} \end{bmatrix} \otimes e^{i\phi/2}\begin{bmatrix}e^{i\phi_1}\cos{\theta}\\-e^{-i\phi_2}\sin{\theta}\end{bmatrix}\\ =e^{i\phi} \begin{bmatrix} e^{i\phi_2}\sin{\theta}e^{i\phi_1}\cos{\theta}\\ -e^{i\phi_2}\sin{\theta}e^{-i\phi_2}\sin{\theta}\\ e^{-i\phi_1}\cos{\theta}e^{i\phi_1}\cos{\theta}\\ -e^{-i\phi_1}\cos{\theta}e^{-i\phi_2}\sin{\theta} \end{bmatrix}\\ =e^{i\phi} \begin{bmatrix} e^{i(\phi_1+\phi_2)}\cos{\theta}\sin{\theta}\\ -\sin^2{\theta}\\\cos^2{\theta}\\-e^{-i(\phi_1+\phi_2)}\cos{\theta}\sin{\theta}\end{bmatrix}. $ Then, $$U|0\rangle \otimes U|1\rangle - U|1\rangle \otimes U|0\rangle\\ =e^{i\phi} \begin{bmatrix} 0\\ \cos^2{\theta}+\sin^2{\theta}\\ -\sin^2{\theta}-\cos^2{\theta}\\0\end{bmatrix}\\ =e^{i\phi} \begin{bmatrix} 0\\1\\ -1\\0\end{bmatrix}. $$ I didn't include the normalization constant of $\frac{1}{\sqrt{2}}$ in these calculations, but clearly, applying $U \otimes U$ to this Bell state results in the same state multiplied by $e^{i\phi}$.
Any unitary matrix can be written as $SU(2)$ matrix discarding global phase. The general form for 2x2 $SU(2)$ is $\begin{pmatrix}a & b\\-b^*&a^*\end{pmatrix}$, where $a$ and $b$ are two complex number satisfying $|a|^2+|b|^2=1$, star stands for complex conjugate.
Replace $|\Psi\rangle = \frac{1}{\sqrt2} (|01\rangle - |10\rangle)$ into $\langle \Psi|U\otimes U|\Psi\rangle$, and change $\langle 0|U|0\rangle = a, \langle 0|U|1\rangle=b,\langle 1|U|0\rangle=-b^*,\langle 1|U|1\rangle=a^*$, we can get $\langle \Psi|U\otimes U|\Psi\rangle=1$ .
Your question is also refer to that the maximal entangled state remain maximal entangled with the operation of LOCC.
First verify $H^{\otimes 2} |\psi_{-}\rangle = |\psi_{-}\rangle$.
Second verify $(Z^t)^{\otimes 2} |\psi_{-}\rangle = |\psi_{-}\rangle$ for all $t$.
This implies $(X^t)^{\otimes 2} |\psi_{-}\rangle = |\psi_{-}\rangle$ because $X^t = H Z^t H$.
This implies $U^{\otimes 2} |\psi_{-}\rangle = e^{i\theta}|\psi_{-}\rangle$ because any $U$ can be decomposed up to global phase into $U \equiv Z^a X^b Z^c$.
The invariance follows readily from Grassman algebras. The answer below consists of a crash course on Grassman algebras followed by a simple calculation to prove that the singlet is preserved under $U\otimes U$. My other answer extracts the key fact from this post to provide an elementary argument.
The singlet $|01\rangle-|10\rangle$ is distinguished among the four Bell states by its relationship to a certain Grassmann algebra. Namely, it is the sole Bell state which is not annihilated by the canonical projection of the two-qubit Hilbert space onto the second exterior power of the single-qubit Hilbert space. For a qubit, the second exterior power is a one-dimensional subspace preserved by $U\otimes U$, so the only possible linear action of $U\otimes U$ on the subspace is multiplication by a scalar.
Let $\mathcal{H}$ denote a vector space over field $K$. Let $T^n(\mathcal{H}):=\mathcal{H}^{\otimes n}$ denote the $n$th tensor power of $\mathcal{H}$ and $T(\mathcal{H}):=\bigoplus_{n=0}^\infty T^n(\mathcal{H})$ the tensor algebra over $\mathcal{H}$. Elements of $T(\mathcal{H})$ can be multiplied by the scalars in $K$, added to each other and multiplied by each other using the tensor product $\otimes$.
Let $I$ denote the two-sided ideal in $T(\mathcal{H})$ generated by all elements of the form $|\psi\rangle\otimes|\psi\rangle$. The Grassman or exterior algebra of $\mathcal{H}$ is the quotient $\bigwedge(\mathcal{H}):=T(\mathcal{H})/I$. Elements of $\bigwedge(\mathcal{H})$ can be multiplied by scalars in $K$ and can be added to each other. They can also be multiplied by each other using the exterior product \begin{align} |\psi\rangle\wedge|\phi\rangle := |\psi\rangle\otimes|\phi\rangle + I.\tag1 \end{align} An easy, but important observation is that \begin{align} |\psi\rangle\wedge|\psi\rangle=0\tag2 \end{align} for every $|\psi\rangle$. This follows from the fact that $|\psi\rangle\otimes|\psi\rangle\in I$ for every $|\psi\rangle$. Taking for $|\psi\rangle$ a superposition of two states, one can generalize $(2)$ to \begin{align} |\psi\rangle\wedge|\phi\rangle=-|\phi\rangle\wedge|\psi\rangle.\tag3 \end{align}
The set $\bigwedge^k(\mathcal{H})$ consisting of all elements of $\bigwedge(\mathcal{H})$ that are exterior products of exactly $k$ elements of $\mathcal{H}$ is a vector space called the $k$th exterior power of $\mathcal{H}$. It is a simple exercise to verify that $\dim\bigwedge^k(\mathcal{H})={\dim\mathcal{H} \choose k}$. The algebra $\bigwedge(\mathcal{H})$ is the direct sum of the subspaces $\bigwedge^k(\mathcal{H})$ for $k=0,1,\ldots$.
The canonical projection $\pi:T(\mathcal{H})\to\bigwedge(\mathcal{H})$ defined by $\pi(|\psi\rangle)=|\psi\rangle+I$ preserves the graded structure of the tensor and exterior algebras, i.e. it sends $T^k(\mathcal{H})$ to $\bigwedge^k(\mathcal{H})$.
Every linear map $A:\mathcal{H}\to\mathcal{H}$ induces a unique map $\bigwedge^k(A):\bigwedge^k(\mathcal{H})\to\bigwedge^k(\mathcal{H})$ which sends $|\psi_1\rangle\wedge\ldots\wedge|\psi_k\rangle$ to $A|\psi_1\rangle\wedge\ldots\wedge A|\psi_k\rangle$. In the special case where $k=\dim\mathcal{H}$ the map acts on the one-dimensional space $\bigwedge^{\dim\mathcal{H}}(\mathcal{H})$ as scalar multiplication by the determinant$^1$ of $A$. In particular, if $U$ is a single-qubit operator, then $U\wedge U$ sends $|\psi\rangle$ to $\det(U)|\psi\rangle$.
End of crash course.
Let $\mathcal{H}:=\mathbb{C}^2$ be the single-qubit Hilbert space and consider the restriction $\pi_2$ of $\pi$ to the second grade of the tensor and exterior algebras. In other words, consider the map $\pi_2:\mathcal{H}^{\otimes 2}\to\bigwedge^2(\mathcal{H})$ given by \begin{align} \pi_2(|\psi\rangle\otimes|\phi\rangle) := |\psi\rangle\wedge|\phi\rangle.\tag4 \end{align} Note that $\dim\bigwedge^2(\mathcal{H})=1$, so by rank-nullity theorem $\pi_2$ has three-dimensional kernel. Indeed, by equation $(2)$, we have $\pi_2(|00\rangle)=\pi_2(|11\rangle)=0$. Consequently, the Bell states $|00\rangle+|11\rangle$ and $|00\rangle-|11\rangle$ are in the kernel of $\pi_2$. In addition, using $(3)$ we find \begin{align} \pi_2(|01\rangle+|10\rangle)&=|0\rangle\wedge|1\rangle+|1\rangle\wedge|0\rangle=\tag5\\ &=|0\rangle\wedge|1\rangle-|0\rangle\wedge|1\rangle=0.\tag6 \end{align} By contrast, \begin{align} \pi_2(|01\rangle-|10\rangle)&=|0\rangle\wedge|1\rangle-|1\rangle\wedge|0\rangle=\tag7\\ &=|0\rangle\wedge|1\rangle+|0\rangle\wedge|1\rangle=\tag8\\ &=2|0\rangle\wedge|1\rangle\ne 0.\tag9 \end{align} Thus, the singlet $|01\rangle-|10\rangle$ spans the preimage of $\bigwedge^2(\mathcal{H})$ via the restricted canonical projection $\pi_2$.
Finally, from equation $(4)$, we have \begin{align} \pi_2\circ(U\otimes U) = (U\wedge U)\circ\pi_2\tag{10} \end{align} but $U\wedge U$ is the scalar multiplication by the determinant of $U$, so the action of $U\otimes U$ on $|\Psi^-\rangle$ is to multiply it by the determinant of $U$.
$^1$ This is the only non-trivial fact whose proof is omitted in the crash course. However, note that we don't actually need to know that $U\wedge U$ is multiplication by $\det U$. All we need is that it is multiplication by some scalar. And this does follow trivially from the fact that the second exterior power of a two-dimensional vector space is one-dimensional.
$U\otimes U$ commutes with the SWAP gate, so it preserves its $+1$ and $-1$ eigenspaces \begin{align} S&:=\mathrm{span}(|00\rangle, |11\rangle, |01\rangle+|10\rangle)\tag1\\ T&:=\mathrm{span}(|01\rangle-|10\rangle).\tag2 \end{align} But $T$ is one-dimensional, so $|\Psi^-\rangle\in T$ is an eigenvector of $U\otimes U$.
I guess someone oughts to mention the generalisation of this statement. Let's start with a few observations:
Let $|\Psi\rangle$ be any maximally entangled two-qubit state. As easily seen e.g. from the Schmidt decomposition, we know we can always write it as $|\Psi\rangle=\frac{1}{\sqrt2} \operatorname{vec}(V)$ for some $2\times 2$ unitary $V\equiv V_\Psi$.
Observe that $$(U\otimes \bar W)|\Psi\rangle=\frac{1}{\sqrt2}\operatorname{vec}(UV W^\dagger).\tag A$$ For any pair of 2x2 matrices $U,W$ (in fact, this also holds more generally when $|\Psi\rangle$ isn't maximally entangled and thus $V$ isn't unitary).
It follows that finding unitaries $U,W$ such that $(U\otimes \bar W)|\Psi\rangle=e^{i\theta}|\Psi\rangle$ is the same as looking for unitaries such that $$UV W^\dagger = e^{i\theta} V \tag B.$$ In other words, we want to find the structure of $W$ such that $e^{i\theta}W=V^\dagger UV$, hoping it gives us a relation between $W$ and $U$ that doesn't depend on $V$. Multiplying by a phase doesn't affect unitarity, so this relation can be simplified to just $W=V^\dagger UV$.
From these we can state the more general result: for any given maximally entangled two-qubit $|\Psi\rangle=\frac{1}{\sqrt2}\operatorname{vec}(V)$, we have for all unitaries $U$ the identity: $$(U\otimes (V^T \bar U\bar V))|\Psi\rangle=|\Psi\rangle.$$ This doesn't look as nice as the well-known statement about the singlet, but it is in fact of exactly the same nature: given a maximally entangled state, there is a recipe to find a unitary $W_U$ from any given $U$ such that $|\Psi\rangle$ is fixed by $U\otimes \bar W_U$. Sometimes the recipe is simple enough, and $W_U=\bar U$ (for the singlet $|\Psi^-\rangle$) or $W_U=U$ (for $|\Phi^+\rangle$), while in other cases it's less trivial.
Let's consider a few examples to show this:
If $|\Psi\rangle\equiv|\Phi^+\rangle=\frac1{\sqrt2} (|00\rangle+|11\rangle)$, then $V=I$ is the identity matrix, and thus from (B) we get the easy solution $W=U$. This corresponds to the easily verifiable statement $$(U\otimes \bar U)|\Phi^+\rangle = |\Phi^+\rangle,\quad \forall \text{ unitaries } U.$$
If $|\Psi\rangle\equiv |\Phi^-\rangle\equiv \frac1{\sqrt2}(|00\rangle-|11\rangle)$, then $V=Z$, and we want $W=ZUZ$. This means to change the sign of the off-diagonal elements of $U$. I don't know if there's a particularly simpler way to write it, but the identity becomes $$(U\otimes Z\bar UZ)|\Phi^-\rangle=|\Phi^-\rangle.$$
If $|\Psi\rangle\equiv|\Psi^+\rangle\equiv\frac{1}{\sqrt2}(|01\rangle+|10\rangle)$, then $V=X$, hence $W=XUX$ (switch elements of both diagonals), and $$(U\otimes X\bar UX)|\Psi^+\rangle=|\Psi^+\rangle.$$
If $|\Psi\rangle\equiv|\Psi^-\rangle\equiv\frac{1}{\sqrt2}(|01\rangle-|10\rangle)$, then $V=iY$, hence $W=(iY)^\dagger U(iY)=YUY$. As also shown in this other answer, we have $W=YUY=\det(U) \bar U$. We thus finally recover the original identity for the singlet, in the form: $$\overline{\det(U)}(U\otimes U)|\Psi^-\rangle=|\Psi^-\rangle,$$ which collecting the phases into a $e^{i\theta}$ factor becomes the simpler $$(U\otimes U)|\Psi^-\rangle=e^{i\theta}|\Psi^-\rangle.$$
This problem inherently involves a representation of unitary matrices that seems like the stumbling block for people asking about this problem here and elsewhere on the internet. So, I'd like to give an answer that is very explicitly in terms of that representation.
Let $r$ be the representation of $U(2)$ on two qubits so that for any qubit states $\vert a \rangle , \vert b\rangle$ the action of $u\in U(2)$ is $$ r(u) (\vert a \rangle \otimes \vert b \rangle) = u\vert a \rangle \otimes u \vert b\rangle \,. $$
An essential property of this representation (and all representations) is $r(u u' ) = r(u)r(u')$.
If $\vert \Psi \rangle$ is any of the 4 Bell states then $$ \begin{array}{c} r(X) \vert \Psi \rangle = (\pm)_X \vert \Psi \rangle\,,\\ r(Z) \vert \Psi \rangle = (\pm)_Z \vert \Psi \rangle \end{array} $$ where the signs $(\pm)_X$ and $(\pm)_Z$ depend on the Bell state.
An arbitrary $u\in U(2)$ has the Euler decomposition $$ u = e^{i\omega} e^{i\theta Z}e^{i\phi X }e^{-i\theta Z} $$ for some $\omega, \theta, \phi $.
Then, acting on any of the 4 Bell states via the representation, $$ \begin{array}{rl} r(u)\vert \psi \rangle &= r(e^{i\omega} e^{i\theta Z}e^{i\phi X } e^{-i\theta Z} )\vert \psi \rangle\\ &= r(e^{i\omega}) r( e^{i\theta Z} ) r( e^{i\phi X } ) r( e^{-i\theta Z} ) \vert \psi \rangle\\ &= r(e^{i\omega}) r( e^{i\theta Z} ) r( e^{i\phi X } ) e^{-i\theta (\pm)_Z} \vert \psi \rangle\\ &= r(e^{i\omega} ) r( e^{i\theta Z} ) e^{i\phi (\pm )_X} ) e^{-i\theta (\pm)_Z} \vert \psi \rangle\\ &= r(e^{i\omega}) e^{i\theta (\pm)_Z} e^{i\phi (\pm )_X} e^{-i\theta (\pm)_Z} \vert \psi \rangle\\ &=r (e^{i\omega}) e^{i\phi (\pm )_X} \vert \psi \rangle \\ &=e^{2 i\omega} e^{i\phi (\pm )_X} \vert \psi \rangle \,. \\ \square \end{array} $$
Here's another, quite neat, way to do it. (Nowhere near as neat as the answer of John Watrous)
First, verify that $|\Psi^-\rangle$ is a $-1$ eigenstate of both $X\otimes X$ and $Z\otimes Z$ (and hence also $Y\otimes Y$). We're going to use these relations to say, for example $$ X\otimes I|\Psi^-\rangle=-I\otimes X|\Psi^-\rangle. $$ So, consider an arbitrary single-qubit unitary (up to a global phase), which can be written as $e^{-\theta\vec{n}\cdot\vec{\sigma}}$, where $\vec{n}$ is a unit vector in $\mathbb{R}^3$. \begin{align*} e^{i\theta\vec{n}\cdot\vec{\sigma}}\otimes I|\Psi^-\rangle&=(I\cos\theta+i\sin\theta\vec{n}\cdot\vec{\sigma})\otimes I|\Psi^-\rangle \\ &=I\otimes (I\cos\theta-i\sin\theta\vec{n}\cdot\vec{\sigma})|\Psi^-\rangle \\ &=I\otimes e^{-i\theta\vec{n}\cdot\vec{\sigma}}|\Psi^-\rangle. \end{align*} Thus, $$ e^{i\theta\vec{n}\cdot\vec{\sigma}}\otimes e^{i\theta\vec{n}\cdot\vec{\sigma}}|\Psi^-\rangle=I\otimes e^{i\theta\vec{n}\cdot\vec{\sigma}}e^{-i\theta\vec{n}\cdot\vec{\sigma}}|\Psi^-\rangle=|\Psi^-\rangle. $$
Another way for fun.
My thought process was $|\psi_-\rangle = \frac{1}{\sqrt{2}} (|x_0x_1\rangle - |x_1x_0\rangle)$ is the same for any pair orthonormal states. To see this lets say $|x_0\rangle = a|0\rangle + b|1\rangle$ and $|x_1\rangle = -b^*|0\rangle + a^*|1\rangle$.
We then have $$|\psi_-\rangle = \frac{1}{\sqrt{2}}((a|0\rangle + b|1\rangle)(-b^*|0\rangle + a^*|1\rangle) - (-b^*|0\rangle + a^*|1\rangle)(a|0\rangle + b|1\rangle))$$ Collecting terms we then have $$|\psi_-\rangle = \frac{1}{\sqrt{2}} ((-ab^* +b^*a)|00\rangle+(aa^* + bb^* )|01\rangle +(-aa^* - bb^* )|10\rangle+(ba^* -a^*b)|11\rangle$$ $$=\frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$$
To answer your initial question now, lets write $|\psi_-\rangle = \frac{1}{\sqrt{2}} (|\theta_0\theta_1\rangle - |\theta_1\theta_0\rangle)$ where $U|\theta_i\rangle = e^{i\theta_i}|\theta_i\rangle $.
We now have that $U\otimes U |\psi_-\rangle = e^{i(\theta_0 + \theta_1)} \frac{1}{\sqrt{2}} (|\theta_0\theta_1\rangle - |\theta_1\theta_0\rangle) = e^{i\theta} |\psi_-\rangle $
