The question below has been posted on Stackexchange few days ago but I decided to share it on MO also. Hope this is not a misuse.
Fix $t\geqslant1$ and define $u_t=\pmatrix{1 & 0\\0 & t}\otimes\pmatrix{1 & 0\\0 & 0}+\pmatrix{1 & 0\\0 & -t}\otimes\pmatrix{0 & 0\\0 & 1}\in\mathcal{M}_2\otimes\mathcal{M}_2$. Calculate $\|u_t\|_{\pi}$ and $\|u_t\|_h$, where these are the (Banach space) projective and the Haagerup tensor norms, respectively.
Here is a computation of the Haagerup norm. It is not particularly instructive, since it relies very heavily on a special feature of the case at hand. I for one would be very glad to learn some more robust techniques for computing in Haagerup tensor products.
The first step, as pointed out by Mateusz Wasilewski, is to use the isometric embedding $\mathbb{C}^2\otimes_h \mathbb{C}^2 \hookrightarrow M_2(\mathbb C)\otimes_h M_2(\mathbb C)$.
The next step is to use the isometric embedding $\mathbb{C}^2\otimes_h \mathbb{C}^2 \hookrightarrow \mathbb{C}^2 \ast \mathbb{C}^2$, where $\ast$ denotes the unital free product of unital $C^*$-algebras. (This embedding extends the obvious multiplication map $\mathbb C^2\otimes_{\mathrm{alg}} \mathbb C^2\to \mathbb C^2 * \mathbb C^2$; see, for instance, Theorem 5.13 of Pisier's book on operator spaces.)
The next step is to identify $\mathbb C^2 *\mathbb C^2 \cong C^*(S_2 *S_2)$, where $S_2$ is the two-element group. This is done by identifying $\mathbb C^2\cong C^*(S_2)$, and comparing the universal properties. If, for the sake of definiteness, we send the element $(1,-1)\in \mathbb C^2$ to the generator of $S_2$, then the isomorphism $\mathbb C^2 * \mathbb C^2 \cong C^*(S_2 * S_2)$ sends $u_t^*u_t$ to $\frac{1}{2}(1+t^2 + a - t^2bab)$, where $a$ and $b$ are the respective generators of the two canonical copies of $S_2$ in $S_2\ast S_2$.
Now comes the special feature: $S_2\ast S_2$ is virtually unique among free products in having a very tractable representation theory. Identifying this group with the infinite dihedral group $\mathbb Z \rtimes S_2$ and applying Mackey's theory for semidirect products, one finds that every irreducible unitary representation of $S_2\ast S_2$ occurs as a subrepresentation of one of the two-dimensional representations $$ \pi_z(a) = \begin{bmatrix} 0 & z \\ \overline{z} & 0\end{bmatrix},\quad \pi_z(b) = \begin{bmatrix} 0 & 1\\ 1& 0\end{bmatrix} $$ where $z$ ranges over the unit circle in $\mathbb C$. (For Mackey's theory, see e.g. Chapter 4, and particularly Theorem 4.40, in Kaniuth and Taylor's book on induced representations of locally compact groups.)
Using the formula for the image of $u_t^*u_t$ in $C^*(S_2*S_2)$ found above, we end up with $$ \|u_t\|_h = \sup_{|z|=1} \|\pi_z(u_t^* u_t)\|^{1/2}=\sup_{|z|=1} \left\| \frac{1}{2}\begin{bmatrix} 1+t^2 & z-t^2 \overline{z} \\ \overline{z}-t^2 z & 1+t^2 \end{bmatrix} \right\|_{M_2(\mathbb C)}^{1/2}. $$ Computing the spectrum of this $2\times 2$ matrix, one finds that the supremum occurs at $z=i$, giving $\|u_t\|_h = (1+t^2)^{1/2}$.
There must be a better way to do this.
Ok, I think I have something that works. It seems to me that the Haagerup tensor norm coincides in this case with the projective norm. If I am not mistaken, the Haagerup tensor norm of a tensor $v$ can be written as infimum of $\sqrt{\sum_{i} \|x_i\|^2}\cdot \sqrt{\sum_{i} \|y_{i}\|^2}$ over all possible decompositions $v=\sum_{i} x_i \otimes y_i$. By working with $\frac{x_i}{c_i}$ and $c_i x_i$ for appropriate numbers $c_i$, we see that it is equal to the projective norm.
EDIT: I will only compute the projective norm.
Since everything is real, we can work with $\mathbb{R}^2 \widehat{\otimes} \mathbb{R}^2$, where $\mathbb{R}^2$ is endowed with the $\ell_{\infty}$-norm. This is not very good, as the projective tensor product interacts nicely with the $\ell_1$ norm. However, $\ell_1^{2}$ and $\ell_{\infty}^2$ are isometric, so we can work with $\ell_1^2$. This isometry is given by $\ell_{\infty}^2 \ni (x_1,x_2) \mapsto (\frac{x_1+x_2}{2}, \frac{x_1-x_2}{2}) \in \ell_{1}^2$. Our tensor is given by $(1,0) \otimes (1,t) + (0,1)\otimes (1,-t)$, so the image under this isometry is given by $(\frac{1}{2}, \frac{1}{2})\otimes (\frac{1+t}{2}, \frac{1-t}{2}) + (\frac{1}{2}, - \frac{1}{2})\otimes (\frac{1-t}{2}, \frac{1+t}{2})$. We want to use the fact that $\ell_1^2\widehat{\otimes} \ell_1^2$ is isometric to a vector valued $\ell_1$ space. We rewrite our tensor as $(1,0)\otimes (\frac{1}{2}, \frac{1}{2}) + (0,1)\otimes (\frac{t}{2}, \frac{t}{2})$ and the $\ell_1$-norm is equal to $1+t$.
So far, I don't know how to handle the case of the Haagerup tensor norm.
