How unique is the Schmidt decomposition? I read the Schmidt basis is unique "up to a phase" or as stated here in  answer given by Norbert Schuch "modulo degeneracies".
If I choose a Bell state 
$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle\otimes|0\rangle + |1\rangle\otimes|1\rangle \right)$
and then define a second basis on each of the two subspaces by
$ |0\rangle \equiv \frac{1}{\sqrt{2}}(|\tilde{0}\rangle +|\tilde{1}\rangle) \quad \quad |1\rangle \equiv \frac{1}{\sqrt{2}}(|\tilde{0}\rangle -|\tilde{1}\rangle) $
then $|\psi\rangle$ becomes
$|\psi\rangle = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}(|\tilde{0}\rangle +|\tilde{1}\rangle) \otimes\frac{1}{\sqrt{2}}(|\tilde{0}\rangle +|\tilde{1}\rangle) + \frac{1}{\sqrt{2}}(|\tilde{0}\rangle -|\tilde{1}\rangle) \otimes\frac{1}{\sqrt{2}}(|\tilde{0}\rangle -|\tilde{1}\rangle) \right) =$
$\frac{1}{2 \sqrt{2}}( 
|\tilde{0}\rangle\otimes|\tilde{0}\rangle + |\tilde{0}\rangle\otimes|\tilde{1}\rangle + |\tilde{1}\rangle\otimes|\tilde{0}\rangle + |\tilde{1}\rangle\otimes|\tilde{1}\rangle +
|\tilde{0}\rangle\otimes|\tilde{0}\rangle - |\tilde{0}\rangle\otimes|\tilde{1}\rangle - |\tilde{1}\rangle\otimes|\tilde{0}\rangle + |\tilde{1}\rangle\otimes|\tilde{1}\rangle 
)$
finally
$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|\tilde{0}\rangle\otimes|\tilde{0}\rangle + |\tilde{1}\rangle\otimes|\tilde{1}\rangle \right)$
which again is a Schmidt decomposition with the same coefficients as above but with different bases. What is my mistake?
 A: You're not doing anything wrong. The decomposition essentially comes from the singular value decomposition of your tensor product element when written as a matrix. 
In this case a scalar matrix (multiple of the identity), so it has every vector as an eigenvector, i.e. we are in a degenerate case and every unitary change of basis will give a valid decomposition. The coefficients will still be uniquely determined. 
EDIT: Remarks on unicity 
In general the Schmidt decomposition is formulated for the tensor product of any two (finite dimensional) state spaces, not necessarily even of the same dimension, and even if they are equal, the (orthonormal) basis elements of both factors can generally not be chosen equal. We have unicity of the components if we additionally require them to be real and nonnegative. 
When the matrix is Hermitian (or more generally, normal), it can be diagonalized by a single unitary transformation, meaning that we can write the general tensor element as
$$\sum a_i|\psi_i\rangle\otimes |\psi_i\rangle$$
where the $a_i$ are general complex numbers. To convert this into a combination of the form
$$\sum A_i|\phi_i\rangle\otimes |\psi_i\rangle$$
where the $A_i$ are nonnegative real numbers, we can simply move the phase into one of the bases or the other without changing unitarity (so now we have bases $|\phi_i\rangle$ and $|\psi_i\rangle$). These $A_i$ are unique, but the bases are only unique up to a phase (if all eigenvalues are different, i.e. no degeneracy). In the first representation the $a_i$ themselves are only unique up to a phase.
In the general case (still equal dimensions of the factors) we cannot unitarily diagonalize and we must use different bases. Here again the $a_i$ are unique when restricting to non-negative real numbers, while we can move phase factors from left to right to change the bases.
A: To give explicitly the general answer: consider a generic bipartite pure state $|\Psi\rangle=\sum_{ij} c_{ij} |i,j\rangle$. So here $c_{ij}$ are the coefficients in some fixed orthonormal basis for the space.
Let $C$ be the matrix whose elements are the coefficients: $C_{ij}=c_{ij}$. This means that $C=\sum_{ij} c_{ij} |i\rangle\!\langle j|$.
We can write concisely the relation between $|\Psi\rangle$ and $C$ via the vectorization operation:
$$|\Psi\rangle = \operatorname{vec}(C),$$
where $\operatorname{vec}$ is a linear map sending operators $|i\rangle\!\langle j|$ to vectors $|i,j\rangle$.
Any property of the Schmidt decomposition of $|\Psi\rangle$ correspond to properties of the singular value decomposition of $C$, so I'll just focus on that.
Uniqueness of the singular value decomposition
The Schmidt decomposition of $|\Psi\rangle$ is the singular value decomposition (SVD) of $C$, the non-unicity of which is also discussed in this math.SE post.
Note that $C$ is not in general a square matrix here. It won't be whenever $|\Psi\rangle$ is a bipartite state across two spaces of different dimensions.
In summary, the SVD reads $$C=UDV^\dagger$$ for some unitaries $U,V$ and diagonal matrix $D$. A few notes about this:

*

*The (diagonal) elements of $D$ are singular values of $C$.

*The columns of $U$ and $V$ are the left and right principal vectors of $C$, respectively.

*If $C$ is singular, meaning there are vanishing singular values, some diagonal elements of $D$ will be zero. In such situations, you could redefine things to have $D$ only include non-zero singular values. If you choose to do this, the matrices $U,V$ won't be unitaries anymore (they wouldn't even be square), but they remain isometries. An advantage of using this notation is that you can meaningfully compute $D^{-1}$, which makes computing pseudo-inverses a bit easier (formally, not computationally).

Given an SVD for $C$, we can write it in bra-ket notation as $C=\sum_k s_k |u_k\rangle\!\langle v_k$, where $s_k$ are the singular values, and $|u_k\rangle$ and $|v_k\rangle$ the left and right principal components, respectively, and we the obtain the corresponding Schmidt decomposition for $|\Psi\rangle$ as $|\Psi\rangle=\sum_k s_k (|u_k\rangle\otimes|v_k\rangle)$.
Now for the non-uniqueness. This comes from the degeneracy of the singular values of $C$ (equivalently, from the degeneracy of the eigenvalues of $CC^\dagger$).
A concise way to characterise the degree of non-uniqueness is to write it as
$$C = (UW^\dagger) D (VW^\dagger)^\dagger,$$
where $W$ is a unitary operator such that $W^\dagger DW=D$.
It is not hard to see that this condition amounts to asking for $W$ to be block diagonal, with vanishing matrix elements correspondingly to each pair of distinct eigenvalues of $D$. More abstractly, this means to ask for $W$ to preserve the eigenspaces of $D$ (and thus of $C$).
In other words, if
$$D= \bigoplus_i D_i,$$
where each $D_i=d_i I_{\mu_i}$ is a multiple of the identity on a subspace, $\mu_i$ is the algebraic multiplicity of the eigenvalue $d_i$, and $d_i\neq d_j$ for all $i\neq j$, then we also have
$$W = \bigoplus_i W_i,$$
with each $W_i$ a unitary acting on the $i$-th eigenspace of $D$.
In fact, we could even more generally use any isometry $W$ satisfying the above condition. Using isometries gives decompositions which are not technically speaking singular value/Schmidt decompositions, but it's an interesting fact nonetheless.
Once you have the new decomposition $C=(UW^\dagger)D(VW^\dagger)^\dagger$, the columns of $UW^\dagger$ and $VW^\dagger$ are your new left and right principal vectors, and you span all the possibilities with this procedure.
Example
It can be useful to work out an explicit example to better understand the above.
Consider
$$|\Psi\rangle = |00\rangle + \frac32(|11\rangle + |22\rangle) - \frac12(|12\rangle+|21\rangle).$$
This state is obviously not normalised, but that doesn't really matter here. Redefine it with a normalisation factor and you'll see everything else works the same.
Working out the singular value decomposition of the corresponding $C$ (which in this case turns out to be Hermitian, so the SVD corresponds to the standard eigenvalue decomposition) we get
$$C = UDU^\dagger,
\quad
U = \begin{pmatrix}0&0&1\\-1/\sqrt2 & 1/\sqrt2 & 0 \\ 1/\sqrt2 & 1/\sqrt2 & 0\end{pmatrix},
\qquad
D = \begin{pmatrix}2&0&0\\0&1&0\\0&0&1\end{pmatrix}.
$$
We thus notice that in this case $D$ is partially degenerate.
We thus see that a first possible Schmidt decomposition is
$$|\Psi\rangle = 2 (|u_1\rangle\otimes|u_1\rangle)
+ (|u_2\rangle\otimes|u_2\rangle) + |0,0\rangle,$$
where
$$|u_1\rangle \equiv \frac1{\sqrt2}(-|1\rangle+|2\rangle),
\qquad
|u_2\rangle \equiv \frac1{\sqrt2}(|1\rangle+|2\rangle).$$
You can now obtain all other possible decompositions via unitaries of the form
$$W^\dagger = \begin{pmatrix}e^{i\alpha} & \mathbf 0_{1\times 2} \\ \mathbf 0_{2\times1} & \tilde U \end{pmatrix},$$
for any $2\times2$ unitary $\tilde U$.
I defined the unitary via $W^\dagger$ here simply for notational convenience. For different choices of $W$ we get different decompositions.
For example, if we use $\tilde U = H$ (the Hadamard matrix), we obtain the Schmidt decomposition
$$|\Psi\rangle = 2 (|u_1\rangle\otimes|u_1\rangle)
\\ + \frac14 [(\sqrt2|0\rangle+|1\rangle+|2\rangle)\otimes (\sqrt2|0\rangle+|1\rangle+|2\rangle)]
\\+ \frac14 [(-\sqrt2|0\rangle+|1\rangle+|2\rangle)\otimes (\sqrt2|0\rangle+|1\rangle+|2\rangle)].$$
