# How to expand in a non-orthogonal basis in an intuitive way?

I have a basis that consists of four non-orthogonal vectors $\{|u_i\rangle\}, 1 \le i \le 4$. Can the formula for an orthonormal expansion be modified so that it holds true for any given basis?

$$|v>=\sum_i |u_i\rangle\langle u_i|v\rangle$$

I could always write $|v\rangle$ as a linear conbination of $\{u_i\}$ and solve the equation system, but I would like to approach the problem from a different perspective. I am not interested in finding an orthonormal basis with the Gram–Schmidt process.

• Given a basis, the expansion of a vector in the basis is unique. However the coefficients won't be that inner product. It seems that you already know how to use the change of basis matrix. You can probably rewrite the change of basis matrix as a combination of inner products, is this what you are asking for? – Calvin Khor Aug 10 at 10:05
• @Calvin Khor no, I don't want to change the basis, I would like to expand a vector in a non-orthogonal basis using a formula similar to the one I provided, if possible. – TheAverageHijano Aug 10 at 10:12
• If you start of with $v$ written in the so-called "standard basis", then expanding a vector in a new basis, orthogonal or not, uses a change of basis. The formula will be exactly equivalent to the change of basis matrix – Calvin Khor Aug 10 at 10:13
• Wouldn't the method you propose involve inner products with the vectors of the canonical base? I'm trying to stick to inner products between $|v\rangle$ and $|u_i\rangle$ and avoiding using matrices. – TheAverageHijano Aug 10 at 10:20

Let me write vectors without using bra-ket notation because I'm not familliar with it. Also let there be just 2 basis vectors, for ease of typing. Let $\langle a,b\rangle$ denote the inner product of $a,b$.

You are asking for the correct coefficients $a_1,a_2$ for the expansion $$v = a_1 u_1 + a_2 u_2$$ without using explicitly a change of basis from an original known basis, expressed solely in terms of $\langle v,u_i\rangle$. By taking inner products with the $u_i$, this is the same problem as trying to solve $$\mathbf v = M\mathbf a$$ where: $$\mathbf v = \binom{\langle u_1,v\rangle}{\langle u_2,v\rangle},\quad \mathbf a = \binom{a_1}{a_2},\quad M =\begin{bmatrix} \langle u_1, u_1\rangle & \langle u_1,u_2\rangle \\ \langle u_2, u_1\rangle & \langle u_2,u_2\rangle \end{bmatrix}.$$ So the coefficients are given by $M^{-1}\mathbf v$. With just 2 vectors, this is easy to write down explicitly, $$\mathbf a = \frac1{\langle u_1, u_1\rangle \langle u_2, u_2\rangle - \langle u_1, u_2\rangle\langle u_2, u_1\rangle } \binom {\phantom{+}\langle u_2, u_2\rangle \langle u_1, v\rangle - \langle u_1, u_2\rangle \langle u_2, v\rangle} {-\langle u_2, u_1\rangle \langle u_1,v\rangle + \langle u_1, u_1\rangle \langle u_2,v\rangle}$$ As requested, this is not written in terms of the canonical basis. It does involve a matrix in the derivation but once derived, the formula is plug-and-play. It is clear how the same idea works for any number of basis vectors.

Remark: this matrix $M$ is known as the Gramian matrix of the vectors $u_1,\dots,u_n$.

• Thanks, this is what I was looking for. For a basis of a different dimension I would only need to calculate $M^{-1}$, and I have a nice expression for $\langle u_i|u_j \rangle$. – TheAverageHijano Aug 10 at 13:24

In co-ordinate-free terms, as long as you define your bras $\langle u_i|$ so that $\langle u_i|u_j \rangle = \delta_{ij}$ then if

$| v \rangle = \sum_j \lambda_j |u_j\rangle$

you have

$\langle u_i| v \rangle = \sum_j \lambda_j \langle u_i|u_j\rangle = \sum_j \lambda_j \delta_{ij} = \lambda_i$

so

$|v \rangle = \sum_i |u_i\rangle \langle u_i| v \rangle$

In co-ordinate terms, if you are expressing your kets $|u_i\rangle$ as co-ordinates $|u_i\rangle = \{u_{ij}|1\le j\le 4\}$ realtive to some basis then expressing the bras $\langle u_i|$ in the same basis is equivalent to finding the inverse of the $4 \times 4$ matrix formed by the $\{u_{ij}\}$.

• The thing is that $|u_i\rangle$ are not orthonormal – TheAverageHijano Aug 10 at 10:42
• @TheAverageHijano - the property of being orthomormal depends on your choice of inner product i.e. on your choice of bras.As long as your kets are linearly independent you have the freedom to choose your bras to make them orthonormal. – gandalf61 Aug 10 at 10:54
• @kvantour Thank you - I have fixed my answer. – gandalf61 Aug 10 at 13:44