# Looking for a basis for C^2

1. Jan 29, 2010

### TimH

I'm teaching myself quantum mechanics and am learning about bra-ket notation. There is a particular operator used, the ket-bra (e.g. |X><X|). To understand it, I'm trying to come up with an orthonormal basis for C^2 as a simple case (i.e., the 2-dimensional vector space over the field of complex numbers). That is, I want two vectors, each with two components, each component a complex number, that span C^2 and are orthonormal. I've tried some combinations like (1,0) (0,i) and such, but no luck. Right now my TI-89 is chugging away looking at a few thousand possible vector combinations, but there has to be a better way.

Can anybody suggest how I would find two such vectors? Thanks.

2. Jan 29, 2010

### Physics Monkey

Hi TimH,

I'm a bit confused by your question. It seems to me that you've already found such a basis. Why do you think $$(1,0)$$ and $$(0,i)$$ don't work?

3. Jan 29, 2010

### Fredrik

Staff Emeritus
(1,0) and (0,1) works too. (z,w)=z(1,0)+w(0,1). (You must have misunderstood some definition).

See this post for more about bra-ket notation.

If $|X\rangle$ is a member of $\mathbb C^2$, then $|X\rangle\langle X|$ is a linear operator on $\mathbb C^2$. To be more precise, it's the projection operator for the one-dimensional subspace spanned by $|X\rangle$. If you write the vectors as $$|V\rangle=\begin{pmatrix}V_1\\ V_2\end{pmatrix}$$, then you can write $$|X\rangle\langle X|=\begin{pmatrix}X_1\\ X_2\end{pmatrix}\begin{pmatrix}X_1 & X_2\end{pmatrix}$$.

But I'm not a big fan of the "kets are column vectors, bras are row vectors" approach to bra-ket notation. It will give you the right intuition about bras and kets, but it doesn't explain why the notation still works when the vector space is infinite-dimensional. (See the post I linked to instead).

Last edited: Jan 29, 2010
4. Jan 29, 2010

### TimH

Okay, I figured out what happened. Thank you for your posts. I first tried (1,0) and (0,i) on my TI-89 calculator, using the "completeness relation," i.e. that the ket-bra |x><x| of the two vectors, when added together, should give the identity matrix. It didn't work on the calculator which is why I didn't think this pair worked. Then I did it by hand and it worked. Then I poked around on the calculator and discovered that when you take the transpose of a complex matrix or vector, it gives you the adjoint, which screwed up my formula. A feature, I guess...

So thank you for persisting and getting me to do it by hand...It would still be nice to find a set of orthonormal vectors in C^2 which don't have any zero-coefficients, i.e. where each component is a full-blown complex number with real and imaginary parts. Are there any well-known examples? I couldn't find anything online.

This is part of my effort to understand the machinery of Hilbert space, even if the space itself isn't visualizable. Thanks.

Last edited: Jan 29, 2010
5. Jan 29, 2010

### hamster143

Just pick any two complex numbers a & b.

You can construct an orthogonal basis that consists of two vectors, (a,b) and (b*,-a*).

You can then normalize it by rescaling those vectors.

6. Jan 29, 2010

### TimH

Thanks Hamster. I figured there was some general form. I'll play around with that.

7. Jan 29, 2010

### Hurkyl

Staff Emeritus
If you're having trouble creating an orthonormal basis, then why not:
• Use an orthonormalization algorithm? (e.g. Graham-Schmidt)
• Use an inner-product-preserving transformation to alter a known orthonormal basis?
• Write down -- and solve -- a system of equations that expresses exactly what you want?

8. Feb 1, 2010

### SpectraCat

Well, I should have thought of this before, but the general solution to the two-state problem is:

$$\left|\alpha\right\rangle=\begin{pmatrix}cos\theta \\ sin\theta e^{i\phi}\end{pmatrix},\left|\beta\right\rangle=\begin{pmatrix}-sin\theta \\ cos\theta e^{i\phi}\end{pmatrix}$$

So $$\left|\alpha\right\rangle$$ and $$\left|\beta\right\rangle$$ are orthonomal for all choices of $$\theta$$ and $$\phi$$.

As I understand it, this works because in the 2-D case, all orthogonal bases can be obtained by rotation of some initial set of diagonal eigenvectors (the ones Fredrik gave in post #3) in the (complex) 2-D Hilbert space. (This explanation may not be strictly correct ... I am still learning the ins and outs of Hilbert spaces).