What is the General Solution for Finding Orthonormal Bases in C^2?

[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.

 

[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?

 

[Fredrik]

(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 &amp; 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).

 

[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.

 

[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.

 

[TimH]

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

 

[Hurkyl]

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?

 

[SpectraCat]

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

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).