# Help with tensor product of two bit operators (in quantum-computing)

I'm having trouble understanding how to put together two one-bit operators to get a two-bit operator. For example, suppose I have two electrons in the spin state:

$$\frac{1}{\sqrt{2}}(|\text{up},\text{up}\rangle+|\text{down},\text{down}\rangle)$$

If I'm understanding things correctly, then to measure the state of the first one (along the z axis) I would tensor $\sigma_z$ with the identity operator. This seems to work.

My problem comes in if I want to measure the state of both of them. I would have thought that I would tensor $\sigma_z$ with itself. But when I do that I get an operator that has degenerate eigenvalues. In other words, I get an eigenvalue of $+1$ for either of the possible outcomes. I'm not sure if the whole approach is wrong, or if I'm just making a mistake somewhere.

I hope this description is clear. I'm somewhat of a novice at this. Also I imagine that there might be a better existing tag for this question than "linear-algebra" but I couldn't find it. Any comments appreciated.

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There was a time when "two-bit" as an adjective meant cheap. Literally "two bits" was 25 cents. – Michael Hardy Jan 26 '12 at 2:25
Yeah, I was actually thinking about that when I typed it in :-) – Mike Witt Jan 26 '12 at 5:44

## 1 Answer

Measuring the state of a quantum system with an operator $A$ will project the system into an eigenspace of $A$. The operator $\sigma_z\otimes\sigma_z$ has two eigenspaces: one is spanned by $\{|\uparrow>\otimes|\uparrow>, |\downarrow>\otimes|\downarrow>\}$ and has eigenvalue $+1$, and the other is spanned by $\{|\uparrow>\otimes|\downarrow>, |\downarrow>\otimes|\uparrow>\}$ and has eigenvalue $-1$. The state vector $(1/\sqrt{2})(|\uparrow>\otimes|\uparrow>+|\downarrow>\otimes|\downarrow>)$ is already in the first of these eigenspaces, so applying $\sigma_z\otimes\sigma_z$ will leave it unchanged and give a measurement of $+1$. Looking at the spanning sets, you can see that $\sigma_z\otimes\sigma_z$ measures the exclusive OR of the two qubits.

If you wanted a measurement with four outcomes, you could either apply an operator with four different eigenvalues (e.g., $(\sigma_z\otimes 1)+\frac12 (1\otimes \sigma_z)$), or perform two successive measurements, each of which has two possible outcomes (e.g., measure $\sigma_z\otimes 1$ and follow this by measuring $1\otimes \sigma_z$.)

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That answers my question. Thanks! – Mike Witt Jan 26 '12 at 5:44