# Operator change of basis (QM / QI)

1. Apr 18, 2005

### James Jackson

Hi there, just doing some basic linear algebra for quantum computation / quantum information theory, and am wondering whether I'm changing the basis of an operator correctly.

If I have two orthogonal basis vectors of space C2 given by (~ = complex conjugate):

S1 = [|0>, |1>]

and S2 = [|u> = a|0> + b|1> and |v> = b~|0> - a~|1>]

(S2 is orthonormal given aa~+bb~=1, easy enough to prove (<u|v>=0))

and the operator, A, given in terms of the basis set S2:

A = |u><u| - |v><v|

(This is from the given fact that A has eigenvectors |u>,|v> with eigenvalues 1,-1 respectively)

To change A into the basis set S1, do I simply do:

A' = UA

where U is the unitary matrix |0><u|+|1><v|

This results in A' = |0><u| - |1><v|

So, if I want to find the probabiliy of a measurement of A on the state |0> I then do:

A'|0> = |0><u|o> - |1><v|0>

As <u|0> = a~ and <v|0> = b this gives

Therefore A'|0> = a~|0> - b|1>

So the probability of this measurement returning 1 is |b|^2
This also means the expectation value of the measurement is 0*p(0)+1*p(1) = 0*|a|^2 + 1*|b|^2 = |b|^2

Is this correct or have I made a mistake somewhere?

Cheers!

2. Apr 18, 2005

### James Jackson

P.S. I'm in my final year physics degree, but haven't really covered linear algebra before. In the above, I'm assuming that this expansion for A'=UA is valid:

A'=UA=(|0><u|+|1><v|)(|u><u|-|v><v|)

= |0><u|u><u|-|1><v|v><v|+|1><v|u><u|-|0><u|v><v|

which as <u|u>, <v|v> = 1 and <u|v>, <v|u> = 0 gives:

A' = UA = |0><u|-|1><v|

3. Apr 18, 2005

### dextercioby

Here's what i get

$$\hat{A}'=:|0\rangle\langle u|-|1\rangle\langle v|$$

U want to compute the expectation value on the state $|1\rangle$

$$\langle \hat{A}'\rangle_{|1\rangle}=:\langle 1|\hat{A}'|1\rangle$$

Okay.

$$\hat{A}'|1\rangle=|0\rangle\langle u|1\rangle - |1\rangle \langle v|1\rangle$$

$$\langle u|1\rangle=a^{*}\langle 0|1\rangle+b^{*}\langle 1|1\rangle=b^{*}$$

$$\langle v|1\rangle=b\langle 0|1\rangle-a\langle 1|1\rangle =-a$$

$$\hat{A}'|1\rangle=b^{*}|0\rangle+a|1\rangle$$

$$\langle 1|\hat{A}'|1\rangle =b^{*}\langle 1|0\rangle+a\langle 1|1\rangle=a$$

U can compute the other expectation value in the same way.

Daniel.

4. Apr 18, 2005

### James Jackson

Thanks for that, could I clarify a few things? I assume that I've got the change of operator basis correct as you haven't commented on that, which is nice. On to the expectation measurement: The question is set as such: A measurement is made with A on the state $$|0\rangle$$. What is the probability of the result being 1? What is the expected value of the outcome?

So, following from you above, the expected value of the measurement A on the state $$|0\rangle$$ would be:

$$\langle0|A'|0\rangle = a^{*}$$

I was under the impression that all measurements returned real numbers (assuming the operator is hermitian. I haven't checked this for A' yet as I haven't learnt how to change from dirac outer product notation to matrix notation for an operator - that's a job for tomorrow!). In both cases (i.e. measurements on $$|0\rangle$$ and $$|1\rangle$$) a complex expectation value is being returned - am I missing something?

Secondly, from the way you have expressed expectation measurements above as an inner product (say, expectation value of |0>): $$\langle 0|A'|0\rangle$$, for a state $$|x\rangle = \alpha|0\rangle + \beta|1\rangle$$, how can one determine the probability (and not just expectation value, which would of course be given by $$\langle x|A'|x\rangle$$) of a measurement of A' on the state $$|x\rangle$$ returning a given state (either $$|0\rangle$$ or $$|1\rangle$$).

From my original post (and backed up by your reply), $$A'|0\rangle = a^{*}|0\rangle - b|1\rangle$$ which imples the probability of the measurement of A on $$|0\rangle$$ returning 1 is $$\|b\|^2$$. Is there a more 'elegant' way or is this completely incorrect anyway?

Many thanks for you help.

Edit: Hmm, found some stuff about measurement probabilities. So, for measurement of a qubit in the computational basis, we have:

$$M_{0}=|0\rangle\langle 0|, M_{1}=|1\rangle\langle 1|$$

so p(m) on a state $$|\phi\rangle$$ equals $$p(m)=\langle\phi |{M_{m}}^{\dagger}M_{m}|\phi\rangle$$

but I can't see how to apply this to a measurement of $$A'|0\rangle$$ unless I put that state equal to $$|\phi\rangle$$ and then use $$M_1$$ in the above to get the probability of the measurement returning 1. Is this correct?

Last edited: Apr 18, 2005
5. Apr 19, 2005

### James Jackson

Right... Done a bit of calculation in the light of day. OK, so if I make some (hopefully correct) definitions:

For a particle in state $$|\phi\rangle$$:

Probability measurements: $$p(k)=\| c_{k}\|^{2}=\|\langle k|\phi\rangle\|^{2}$$ for k being the eigenstates of an operator A which must be the same as the basis for $$|\phi\rangle$$.

Expectation values: $$E(A)=\langle\phi |A|\phi \rangle$$ for operator A.

So, define $$|u\rangle = \alpha |0\rangle + \beta |1\rangle; |v\rangle = \beta^{*} |0\rangle - \alpha^{*} |1\rangle$$ as an orthonormal basis on $$C^2$$

We then are given that an operator, A, has eigenvectors $$|u\rangle$$ and $$|v\rangle$$ with eigenvalues 1 and -1 respectivaly. This operator, in Dirac form, is therefore:

$$A=|u\rangle\langle u|-|v\rangle\langle v|$$

To transform this into the $$|0\rangle ;|1\rangle$$ basis I apply the unitary transform U to A, given by:

$$U=|0\rangle\langle u|+|1\rangle\langle v|$$

So I get the new operator A':

$$A'=UA=|0\rangle\langle u|-|1\rangle\langle v|$$

So now, as A' is expressed in the basis $$|0\rangle ;|1\rangle$$, any results of the measurement are going to be either of this orthonormal set. So to get the probability of of measuring the state $$|1\rangle$$ when A' is applied to $$|0\rangle$$ is:

$$P_{1}=\|\langle 1|A'|0\rangle\|^{2}=\|\beta\|^2$$

The expectation value of a measurement of A' on $$|0\rangle$$ is given by:

$$E(A'|0\rangle )=\langle 0|A'|0\rangle =\alpha^{*}$$

This makes sense to me, but still the fact that the expectation value is a complex number confuses me. Surely it must be real (unless, of course, A' isn't Hamiltonian)?

Last edited: Apr 19, 2005
6. Apr 19, 2005

### dextercioby

There's a big problem in everything that u did.The basis thansformation,unlike the operator one,was not unitary.Therefore,the new operator,even if unitarily transformed,is not selfadjoint.Not even hermitean,i think.So that would account for complex eigenvalues,even,and complex expectation values.

Daniel.

7. Apr 19, 2005

### James Jackson

Sorry, getting confused with nomclementure - what you say makes sense but I'm losing something somewhere. Surely my basis change was the 'operator one' (i.e. applying U to A), so by that as A is Hamiltonian over $$|u\rangle ;|v\rangle$$ and U is, by definition, unitary then UA is also Hamiltonian, or is this the wrong assumption I am making?