Operator change of basis (QM / QI)

[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!

 

[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|

 

[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.

 

[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 learned 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?

 

[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)?

 

[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.

 

[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?

Thanks for your time!