# Operator change of basis (QM / QI)

• James Jackson
In summary, the conversation is discussing a problem in linear algebra for quantum computation and quantum information theory. The main focus is on changing the basis of an operator and calculating probabilities and expectation values for measurements on different states. There is some confusion about the unitarity of the basis transformation and the resulting operator being self-adjoint or Hermitian, which could explain the complex values in the calculations.
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!

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|

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.

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?

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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)?

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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.

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?

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## What is an operator change of basis in quantum mechanics/quantum information?

An operator change of basis is a mathematical operation that transforms a quantum mechanical operator from one basis to another. It is used to analyze and manipulate quantum states and their corresponding measurements in different bases.

## Why is operator change of basis important in quantum mechanics/quantum information?

Operator change of basis is important because it allows us to study and understand the behavior of quantum systems in different bases, providing us with a more complete understanding of their properties and allowing for more precise and efficient calculations and measurements.

## How is an operator change of basis performed?

An operator change of basis is performed by first expressing the original operator in terms of the new basis vectors, and then applying a transformation matrix to map the original operator to the new basis. This transformation matrix is typically a unitary matrix, which ensures that the operator remains Hermitian and preserves the physical interpretation of the quantum system.

## What are some applications of operator change of basis in quantum mechanics/quantum information?

Operator change of basis has many applications in quantum mechanics and quantum information, including quantum state tomography, quantum error correction, and quantum computing algorithms. It is also used in the analysis of quantum entanglement and quantum teleportation.

## Are there any limitations or challenges associated with operator change of basis?

One potential limitation of operator change of basis is that it can be computationally intensive, especially for large quantum systems with many basis states. Additionally, the choice of basis can greatly affect the complexity and efficiency of calculations, making it important to carefully select the most appropriate basis for a given problem.

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