Unitary Operator: Can X Make L(A) Unitary?

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In summary, the operator is unitary if and only if it has the standard inner product as its defined operation.
  • #1
nojojo121
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TL;DR Summary
Unitary operator in complex matrix space equipped with standard inner product
Unitary operator

If an operator L(A) = [X, A], are there matrix X for which the operator is unitary? Keep in mind it is in a complex matrix space with standard inner product (A, B) = tr(A*B)
 
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  • #2
nojojo121 said:
TL;DR Summary: Unitary operator in complex matrix space equipped with standard inner product

Unitary operator

If an operator L(A) = [X, A],...
How is ##[\cdot,\cdot]## defined?
And are we talking about a linear operator?

The use of unitary suggests that we speak about groups, in which case it would be ##[X,A]=X^{-1}\cdot A^{-1}\cdot X\cdot A##. However, in that case, ##L## wouldn't be a linear operator.

It is also possible, that you meant ##L(A)=X^{-1}AX,## the adjoint operator ##\operatorname{Ad}(X)## of a matrix group. This would at least be a linear operator.

However, experience plus the term operator suggests that you use it as ##L(A)=\mathfrak{ad}(X)(A)=[X,A]=X\cdot A-A\cdot X## in which case the corresponding property would be skew-hermitian.
nojojo121 said:
... are there matrix X for which the operator is unitary?
So you want to know whether there is an operator ##L\, : \,A\longmapsto [X,A]## such that
$$
\langle L(A),L(B) \rangle =\langle [X,A],[X,B] \rangle =\langle A,B \rangle
$$
I am not sure whether there is such a general answer to it, that it is completely irrelevant where ##A,B## are taken from and how ##[\cdot,\cdot]## is defined. My suspicion is a no. So please define the commutator and the domain the matrices are taken from.

In case ##L(A)=[X,A]=X^{-1}A^{-1}XA## the answer is no. ##L## isn't linear in that case and unitary requires linearity.
In case ##L(A)=[X,A]=\operatorname{Ad}(X)(A)=X^{-1}AX##, the answer is yes, e.g. ##L=X=1.##
In case ##L(A)=\mathfrak{ad}X## then ##L=X=0## is unitäry on all complex vector spaces.

nojojo121 said:
Keep in mind it is in a complex matrix space with standard inner product (A, B) = tr(A*B)

This is neither standard nor generally an inner product at all. It suggests that you mean the Killing-form ##(A,B)\longmapsto \operatorname{tr}(\mathfrak{ad}(A),\mathfrak{ad}(B)).## This is an inner product in case ##A,B## are taken from a semisimple, in your case complex, Lie algebra.

Here is an overview that should help you navigate through the various set-ups:
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-basics/ usage of the terms.
 
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  • #3
fresh_42 said:
How is ##[\cdot,\cdot]## defined?
And are we talking about a linear operator?

The use of unitary suggests that we speak about groups, in which case it would be ##[X,A]=X^{-1}\cdot A^{-1}\cdot X\cdot A##. However, in that case, ##L## wouldn't be a linear operator.

It is also possible, that you meant ##L(A)=X^{-1}AX,## the adjoint operator ##\operatorname{Ad}(X)## of a matrix group. This would at least be a linear operator.

However, experience plus the term operator suggests that you use it as ##L(A)=\mathfrak{ad}(X)(A)=[X,A]=X\cdot A-A\cdot X## in which case the corresponding property would be skew-hermitian.

So you want to know whether there is an operator ##L\, : \,A\longmapsto [X,A]## such that
$$
\langle L(A),L(B) \rangle =\langle [X,A],[X,B] \rangle =\langle A,B \rangle
$$
I am not sure whether there is such a general answer to it, that it is completely irrelevant where ##A,B## are taken from and how ##[\cdot,\cdot]## is defined. My suspicion is a no. So please define the commutator and the domain the matrices are taken from.

In case ##L(A)=[X,A]=X^{-1}A^{-1}XA## the answer is no. ##L## isn't linear in that case and unitary requires linearity.
In case ##L(A)=[X,A]=\operatorname{Ad}(X)(A)=X^{-1}AX##, the answer is yes, e.g. ##L=X=1.##
In case ##L(A)=\mathfrak{ad}X## then ##L=X=0## is unitäry on all complex vector spaces.
This is neither standard nor generally an inner product at all. It suggests that you mean the Killing-form ##(A,B)\longmapsto \operatorname{tr}(\mathfrak{ad}(A),\mathfrak{ad}(B)).## This is an inner product in case ##A,B## are taken from a semisimple, in your case complex, Lie algebra.

Here is an overview that should help you navigate through the various set-ups:
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-basics/ usage of the terms.
Thanks for the reply. I should have been more specific. My definition for the operator is like you guessed, that is ##L(A)=\mathfrak{ad}(X)(A)=[X,A]=X\cdot A-A\cdot X## and ##L\, : \,M{_{n\times n} } \longmapsto M{_{n\times n}} ##. The inner product is ##(A,B)= \operatorname{tr}(A{^\dagger} B). ## which apparently is standard according to Wikipedia. Nevertheless I am searching for X for which the defined operator is unitary.
 
  • #4
nojojo121 said:
Thanks for the reply. I should have been more specific. My definition for the operator is like you guessed, that is ##L(A)=\mathfrak{ad}(X)(A)=[X,A]=X\cdot A-A\cdot X## and ##L\, : \,M{_{n\times n} } \longmapsto M{_{n\times n}} ##. The inner product is ##(A,B)= \operatorname{tr}(A{^\dagger} B). ## which apparently is standard according to Wikipedia.
Reference?
nojojo121 said:
Nevertheless I am searching for X for which the defined operator is unitary.
##X=0## does the job, and ##(A,B)\longmapsto \operatorname{tr}(A^\dagger B)## is definitely no inner product.
$$
\operatorname{tr}\left(\underbrace{\begin{bmatrix}1&0\\0&1\end{bmatrix}^\dagger}_{\neq 0} \cdot \underbrace{\begin{bmatrix}0&1\\1&0\end{bmatrix}}_{\neq 0}\right)=\operatorname{tr}\left(\begin{bmatrix}0&1\\1&0\end{bmatrix}\right)=0
$$
 
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  • #5
@fresh_42 I'm confused, aren't there always lots of pairs of vectors whose inner product is zero? Those are just orthogonal vectors.

I think this is just the standard entry-wise dot product written to look fancy
 
  • #6
Office_Shredder said:
@fresh_42 I'm confused, aren't there always lots of pairs of vectors whose inner product is zero? Those are just orthogonal vectors.

I think this is just the standard entry-wise dot product written to look fancy
Yes, my bad. I was lost in translation. We use the term inner product to describe the dot product, and scalar product to describe the inner product.

Anyway, ##L=X=0## is still an answer.

Edit: ... and I automatically think about Lie algebras if someone uses ##[\cdot,\cdot]## and unitary.
 
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  • #7
I'm confused- unitary operators are supposed to be invertible, but ##L(X)=[X,X]=0## represents a nontrivial kernel (unless ##X=0## but that case is just as easily dismissed). Am I missing something?
 
  • #8
Infrared said:
I'm confused- unitary operators are supposed to be invertible, but ##L(X)=[X,X]=0## represents a nontrivial kernel (unless ##X=0## but that case is just as easily dismissed). Am I missing something?
My very first statement was: define ##[X,Y]## since unitary belongs to groups and skew-hermitian is the corresponding non-regular property.

As I said, I associated Lie algebras since this is the most common use of ##[\cdot,\cdot]## on PF. Rarely anybody ever uses the group commutator. And the corresponding Lie group operation is the adjoint representation ##\operatorname{Ad}## on its tangent space. In that case, the group elements are commonly written by minuscules.

As long as unitary and presumably ##[X,Y]=XY-YX## as an operator occurs in one sentence, as long will I allow myself to consider the trivial operation.

However, you are all right. So I herewith answer: Yes. Set ##L=1## and ##V=\mathbb{M}(1,\mathbb{C}).##

I only wanted to emphasize that the question was badly posed.
 
  • #9
fresh_42 said:
Reference?

##X=0## does the job, and ##(A,B)\longmapsto \operatorname{tr}(A^\dagger B)## is definitely no inner product.
$$
\operatorname{tr}\left(\underbrace{\begin{bmatrix}1&0\\0&1\end{bmatrix}^\dagger}_{\neq 0} \cdot \underbrace{\begin{bmatrix}0&1\\1&0\end{bmatrix}}_{\neq 0}\right)=\operatorname{tr}\left(\begin{bmatrix}0&1\\1&0\end{bmatrix}\right)=0
$$
I poorly worded what is being searched. I need to find all the matrices ##X## that make the operator ##L## unitary. As someone else mentioned above I think there is a loss in translation but what I call inner product you call dot product. So essentially I have to show for which ##X## the operator ##L## is an isometry.
 
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  • #10
nojojo121 said:
I poorly worded what is being searched. I need to find all the matrices ##X## that make the operator ##L## unitary. As someone else mentioned above I think there is a loss in translation but what I call inner product you call dot product. So essentially I have to show for which ##X## the operator ##L## is an isometry.
The important information is where ##X## and ##A,B## are taken from. You said that all complex matrices for ##A## and ##B## are allowed. And you said that the inner product in the definition of the unitary operator ##L_X(A)=XA-AX## is ##\langle A,B\rangle = \operatorname{tr}(A^\dagger B).## This runs down to (if I didn't make a mistake)
$$
\operatorname{tr}(A^\dagger X^\dagger XB-A^\dagger X^\dagger BX-X^\dagger A^\dagger XB+X^\dagger A^\dagger BX-A^\dagger B)=0\quad \forall \;A,B\in \mathbb{M}(n,\mathbb{C})
$$
This is a linear equation system in the coordinates ##X=(x_{ij}).## The solution is therefore a complex vector space. Since you are only interested in a bijective operator, we finally have to calculate all ##X## such that ##A\longmapsto XA-AX## is regular for all solutions ##X.##

You can set ##A=E_{ij},B=E_{kl}## where ##E_{rs}## is the matrix with a ##1## at position ##(r,s)## and ##0## elsewhere in order to solve the system. Then see if ##i \cdot E_{rs}## gives even more equations.

I would start with ##n=2,3## and see how it goes. And I still think that there is something wrong here since you basically ask for regular elements in a Lie algebra. This is normally not of any interest.
 
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  • #11
fresh_42 said:
The important information is where ##X## and ##A,B## are taken from. You said that all complex matrices for ##A## and ##B## are allowed. And you said that the inner product in the definition of the unitary operator ##L_X(A)=XA-AX## is ##\langle A,B\rangle = \operatorname{tr}(A^\dagger B).## This runs down to (if I didn't make a mistake)
$$
\operatorname{tr}(A^\dagger X^\dagger XB-A^\dagger X^\dagger BX-X^\dagger A^\dagger XB+X^\dagger A^\dagger BX-A^\dagger B)=0\quad \forall \;A,B\in \mathbb{M}(n,\mathbb{C})
$$
This is a linear equation system in the coordinates ##X=(x_{ij}).## The solution is therefore a complex vector space. Since you are only interested in a bijective operator, we finally have to calculate all ##X## such that ##A\longmapsto XA-AX## is regular for all solutions ##X.##

You can set ##A=E_{ij},B=E_{kl}## where ##E_{rs}## is the matrix with a ##1## at position ##(r,s)## and ##0## elsewhere in order to solve the system. Then see if ##i \cdot E_{rs}## gives even more equations.

I would start with ##n=2,3## and see how it goes. And I still think that there is something wrong here since you basically ask for regular elements in a Lie algebra. This is normally not of any interest.
I think you are misunderstanding my usage of ##B##, which is just an arbitrary matrix used to define the inner product. This is essentially the linear equation I am trying to solve:
$$
\operatorname{tr}(A^\dagger X^\dagger XA-A^\dagger X^\dagger AX-X^\dagger A^\dagger XA+X^\dagger A^\dagger AX)=\operatorname{tr}(A^\dagger A)
$$
 
  • #12
nojojo121 said:
I think you are misunderstanding my usage of ##B##, which is just an arbitrary matrix used to define the inner product. This is essentially the linear equation I am trying to solve:
$$
\operatorname{tr}(A^\dagger X^\dagger XA-A^\dagger X^\dagger AX-X^\dagger A^\dagger XA+X^\dagger A^\dagger AX)=\operatorname{tr}(A^\dagger A)
$$

\begin{align*}
\langle L(A),L(B) \rangle &=\langle A,B \rangle\\
\operatorname{tr}(L(A)^\dagger \cdot L(B))&=\operatorname{tr}(A^\dagger\cdot B)\\
\operatorname{tr}([X,A]^\dagger \cdot [X,B])&=\operatorname{tr}(A^\dagger \cdot B)\\
\operatorname{tr}((XA-AX)^\dagger \cdot (XB-BX))&=\operatorname{tr}(A^\dagger \cdot B)\\
\ldots &\phantom{=} \ldots
\end{align*}
https://en.wikipedia.org/wiki/Unitary_operator

Where am I wrong?
 
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  • #13
fresh_42 said:
\begin{align*}
\langle L(A),L(B) \rangle &=\langle A,B \rangle\\
\operatorname{tr}(L(A)^\dagger \cdot L(B))&=\operatorname{tr}(A^\dagger\cdot B)\\
\operatorname{tr}([X,A]^\dagger \cdot [X,B])&=\operatorname{tr}(A^\dagger \cdot B)\\
\operatorname{tr}((XA-AX)^\dagger \cdot (XB-BX))&=\operatorname{tr}(A^\dagger \cdot B)\\
\ldots &\phantom{=} \ldots
\end{align*}
https://en.wikipedia.org/wiki/Unitary_operator

Where am I wrong?
My mistake. You are completely right, unitary operators must preserve the inner product and not just the norm.
 
  • #14
I'm not sure about the infinite-dimensional case. In finite dimensions, we have the following situation.
If ##X,A,B \in \mathbb{M}(n,\mathbb{C}),## i.e. complex linear transformations of ##\mathbb{C}^n## then we have ##L(A)=(\mathfrak{ad}(X))(A)## and ##L\in \mathfrak{ad}(\mathfrak{gl}(\mathbb{C}^n))=\mathfrak{ad}(\mathfrak{sl}(\mathbb{C}^n)\oplus \mathbb{C}\cdot \operatorname{Id}_{\mathbb{C}^n})=\mathfrak{ad}(\mathfrak{sl}(\mathbb{C}^n))\cong \mathfrak{sl}(\mathbb{C}^n).## Unitarity demands that ##L\in \operatorname{SU}(\mathfrak{gl}(\mathbb{C}^n)).##

You are looking for matrices ##X## that have determinant ##1## and trace ##0## at the same time. This is unusual. May I ask about the context of your research?
 
  • #15
fresh_42 said:
The important information is where ##X## and ##A,B## are taken from. You said that all complex matrices for ##A## and ##B## are allowed. And you said that the inner product in the definition of the unitary operator ##L_X(A)=XA-AX## is ##\langle A,B\rangle = \operatorname{tr}(A^\dagger B).## This runs down to (if I didn't make a mistake)
$$
\operatorname{tr}(A^\dagger X^\dagger XB-A^\dagger X^\dagger BX-X^\dagger A^\dagger XB+X^\dagger A^\dagger BX-A^\dagger B)=0\quad \forall \;A,B\in \mathbb{M}(n,\mathbb{C})
$$
This is a linear equation system in the coordinates ##X=(x_{ij}).## The solution is therefore a complex vector space. Since you are only interested in a bijective operator, we finally have to calculate all ##X## such that ##A\longmapsto XA-AX## is regular for all solutions ##X.##

You can set ##A=E_{ij},B=E_{kl}## where ##E_{rs}## is the matrix with a ##1## at position ##(r,s)## and ##0## elsewhere in order to solve the system. Then see if ##i \cdot E_{rs}## gives even more equations.

I would start with ##n=2,3## and see how it goes. And I still think that there is something wrong here since you basically ask for regular elements in a Lie algebra. This is normally not of any interest.
@Infrared already said that ##L_X## always has a kernel, hence cannot be bijective.
 
  • #16
martinbn said:
@Infrared already said that ##L_X## always has a kernel, hence cannot be bijective.

The kernel of ##L## consists of the diagonal matrices with only one entry, complex multiples of ##1.##
$$
\operatorname{ker}(L)=\{A\in \mathfrak{gl}(\mathbb{C}^n)\,|\,L(A)=0\}=\{A\,|\,XA=AX\;\forall \,X\in\mathfrak{gl}(\mathbb{C}^n)\} = \mathbb{C}\cdot \operatorname{Id}_{\mathbb{C}^n}
$$
Every choice of ##X=S+\lambda\cdot \operatorname{Id}_{\mathbb{C}^n}## with ##S\in\mathfrak{ad}(\mathfrak{sl}(\mathbb{C}^n))## leads to ##L(A)=\mathfrak{ad}(X)(A)=[X,A]=[S,A]=SA-AS.## Thus ##L(A)## has a non-trivial kernel, the operator ##L=\mathfrak{ad}S## itself can be bijective.
 
  • #17
fresh_42 said:
The kernel of ##L## consists of the diagonal matrices with only one entry, complex multiples of ##1.##
$$
\operatorname{ker}(L)=\{A\in \mathfrak{gl}(\mathbb{C}^n)\,|\,L(A)=0\}=\{A\,|\,XA=AX\;\forall \,X\in\mathfrak{gl}(\mathbb{C}^n)\} = \mathbb{C}\cdot \operatorname{Id}_{\mathbb{C}^n}
$$
Every choice of ##X=S+\lambda\cdot \operatorname{Id}_{\mathbb{C}^n}## with ##S\in\mathfrak{ad}(\mathfrak{sl}(\mathbb{C}^n))## leads to ##L(A)=\mathfrak{ad}(X)(A)=[X,A]=[S,A]=SA-AS.## Thus ##L(A)## has a non-trivial kernel, the operator ##L=\mathfrak{ad}S## itself can be bijective.
No, because ##L_X(X)=0##. If ##S=X-\lambda##, then still ##[S,X]=0##.
 
  • #18
martinbn said:
No, because ##L_X(X)=0##. If ##S=X-\lambda##, then still ##[S,X]=0##.
Yes, I see it now. I'm getting too old for so little sleep. I was confused by the fact that ##\mathfrak{sl}(V)## contains regular matrices.

It had to be helpless. ##\det = 1## and ##\operatorname{trace}=0## doesn't fit very well.
 
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Related to Unitary Operator: Can X Make L(A) Unitary?

What is a unitary operator?

A unitary operator is a type of linear operator in mathematics that preserves the inner product of vectors. This means that the length and angle between two vectors remain the same after the operation is applied.

Why is it important for L(A) to be unitary?

Having L(A) be unitary is important because it guarantees that the operator is reversible and that it preserves the norm of the vectors. This is useful in many applications, such as quantum mechanics and signal processing.

What are the conditions for a unitary operator?

A unitary operator must satisfy two conditions: it must be linear and it must preserve the inner product of vectors. This means that the operator must be both one-to-one and onto, and it must also preserve the length and angle between vectors.

Can any operator X make L(A) unitary?

No, not all operators can make L(A) unitary. In order for L(A) to be unitary, the operator X must satisfy certain conditions, such as being linear and preserving the inner product of vectors. If these conditions are not met, then L(A) will not be unitary.

What are some applications of unitary operators?

Unitary operators have many applications in mathematics and physics. In mathematics, they are used in areas such as linear algebra and functional analysis. In physics, they are used in quantum mechanics, signal processing, and other areas where preserving the norm and inner product of vectors is important.

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