Maximize product of a vector with two vectors
$begingroup$
Say we're in a (complex) Hilbert space $H$ where we're given some two elements $a,bin H$ of norm 1.
The question is to find an element $psiin H$ (of norm 1) that maximizes $|langle psi,arangle langle b,psirangle|$ (or $|langle psi,arangle langle b,psirangle|^2$ if you prefer).
Mainly what I want to prove is one of the following inequalities:
$$
|langle psi,arangle langle b,psirangle|leq |langle a,b rangle|
$$
$$
|langle psi,arangle langle b,psirangle|leqbigg|frac{1+|langle a,brangle| }{2}bigg|
$$
Intuitively, my choice is to say that this maximizes when $psi = frac{a+b}{2}$, modulo the norm. And in fact one can get close to proving the inequalities by plugging that in. But I can't prove why this should be the case.
It is obvious that $psi$ should be in the span of $a,b$, since any component that's outside of this subspace will be wasted. So we can write $psi = xa + yb$ for some complex $x$ and $y$. But $a$ and $b$ need not be orthogonal, so it's not so simple.
Is there an inequality that one can use here to prove one of the above statements, or a quick way to find $psi$?
inequality hilbert-spaces inner-product-space cauchy-schwarz-inequality
asked Dec 20 '18 at 15:54
KoljaKolja
590310
$endgroup$
add a comment |
$begingroup$
Say we're in a (complex) Hilbert space $H$ where we're given some two elements $a,bin H$ of norm 1.
The question is to find an element $psiin H$ (of norm 1) that maximizes $|langle psi,arangle langle b,psirangle|$ (or $|langle psi,arangle langle b,psirangle|^2$ if you prefer).
Mainly what I want to prove is one of the following inequalities:
$$
|langle psi,arangle langle b,psirangle|leq |langle a,b rangle|
$$
$$
|langle psi,arangle langle b,psirangle|leqbigg|frac{1+|langle a,brangle| }{2}bigg|
$$
Intuitively, my choice is to say that this maximizes when $psi = frac{a+b}{2}$, modulo the norm. And in fact one can get close to proving the inequalities by plugging that in. But I can't prove why this should be the case.
It is obvious that $psi$ should be in the span of $a,b$, since any component that's outside of this subspace will be wasted. So we can write $psi = xa + yb$ for some complex $x$ and $y$. But $a$ and $b$ need not be orthogonal, so it's not so simple.
Is there an inequality that one can use here to prove one of the above statements, or a quick way to find $psi$?
inequality hilbert-spaces inner-product-space cauchy-schwarz-inequality
asked Dec 20 '18 at 15:54
KoljaKolja
590310
$endgroup$
2
$begingroup$
Why don't you use $psi=x(a+b)+y(a-b)$ instead? The reason is that $(a+b)$ and $(a-b)$ are orthogonal.
$endgroup$
– Andrei
Dec 20 '18 at 16:08
1
$begingroup$
We have $langle a+b,a-brangle = 2mathop{Im}{langle b,arangle } $, and this doesn't have to be zero. In fact $frac{a+b}{2}$ isn't always the correct element that maximizes that product, there can be a rotation added to $a$ or $b$. At least that's what I think.
$endgroup$
– Kolja
Dec 20 '18 at 17:26
1
$begingroup$
All I see after a quick glance is that the first inequality is wrong. Take $H=mathbb C^2$, $a=(1,0)$, $b=(0,1)$ and $psi=frac1{sqrt{2}}(1,1)$. Then $frac12=|langle psi,aranglelangle b,psirangle|notleq |langle a,brangle|=0$. This however does not violate the second inequality you stated so that one has a good chance of being true maybe?
$endgroup$
– Frederik vom Ende
Dec 21 '18 at 8:49
1
$begingroup$
The first inequality counterexample was so obvious, I find it strange that I didn't think of it. Thanks !
$endgroup$
– Kolja
Dec 23 '18 at 11:49
add a comment |
$begingroup$
Say we're in a (complex) Hilbert space $H$ where we're given some two elements $a,bin H$ of norm 1.
The question is to find an element $psiin H$ (of norm 1) that maximizes $|langle psi,arangle langle b,psirangle|$ (or $|langle psi,arangle langle b,psirangle|^2$ if you prefer).
Mainly what I want to prove is one of the following inequalities:
$$
|langle psi,arangle langle b,psirangle|leq |langle a,b rangle|
$$
$$
|langle psi,arangle langle b,psirangle|leqbigg|frac{1+|langle a,brangle| }{2}bigg|
$$
Intuitively, my choice is to say that this maximizes when $psi = frac{a+b}{2}$, modulo the norm. And in fact one can get close to proving the inequalities by plugging that in. But I can't prove why this should be the case.
It is obvious that $psi$ should be in the span of $a,b$, since any component that's outside of this subspace will be wasted. So we can write $psi = xa + yb$ for some complex $x$ and $y$. But $a$ and $b$ need not be orthogonal, so it's not so simple.
Is there an inequality that one can use here to prove one of the above statements, or a quick way to find $psi$?
inequality hilbert-spaces inner-product-space cauchy-schwarz-inequality
asked Dec 20 '18 at 15:54
KoljaKolja
590310
$endgroup$
Say we're in a (complex) Hilbert space $H$ where we're given some two elements $a,bin H$ of norm 1.
The question is to find an element $psiin H$ (of norm 1) that maximizes $|langle psi,arangle langle b,psirangle|$ (or $|langle psi,arangle langle b,psirangle|^2$ if you prefer).
Mainly what I want to prove is one of the following inequalities:
$$
|langle psi,arangle langle b,psirangle|leq |langle a,b rangle|
$$
$$
|langle psi,arangle langle b,psirangle|leqbigg|frac{1+|langle a,brangle| }{2}bigg|
$$
Intuitively, my choice is to say that this maximizes when $psi = frac{a+b}{2}$, modulo the norm. And in fact one can get close to proving the inequalities by plugging that in. But I can't prove why this should be the case.
It is obvious that $psi$ should be in the span of $a,b$, since any component that's outside of this subspace will be wasted. So we can write $psi = xa + yb$ for some complex $x$ and $y$. But $a$ and $b$ need not be orthogonal, so it's not so simple.
Is there an inequality that one can use here to prove one of the above statements, or a quick way to find $psi$?
inequality hilbert-spaces inner-product-space cauchy-schwarz-inequality
inequality hilbert-spaces inner-product-space cauchy-schwarz-inequality
asked Dec 20 '18 at 15:54
KoljaKolja
590310
asked Dec 20 '18 at 15:54
KoljaKolja
590310
edited Dec 20 '18 at 16:02
Kolja
asked Dec 20 '18 at 15:54
KoljaKolja
590310
asked Dec 20 '18 at 15:54
KoljaKolja
590310
asked Dec 20 '18 at 15:54
KoljaKolja
590310
590310
2
$begingroup$
Why don't you use $psi=x(a+b)+y(a-b)$ instead? The reason is that $(a+b)$ and $(a-b)$ are orthogonal.
$endgroup$
– Andrei
Dec 20 '18 at 16:08
1
$begingroup$
We have $langle a+b,a-brangle = 2mathop{Im}{langle b,arangle } $, and this doesn't have to be zero. In fact $frac{a+b}{2}$ isn't always the correct element that maximizes that product, there can be a rotation added to $a$ or $b$. At least that's what I think.
$endgroup$
– Kolja
Dec 20 '18 at 17:26
1
$begingroup$
All I see after a quick glance is that the first inequality is wrong. Take $H=mathbb C^2$, $a=(1,0)$, $b=(0,1)$ and $psi=frac1{sqrt{2}}(1,1)$. Then $frac12=|langle psi,aranglelangle b,psirangle|notleq |langle a,brangle|=0$. This however does not violate the second inequality you stated so that one has a good chance of being true maybe?
$endgroup$
– Frederik vom Ende
Dec 21 '18 at 8:49
1
$begingroup$
The first inequality counterexample was so obvious, I find it strange that I didn't think of it. Thanks !
$endgroup$
– Kolja
Dec 23 '18 at 11:49
add a comment |
2
$begingroup$
Why don't you use $psi=x(a+b)+y(a-b)$ instead? The reason is that $(a+b)$ and $(a-b)$ are orthogonal.
$endgroup$
– Andrei
Dec 20 '18 at 16:08
1
$begingroup$
We have $langle a+b,a-brangle = 2mathop{Im}{langle b,arangle } $, and this doesn't have to be zero. In fact $frac{a+b}{2}$ isn't always the correct element that maximizes that product, there can be a rotation added to $a$ or $b$. At least that's what I think.
$endgroup$
– Kolja
Dec 20 '18 at 17:26
1
$begingroup$
All I see after a quick glance is that the first inequality is wrong. Take $H=mathbb C^2$, $a=(1,0)$, $b=(0,1)$ and $psi=frac1{sqrt{2}}(1,1)$. Then $frac12=|langle psi,aranglelangle b,psirangle|notleq |langle a,brangle|=0$. This however does not violate the second inequality you stated so that one has a good chance of being true maybe?
$endgroup$
– Frederik vom Ende
Dec 21 '18 at 8:49
1
$begingroup$
The first inequality counterexample was so obvious, I find it strange that I didn't think of it. Thanks !
$endgroup$
– Kolja
Dec 23 '18 at 11:49
2
2
$begingroup$
Why don't you use $psi=x(a+b)+y(a-b)$ instead? The reason is that $(a+b)$ and $(a-b)$ are orthogonal.
$endgroup$
– Andrei
Dec 20 '18 at 16:08
$begingroup$
Why don't you use $psi=x(a+b)+y(a-b)$ instead? The reason is that $(a+b)$ and $(a-b)$ are orthogonal.
$endgroup$
– Andrei
Dec 20 '18 at 16:08
1
1
$begingroup$
We have $langle a+b,a-brangle = 2mathop{Im}{langle b,arangle } $, and this doesn't have to be zero. In fact $frac{a+b}{2}$ isn't always the correct element that maximizes that product, there can be a rotation added to $a$ or $b$. At least that's what I think.
$endgroup$
– Kolja
Dec 20 '18 at 17:26
$begingroup$
We have $langle a+b,a-brangle = 2mathop{Im}{langle b,arangle } $, and this doesn't have to be zero. In fact $frac{a+b}{2}$ isn't always the correct element that maximizes that product, there can be a rotation added to $a$ or $b$. At least that's what I think.
$endgroup$
– Kolja
Dec 20 '18 at 17:26
1
1
$begingroup$
All I see after a quick glance is that the first inequality is wrong. Take $H=mathbb C^2$, $a=(1,0)$, $b=(0,1)$ and $psi=frac1{sqrt{2}}(1,1)$. Then $frac12=|langle psi,aranglelangle b,psirangle|notleq |langle a,brangle|=0$. This however does not violate the second inequality you stated so that one has a good chance of being true maybe?
$endgroup$
– Frederik vom Ende
Dec 21 '18 at 8:49
$begingroup$
All I see after a quick glance is that the first inequality is wrong. Take $H=mathbb C^2$, $a=(1,0)$, $b=(0,1)$ and $psi=frac1{sqrt{2}}(1,1)$. Then $frac12=|langle psi,aranglelangle b,psirangle|notleq |langle a,brangle|=0$. This however does not violate the second inequality you stated so that one has a good chance of being true maybe?
$endgroup$
– Frederik vom Ende
Dec 21 '18 at 8:49
1
1
$begingroup$
The first inequality counterexample was so obvious, I find it strange that I didn't think of it. Thanks !
$endgroup$
– Kolja
Dec 23 '18 at 11:49
$begingroup$
The first inequality counterexample was so obvious, I find it strange that I didn't think of it. Thanks !
$endgroup$
– Kolja
Dec 23 '18 at 11:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
What you consider here is the numerical radius of the rank-1 operator $A:Hto H$, $hmapsto langle b,hrangle a$. To put things into context, the numerical range of $A$ is given by
$$
W(A)overset{text{Def.}}=lbracelangle psi,Apsirangle,|,psiin H,|psi|_H=1rbrace=lbracelangle psi,aranglelangle b,psirangle,|,psiin H,|psi|_H=1rbrace
$$
and the numerical radius of $A$ by
$$
r(A)=suplbrace |lambda|,|,lambdain W(A)rbrace=sup_{psiin H,|psi|_H=1}|langle psi,aranglelangle b,psirangle|,.
$$
As $A$ is of rank one, Lemma 2.1 in this paper (click here for accessible pdf) yields
$$
r(A)=frac{|A|+|operatorname{tr}(A)|}2,.
$$
One readily verifies that $A$ has operator norm $1$ and $operatorname{tr}(A)=langle b,arangle$ so
$$
|langlepsi,aranglelangle b,psirangle|leq sup_{psiin H,|psi|_H=1}|langle psi,aranglelangle b,psirangle|=r(A)=frac{1+|langle a,brangle|}2
$$
for all $psiin H,|psi|_H=1$ which is the desired inequality.
answered Dec 21 '18 at 11:50
Frederik vom EndeFrederik vom Ende
9821322
$endgroup$
add a comment |
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$begingroup$
What you consider here is the numerical radius of the rank-1 operator $A:Hto H$, $hmapsto langle b,hrangle a$. To put things into context, the numerical range of $A$ is given by
$$
W(A)overset{text{Def.}}=lbracelangle psi,Apsirangle,|,psiin H,|psi|_H=1rbrace=lbracelangle psi,aranglelangle b,psirangle,|,psiin H,|psi|_H=1rbrace
$$
and the numerical radius of $A$ by
$$
r(A)=suplbrace |lambda|,|,lambdain W(A)rbrace=sup_{psiin H,|psi|_H=1}|langle psi,aranglelangle b,psirangle|,.
$$
As $A$ is of rank one, Lemma 2.1 in this paper (click here for accessible pdf) yields
$$
r(A)=frac{|A|+|operatorname{tr}(A)|}2,.
$$
One readily verifies that $A$ has operator norm $1$ and $operatorname{tr}(A)=langle b,arangle$ so
$$
|langlepsi,aranglelangle b,psirangle|leq sup_{psiin H,|psi|_H=1}|langle psi,aranglelangle b,psirangle|=r(A)=frac{1+|langle a,brangle|}2
$$
for all $psiin H,|psi|_H=1$ which is the desired inequality.
answered Dec 21 '18 at 11:50
Frederik vom EndeFrederik vom Ende
9821322
$endgroup$
add a comment |
$begingroup$
What you consider here is the numerical radius of the rank-1 operator $A:Hto H$, $hmapsto langle b,hrangle a$. To put things into context, the numerical range of $A$ is given by
$$
W(A)overset{text{Def.}}=lbracelangle psi,Apsirangle,|,psiin H,|psi|_H=1rbrace=lbracelangle psi,aranglelangle b,psirangle,|,psiin H,|psi|_H=1rbrace
$$
and the numerical radius of $A$ by
$$
r(A)=suplbrace |lambda|,|,lambdain W(A)rbrace=sup_{psiin H,|psi|_H=1}|langle psi,aranglelangle b,psirangle|,.
$$
As $A$ is of rank one, Lemma 2.1 in this paper (click here for accessible pdf) yields
$$
r(A)=frac{|A|+|operatorname{tr}(A)|}2,.
$$
One readily verifies that $A$ has operator norm $1$ and $operatorname{tr}(A)=langle b,arangle$ so
$$
|langlepsi,aranglelangle b,psirangle|leq sup_{psiin H,|psi|_H=1}|langle psi,aranglelangle b,psirangle|=r(A)=frac{1+|langle a,brangle|}2
$$
for all $psiin H,|psi|_H=1$ which is the desired inequality.
answered Dec 21 '18 at 11:50
Frederik vom EndeFrederik vom Ende
9821322
$endgroup$
add a comment |
$begingroup$
What you consider here is the numerical radius of the rank-1 operator $A:Hto H$, $hmapsto langle b,hrangle a$. To put things into context, the numerical range of $A$ is given by
$$
W(A)overset{text{Def.}}=lbracelangle psi,Apsirangle,|,psiin H,|psi|_H=1rbrace=lbracelangle psi,aranglelangle b,psirangle,|,psiin H,|psi|_H=1rbrace
$$
and the numerical radius of $A$ by
$$
r(A)=suplbrace |lambda|,|,lambdain W(A)rbrace=sup_{psiin H,|psi|_H=1}|langle psi,aranglelangle b,psirangle|,.
$$
As $A$ is of rank one, Lemma 2.1 in this paper (click here for accessible pdf) yields
$$
r(A)=frac{|A|+|operatorname{tr}(A)|}2,.
$$
One readily verifies that $A$ has operator norm $1$ and $operatorname{tr}(A)=langle b,arangle$ so
$$
|langlepsi,aranglelangle b,psirangle|leq sup_{psiin H,|psi|_H=1}|langle psi,aranglelangle b,psirangle|=r(A)=frac{1+|langle a,brangle|}2
$$
for all $psiin H,|psi|_H=1$ which is the desired inequality.
answered Dec 21 '18 at 11:50
Frederik vom EndeFrederik vom Ende
9821322
$endgroup$
What you consider here is the numerical radius of the rank-1 operator $A:Hto H$, $hmapsto langle b,hrangle a$. To put things into context, the numerical range of $A$ is given by
$$
W(A)overset{text{Def.}}=lbracelangle psi,Apsirangle,|,psiin H,|psi|_H=1rbrace=lbracelangle psi,aranglelangle b,psirangle,|,psiin H,|psi|_H=1rbrace
$$
and the numerical radius of $A$ by
$$
r(A)=suplbrace |lambda|,|,lambdain W(A)rbrace=sup_{psiin H,|psi|_H=1}|langle psi,aranglelangle b,psirangle|,.
$$
As $A$ is of rank one, Lemma 2.1 in this paper (click here for accessible pdf) yields
$$
r(A)=frac{|A|+|operatorname{tr}(A)|}2,.
$$
One readily verifies that $A$ has operator norm $1$ and $operatorname{tr}(A)=langle b,arangle$ so
$$
|langlepsi,aranglelangle b,psirangle|leq sup_{psiin H,|psi|_H=1}|langle psi,aranglelangle b,psirangle|=r(A)=frac{1+|langle a,brangle|}2
$$
for all $psiin H,|psi|_H=1$ which is the desired inequality.
answered Dec 21 '18 at 11:50
Frederik vom EndeFrederik vom Ende
9821322
answered Dec 21 '18 at 11:50
Frederik vom EndeFrederik vom Ende
9821322
answered Dec 21 '18 at 11:50
Frederik vom EndeFrederik vom Ende
9821322
answered Dec 21 '18 at 11:50
Frederik vom EndeFrederik vom Ende
9821322
9821322
add a comment |
add a comment |
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2
$begingroup$
Why don't you use $psi=x(a+b)+y(a-b)$ instead? The reason is that $(a+b)$ and $(a-b)$ are orthogonal.
$endgroup$
– Andrei
Dec 20 '18 at 16:08
1
$begingroup$
We have $langle a+b,a-brangle = 2mathop{Im}{langle b,arangle } $, and this doesn't have to be zero. In fact $frac{a+b}{2}$ isn't always the correct element that maximizes that product, there can be a rotation added to $a$ or $b$. At least that's what I think.
$endgroup$
– Kolja
Dec 20 '18 at 17:26
1
$begingroup$
All I see after a quick glance is that the first inequality is wrong. Take $H=mathbb C^2$, $a=(1,0)$, $b=(0,1)$ and $psi=frac1{sqrt{2}}(1,1)$. Then $frac12=|langle psi,aranglelangle b,psirangle|notleq |langle a,brangle|=0$. This however does not violate the second inequality you stated so that one has a good chance of being true maybe?
$endgroup$
– Frederik vom Ende
Dec 21 '18 at 8:49
1
$begingroup$
The first inequality counterexample was so obvious, I find it strange that I didn't think of it. Thanks !
$endgroup$
– Kolja
Dec 23 '18 at 11:49