Maximize product of a vector with two vectors












2












$begingroup$


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 $psi$?










share|cite|improve this question











$endgroup$








  • 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
















2












$begingroup$


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 $psi$?










share|cite|improve this question











$endgroup$








  • 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














2












2








2





$begingroup$


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 $psi$?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 16:02







Kolja

















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














  • 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










1 Answer
1






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oldest

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1












$begingroup$

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.






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    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 11:50









        Frederik vom EndeFrederik vom Ende

        9821322




        9821322






























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