Mixing
Hilbert space and the probability to find the particle right of the origin
$begingroup$
We study a quantum particle confined to a one-dimensional box with walls
at the positions $pm 1$. The Hilbert-space of this system in the Schrödinger
representation is hence given by $L_2([-1; 1])$ with the scalar product
$$
left langle f,g right rangle=int_{-1}^{1}overline{f(x)}g(x) ,mathrm{d}x,quad f,gin L_2([-1,1])
$$
In this Hilbert space, we consider the two functions $f_0(x) := (1+i) exp(ipi x)$ and $f_1(x):=exp(i2pi x)$.
Now the problems are as follows
1) We now consider a general superposition $Psi=alphaphi_0+betaphi_1$ of the two states $phi_0:=f_0/2$ and $phi_1:=f_1/sqrt{2}$. Show that in order for $Psi$ to be normalized we need $|alpha|^2+|beta|^2=1$.
2) Determine the probability to find the particle right of the origin if it is in the state $Psi$, depending on the two parameters $alpha=left langle phi_0,Psi right rangle$ and $beta=left langle phi_1,Psi right rangle$
3) Find values for $alpha$ and $beta$ that maximize the probability to find the
particle right of the origin.
Hint: By using that $Psi$ and $e^{ialpha}Psi$ represent the same physical state of the system, we can choose $0leq alphaleq 1$ and $beta=sqrt{1-alpha^2}e^{ieta}$.
I have no problem with number one. I have tried with the second one, but I am not sure if I have reached the right conclusion. Because of that, I can not continue with the last problem.
2) We have to determine $int_{0}^{1}|Psi(x)|^2,mathrm{d}x$. First, we observe that
$$
|Psi|^2=|alpha|^2|phi_0|^2+alpha bar{beta}overline{phi_1}phi_0+ bar{alpha}betaoverline{phi_0}phi_1+|beta|^2|phi_1|^2
$$
then
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=|alpha|^2int_{0}^{1}|phi_0(x)|^2,mathrm{d}x+alpha bar{beta}int_{0}^{1}overline{phi_1(x)}phi_0(x),mathrm{d}x\+ bar{alpha}betaint_{0}^{1}overline{phi_0(x)}phi_1(x),mathrm{d}x+|beta|^2int_{0}^{1}|phi_1(x)|^2,mathrm{d}x
$$
The first and the last integrals are $1/2$. The two other are
$$
int_{0}^{1}overline{phi_1(x)}phi_0(x),mathrm{d}x=frac{1+i}{2sqrt{2}}int_{0}^{1} e^{-ipi x},mathrm{d}x=frac{1-i}{sqrt{2}pi},
$$
$$
int_{0}^{1}overline{phi_0(x)}phi_1(x),mathrm{d}x=frac{1-i}{2sqrt{2}}int_{0}^{1} e^{ipi x},mathrm{d}x=frac{i+1}{sqrt{2}pi}
$$
Altogether, we get
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=frac{1}{2}left ( |alpha|^2+|beta|^2 right )+alpha bar{beta}left ( frac{1-i}{sqrt{2}pi} right )+ bar{alpha}betaleft ( frac{i+1}{sqrt{2}pi} right ).
$$
Since
$$
alpha bar{beta}(1-i)+ bar{alpha}beta(1+i) =alpha bar{beta}(1-i)+ overline{alphabar{beta}(1-i)} =2Re(alpha bar{beta}(1-i))
$$
and with 1), we then get
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=frac{1}{2}+frac{2}{sqrt{2}pi}Re(alpha bar{beta}(1-i)).
$$
Is this correct, or am I missing something?
hilbert-spaces mathematical-physics
asked Jan 30 at 19:55
UnknownWUnknownW
1,045922
$endgroup$
add a comment |
$begingroup$
We study a quantum particle confined to a one-dimensional box with walls
at the positions $pm 1$. The Hilbert-space of this system in the Schrödinger
representation is hence given by $L_2([-1; 1])$ with the scalar product
$$
left langle f,g right rangle=int_{-1}^{1}overline{f(x)}g(x) ,mathrm{d}x,quad f,gin L_2([-1,1])
$$
In this Hilbert space, we consider the two functions $f_0(x) := (1+i) exp(ipi x)$ and $f_1(x):=exp(i2pi x)$.
Now the problems are as follows
1) We now consider a general superposition $Psi=alphaphi_0+betaphi_1$ of the two states $phi_0:=f_0/2$ and $phi_1:=f_1/sqrt{2}$. Show that in order for $Psi$ to be normalized we need $|alpha|^2+|beta|^2=1$.
2) Determine the probability to find the particle right of the origin if it is in the state $Psi$, depending on the two parameters $alpha=left langle phi_0,Psi right rangle$ and $beta=left langle phi_1,Psi right rangle$
3) Find values for $alpha$ and $beta$ that maximize the probability to find the
particle right of the origin.
Hint: By using that $Psi$ and $e^{ialpha}Psi$ represent the same physical state of the system, we can choose $0leq alphaleq 1$ and $beta=sqrt{1-alpha^2}e^{ieta}$.
I have no problem with number one. I have tried with the second one, but I am not sure if I have reached the right conclusion. Because of that, I can not continue with the last problem.
2) We have to determine $int_{0}^{1}|Psi(x)|^2,mathrm{d}x$. First, we observe that
$$
|Psi|^2=|alpha|^2|phi_0|^2+alpha bar{beta}overline{phi_1}phi_0+ bar{alpha}betaoverline{phi_0}phi_1+|beta|^2|phi_1|^2
$$
then
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=|alpha|^2int_{0}^{1}|phi_0(x)|^2,mathrm{d}x+alpha bar{beta}int_{0}^{1}overline{phi_1(x)}phi_0(x),mathrm{d}x\+ bar{alpha}betaint_{0}^{1}overline{phi_0(x)}phi_1(x),mathrm{d}x+|beta|^2int_{0}^{1}|phi_1(x)|^2,mathrm{d}x
$$
The first and the last integrals are $1/2$. The two other are
$$
int_{0}^{1}overline{phi_1(x)}phi_0(x),mathrm{d}x=frac{1+i}{2sqrt{2}}int_{0}^{1} e^{-ipi x},mathrm{d}x=frac{1-i}{sqrt{2}pi},
$$
$$
int_{0}^{1}overline{phi_0(x)}phi_1(x),mathrm{d}x=frac{1-i}{2sqrt{2}}int_{0}^{1} e^{ipi x},mathrm{d}x=frac{i+1}{sqrt{2}pi}
$$
Altogether, we get
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=frac{1}{2}left ( |alpha|^2+|beta|^2 right )+alpha bar{beta}left ( frac{1-i}{sqrt{2}pi} right )+ bar{alpha}betaleft ( frac{i+1}{sqrt{2}pi} right ).
$$
Since
$$
alpha bar{beta}(1-i)+ bar{alpha}beta(1+i) =alpha bar{beta}(1-i)+ overline{alphabar{beta}(1-i)} =2Re(alpha bar{beta}(1-i))
$$
and with 1), we then get
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=frac{1}{2}+frac{2}{sqrt{2}pi}Re(alpha bar{beta}(1-i)).
$$
Is this correct, or am I missing something?
hilbert-spaces mathematical-physics
asked Jan 30 at 19:55
UnknownWUnknownW
1,045922
$endgroup$
1
$begingroup$
It looks right to me!
$endgroup$
– Adrian Keister
Jan 30 at 20:51
$begingroup$
@AdrianKeister I'm glad to hear it... If you have time, could you please show me a way for the last problem? The hint doesn't seem helpful to me.
$endgroup$
– UnknownW
Jan 30 at 21:24
$begingroup$
Hmm. Well, the maximization boils down to maximizing $operatorname{Re}(alphaoverline{beta}(1-i)),$ subject to $|alpha|^2+|beta|^2=1.$
$endgroup$
– Adrian Keister
Jan 30 at 21:35
$begingroup$
@AdrianKeister That's true. I could write $|beta|=sqrt{1-|alpha|^2}$, which implies $beta=sqrt{1-|alpha|^2}e^{ieta}$ for some number $eta$. Should I then calculate the real part of $alpha sqrt{1-|alpha|^2}e^{-ieta}(1-i)$ and then finding the maximum of it with respect to $ alpha$, as long as I assume $alpha$ is a real number?
$endgroup$
– UnknownW
Jan 30 at 21:53
add a comment |
$begingroup$
We study a quantum particle confined to a one-dimensional box with walls
at the positions $pm 1$. The Hilbert-space of this system in the Schrödinger
representation is hence given by $L_2([-1; 1])$ with the scalar product
$$
left langle f,g right rangle=int_{-1}^{1}overline{f(x)}g(x) ,mathrm{d}x,quad f,gin L_2([-1,1])
$$
In this Hilbert space, we consider the two functions $f_0(x) := (1+i) exp(ipi x)$ and $f_1(x):=exp(i2pi x)$.
Now the problems are as follows
1) We now consider a general superposition $Psi=alphaphi_0+betaphi_1$ of the two states $phi_0:=f_0/2$ and $phi_1:=f_1/sqrt{2}$. Show that in order for $Psi$ to be normalized we need $|alpha|^2+|beta|^2=1$.
2) Determine the probability to find the particle right of the origin if it is in the state $Psi$, depending on the two parameters $alpha=left langle phi_0,Psi right rangle$ and $beta=left langle phi_1,Psi right rangle$
3) Find values for $alpha$ and $beta$ that maximize the probability to find the
particle right of the origin.
Hint: By using that $Psi$ and $e^{ialpha}Psi$ represent the same physical state of the system, we can choose $0leq alphaleq 1$ and $beta=sqrt{1-alpha^2}e^{ieta}$.
I have no problem with number one. I have tried with the second one, but I am not sure if I have reached the right conclusion. Because of that, I can not continue with the last problem.
2) We have to determine $int_{0}^{1}|Psi(x)|^2,mathrm{d}x$. First, we observe that
$$
|Psi|^2=|alpha|^2|phi_0|^2+alpha bar{beta}overline{phi_1}phi_0+ bar{alpha}betaoverline{phi_0}phi_1+|beta|^2|phi_1|^2
$$
then
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=|alpha|^2int_{0}^{1}|phi_0(x)|^2,mathrm{d}x+alpha bar{beta}int_{0}^{1}overline{phi_1(x)}phi_0(x),mathrm{d}x\+ bar{alpha}betaint_{0}^{1}overline{phi_0(x)}phi_1(x),mathrm{d}x+|beta|^2int_{0}^{1}|phi_1(x)|^2,mathrm{d}x
$$
The first and the last integrals are $1/2$. The two other are
$$
int_{0}^{1}overline{phi_1(x)}phi_0(x),mathrm{d}x=frac{1+i}{2sqrt{2}}int_{0}^{1} e^{-ipi x},mathrm{d}x=frac{1-i}{sqrt{2}pi},
$$
$$
int_{0}^{1}overline{phi_0(x)}phi_1(x),mathrm{d}x=frac{1-i}{2sqrt{2}}int_{0}^{1} e^{ipi x},mathrm{d}x=frac{i+1}{sqrt{2}pi}
$$
Altogether, we get
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=frac{1}{2}left ( |alpha|^2+|beta|^2 right )+alpha bar{beta}left ( frac{1-i}{sqrt{2}pi} right )+ bar{alpha}betaleft ( frac{i+1}{sqrt{2}pi} right ).
$$
Since
$$
alpha bar{beta}(1-i)+ bar{alpha}beta(1+i) =alpha bar{beta}(1-i)+ overline{alphabar{beta}(1-i)} =2Re(alpha bar{beta}(1-i))
$$
and with 1), we then get
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=frac{1}{2}+frac{2}{sqrt{2}pi}Re(alpha bar{beta}(1-i)).
$$
Is this correct, or am I missing something?
hilbert-spaces mathematical-physics
asked Jan 30 at 19:55
UnknownWUnknownW
1,045922
$endgroup$
We study a quantum particle confined to a one-dimensional box with walls
at the positions $pm 1$. The Hilbert-space of this system in the Schrödinger
representation is hence given by $L_2([-1; 1])$ with the scalar product
$$
left langle f,g right rangle=int_{-1}^{1}overline{f(x)}g(x) ,mathrm{d}x,quad f,gin L_2([-1,1])
$$
In this Hilbert space, we consider the two functions $f_0(x) := (1+i) exp(ipi x)$ and $f_1(x):=exp(i2pi x)$.
Now the problems are as follows
1) We now consider a general superposition $Psi=alphaphi_0+betaphi_1$ of the two states $phi_0:=f_0/2$ and $phi_1:=f_1/sqrt{2}$. Show that in order for $Psi$ to be normalized we need $|alpha|^2+|beta|^2=1$.
2) Determine the probability to find the particle right of the origin if it is in the state $Psi$, depending on the two parameters $alpha=left langle phi_0,Psi right rangle$ and $beta=left langle phi_1,Psi right rangle$
3) Find values for $alpha$ and $beta$ that maximize the probability to find the
particle right of the origin.
Hint: By using that $Psi$ and $e^{ialpha}Psi$ represent the same physical state of the system, we can choose $0leq alphaleq 1$ and $beta=sqrt{1-alpha^2}e^{ieta}$.
I have no problem with number one. I have tried with the second one, but I am not sure if I have reached the right conclusion. Because of that, I can not continue with the last problem.
2) We have to determine $int_{0}^{1}|Psi(x)|^2,mathrm{d}x$. First, we observe that
$$
|Psi|^2=|alpha|^2|phi_0|^2+alpha bar{beta}overline{phi_1}phi_0+ bar{alpha}betaoverline{phi_0}phi_1+|beta|^2|phi_1|^2
$$
then
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=|alpha|^2int_{0}^{1}|phi_0(x)|^2,mathrm{d}x+alpha bar{beta}int_{0}^{1}overline{phi_1(x)}phi_0(x),mathrm{d}x\+ bar{alpha}betaint_{0}^{1}overline{phi_0(x)}phi_1(x),mathrm{d}x+|beta|^2int_{0}^{1}|phi_1(x)|^2,mathrm{d}x
$$
The first and the last integrals are $1/2$. The two other are
$$
int_{0}^{1}overline{phi_1(x)}phi_0(x),mathrm{d}x=frac{1+i}{2sqrt{2}}int_{0}^{1} e^{-ipi x},mathrm{d}x=frac{1-i}{sqrt{2}pi},
$$
$$
int_{0}^{1}overline{phi_0(x)}phi_1(x),mathrm{d}x=frac{1-i}{2sqrt{2}}int_{0}^{1} e^{ipi x},mathrm{d}x=frac{i+1}{sqrt{2}pi}
$$
Altogether, we get
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=frac{1}{2}left ( |alpha|^2+|beta|^2 right )+alpha bar{beta}left ( frac{1-i}{sqrt{2}pi} right )+ bar{alpha}betaleft ( frac{i+1}{sqrt{2}pi} right ).
$$
Since
$$
alpha bar{beta}(1-i)+ bar{alpha}beta(1+i) =alpha bar{beta}(1-i)+ overline{alphabar{beta}(1-i)} =2Re(alpha bar{beta}(1-i))
$$
and with 1), we then get
$$
int_{0}^{1} |Psi(x)|^2,mathrm{d}x=frac{1}{2}+frac{2}{sqrt{2}pi}Re(alpha bar{beta}(1-i)).
$$
Is this correct, or am I missing something?
hilbert-spaces mathematical-physics
hilbert-spaces mathematical-physics
asked Jan 30 at 19:55
UnknownWUnknownW
1,045922
asked Jan 30 at 19:55
UnknownWUnknownW
1,045922
asked Jan 30 at 19:55
UnknownWUnknownW
1,045922
asked Jan 30 at 19:55
UnknownWUnknownW
1,045922
asked Jan 30 at 19:55
UnknownWUnknownW
1,045922
1,045922
1
$begingroup$
It looks right to me!
$endgroup$
– Adrian Keister
Jan 30 at 20:51
$begingroup$
@AdrianKeister I'm glad to hear it... If you have time, could you please show me a way for the last problem? The hint doesn't seem helpful to me.
$endgroup$
– UnknownW
Jan 30 at 21:24
$begingroup$
Hmm. Well, the maximization boils down to maximizing $operatorname{Re}(alphaoverline{beta}(1-i)),$ subject to $|alpha|^2+|beta|^2=1.$
$endgroup$
– Adrian Keister
Jan 30 at 21:35
$begingroup$
@AdrianKeister That's true. I could write $|beta|=sqrt{1-|alpha|^2}$, which implies $beta=sqrt{1-|alpha|^2}e^{ieta}$ for some number $eta$. Should I then calculate the real part of $alpha sqrt{1-|alpha|^2}e^{-ieta}(1-i)$ and then finding the maximum of it with respect to $ alpha$, as long as I assume $alpha$ is a real number?
$endgroup$
– UnknownW
Jan 30 at 21:53
add a comment |
1
$begingroup$
It looks right to me!
$endgroup$
– Adrian Keister
Jan 30 at 20:51
$begingroup$
@AdrianKeister I'm glad to hear it... If you have time, could you please show me a way for the last problem? The hint doesn't seem helpful to me.
$endgroup$
– UnknownW
Jan 30 at 21:24
$begingroup$
Hmm. Well, the maximization boils down to maximizing $operatorname{Re}(alphaoverline{beta}(1-i)),$ subject to $|alpha|^2+|beta|^2=1.$
$endgroup$
– Adrian Keister
Jan 30 at 21:35
$begingroup$
@AdrianKeister That's true. I could write $|beta|=sqrt{1-|alpha|^2}$, which implies $beta=sqrt{1-|alpha|^2}e^{ieta}$ for some number $eta$. Should I then calculate the real part of $alpha sqrt{1-|alpha|^2}e^{-ieta}(1-i)$ and then finding the maximum of it with respect to $ alpha$, as long as I assume $alpha$ is a real number?
$endgroup$
– UnknownW
Jan 30 at 21:53
1
1
$begingroup$
It looks right to me!
$endgroup$
– Adrian Keister
Jan 30 at 20:51
$begingroup$
It looks right to me!
$endgroup$
– Adrian Keister
Jan 30 at 20:51
$begingroup$
@AdrianKeister I'm glad to hear it... If you have time, could you please show me a way for the last problem? The hint doesn't seem helpful to me.
$endgroup$
– UnknownW
Jan 30 at 21:24
$begingroup$
@AdrianKeister I'm glad to hear it... If you have time, could you please show me a way for the last problem? The hint doesn't seem helpful to me.
$endgroup$
– UnknownW
Jan 30 at 21:24
$begingroup$
Hmm. Well, the maximization boils down to maximizing $operatorname{Re}(alphaoverline{beta}(1-i)),$ subject to $|alpha|^2+|beta|^2=1.$
$endgroup$
– Adrian Keister
Jan 30 at 21:35
$begingroup$
Hmm. Well, the maximization boils down to maximizing $operatorname{Re}(alphaoverline{beta}(1-i)),$ subject to $|alpha|^2+|beta|^2=1.$
$endgroup$
– Adrian Keister
Jan 30 at 21:35
$begingroup$
@AdrianKeister That's true. I could write $|beta|=sqrt{1-|alpha|^2}$, which implies $beta=sqrt{1-|alpha|^2}e^{ieta}$ for some number $eta$. Should I then calculate the real part of $alpha sqrt{1-|alpha|^2}e^{-ieta}(1-i)$ and then finding the maximum of it with respect to $ alpha$, as long as I assume $alpha$ is a real number?
$endgroup$
– UnknownW
Jan 30 at 21:53
$begingroup$
@AdrianKeister That's true. I could write $|beta|=sqrt{1-|alpha|^2}$, which implies $beta=sqrt{1-|alpha|^2}e^{ieta}$ for some number $eta$. Should I then calculate the real part of $alpha sqrt{1-|alpha|^2}e^{-ieta}(1-i)$ and then finding the maximum of it with respect to $ alpha$, as long as I assume $alpha$ is a real number?
$endgroup$
– UnknownW
Jan 30 at 21:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For problem 3), we need to maximize $operatorname{Re}(alphaoverline{beta}(1-i))$ subject to $|alpha|^2+|beta|^2=1$. Let
begin{align*}
alpha&=r_{alpha}e^{itheta_{alpha}} \
beta&=r_{beta}e^{itheta_{beta}}, \
end{align*}
which implies
$|alpha|^2+|beta|^2=r_{alpha}^2+r_{beta}^2=1,$ and we're trying to maximize
begin{align*}
operatorname{Re}(alphaoverline{beta}(1-i))&=operatorname{Re}(r_{alpha}e^{itheta_{alpha}}r_{beta}e^{-itheta_{beta}}sqrt{2},e^{-ipi/4}) \
&=sqrt{2},r_{alpha}r_{beta}operatorname{Re}left[e^{i(theta_{alpha}-theta_{beta}-pi/4)}right] \
&=sqrt{2},r_{alpha}r_{beta}cos(theta_{alpha}-theta_{beta}-pi/4).
end{align*}
As the $r$'s show up symmetrically, choose $r_{alpha}=r_{beta}=1/sqrt{2},$ and choose
begin{align*}
theta_{alpha}-theta_{beta}-frac{pi}{4}&=0 \
theta_{alpha}-theta_{beta}&=frac{pi}{4}
end{align*}
to maximize the cosine.
One final point: once you do all this, you should definitely back-calculate the actual probability, and make sure it isn't bigger than $1.$ If it is, you can always dial back something (the angles would be the easiest) to make it smaller.
A quick calculation:
$$frac12+frac{2}{sqrt{2},pi}(sqrt{2}(1/2))=frac12+frac{1}{pi}<1,$$
so you're safe.
answered Jan 30 at 21:55
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
$begingroup$
Wow, thank you. Quick question: Is it possible to choose, say $r_alpha=1/2$, so that we have $r_beta=1-1/4 = 3/4$? (I do not understand the symmetry part.)
$endgroup$
– UnknownW
Jan 30 at 23:43
$begingroup$
Well, you could, but the final result would be smaller, and hence not the maximum. The symmetry argument has to do with how the variables are showing up: you could swap them, and you couldn't tell the difference. Often in situations like that, you'll find that the variables being equal is the extremum.
$endgroup$
– Adrian Keister
Jan 31 at 0:45
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
For problem 3), we need to maximize $operatorname{Re}(alphaoverline{beta}(1-i))$ subject to $|alpha|^2+|beta|^2=1$. Let
begin{align*}
alpha&=r_{alpha}e^{itheta_{alpha}} \
beta&=r_{beta}e^{itheta_{beta}}, \
end{align*}
which implies
$|alpha|^2+|beta|^2=r_{alpha}^2+r_{beta}^2=1,$ and we're trying to maximize
begin{align*}
operatorname{Re}(alphaoverline{beta}(1-i))&=operatorname{Re}(r_{alpha}e^{itheta_{alpha}}r_{beta}e^{-itheta_{beta}}sqrt{2},e^{-ipi/4}) \
&=sqrt{2},r_{alpha}r_{beta}operatorname{Re}left[e^{i(theta_{alpha}-theta_{beta}-pi/4)}right] \
&=sqrt{2},r_{alpha}r_{beta}cos(theta_{alpha}-theta_{beta}-pi/4).
end{align*}
As the $r$'s show up symmetrically, choose $r_{alpha}=r_{beta}=1/sqrt{2},$ and choose
begin{align*}
theta_{alpha}-theta_{beta}-frac{pi}{4}&=0 \
theta_{alpha}-theta_{beta}&=frac{pi}{4}
end{align*}
to maximize the cosine.
One final point: once you do all this, you should definitely back-calculate the actual probability, and make sure it isn't bigger than $1.$ If it is, you can always dial back something (the angles would be the easiest) to make it smaller.
A quick calculation:
$$frac12+frac{2}{sqrt{2},pi}(sqrt{2}(1/2))=frac12+frac{1}{pi}<1,$$
so you're safe.
answered Jan 30 at 21:55
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
$begingroup$
Wow, thank you. Quick question: Is it possible to choose, say $r_alpha=1/2$, so that we have $r_beta=1-1/4 = 3/4$? (I do not understand the symmetry part.)
$endgroup$
– UnknownW
Jan 30 at 23:43
$begingroup$
Well, you could, but the final result would be smaller, and hence not the maximum. The symmetry argument has to do with how the variables are showing up: you could swap them, and you couldn't tell the difference. Often in situations like that, you'll find that the variables being equal is the extremum.
$endgroup$
– Adrian Keister
Jan 31 at 0:45
add a comment |
$begingroup$
For problem 3), we need to maximize $operatorname{Re}(alphaoverline{beta}(1-i))$ subject to $|alpha|^2+|beta|^2=1$. Let
begin{align*}
alpha&=r_{alpha}e^{itheta_{alpha}} \
beta&=r_{beta}e^{itheta_{beta}}, \
end{align*}
which implies
$|alpha|^2+|beta|^2=r_{alpha}^2+r_{beta}^2=1,$ and we're trying to maximize
begin{align*}
operatorname{Re}(alphaoverline{beta}(1-i))&=operatorname{Re}(r_{alpha}e^{itheta_{alpha}}r_{beta}e^{-itheta_{beta}}sqrt{2},e^{-ipi/4}) \
&=sqrt{2},r_{alpha}r_{beta}operatorname{Re}left[e^{i(theta_{alpha}-theta_{beta}-pi/4)}right] \
&=sqrt{2},r_{alpha}r_{beta}cos(theta_{alpha}-theta_{beta}-pi/4).
end{align*}
As the $r$'s show up symmetrically, choose $r_{alpha}=r_{beta}=1/sqrt{2},$ and choose
begin{align*}
theta_{alpha}-theta_{beta}-frac{pi}{4}&=0 \
theta_{alpha}-theta_{beta}&=frac{pi}{4}
end{align*}
to maximize the cosine.
One final point: once you do all this, you should definitely back-calculate the actual probability, and make sure it isn't bigger than $1.$ If it is, you can always dial back something (the angles would be the easiest) to make it smaller.
A quick calculation:
$$frac12+frac{2}{sqrt{2},pi}(sqrt{2}(1/2))=frac12+frac{1}{pi}<1,$$
so you're safe.
answered Jan 30 at 21:55
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
$begingroup$
Wow, thank you. Quick question: Is it possible to choose, say $r_alpha=1/2$, so that we have $r_beta=1-1/4 = 3/4$? (I do not understand the symmetry part.)
$endgroup$
– UnknownW
Jan 30 at 23:43
$begingroup$
Well, you could, but the final result would be smaller, and hence not the maximum. The symmetry argument has to do with how the variables are showing up: you could swap them, and you couldn't tell the difference. Often in situations like that, you'll find that the variables being equal is the extremum.
$endgroup$
– Adrian Keister
Jan 31 at 0:45
add a comment |
$begingroup$
For problem 3), we need to maximize $operatorname{Re}(alphaoverline{beta}(1-i))$ subject to $|alpha|^2+|beta|^2=1$. Let
begin{align*}
alpha&=r_{alpha}e^{itheta_{alpha}} \
beta&=r_{beta}e^{itheta_{beta}}, \
end{align*}
which implies
$|alpha|^2+|beta|^2=r_{alpha}^2+r_{beta}^2=1,$ and we're trying to maximize
begin{align*}
operatorname{Re}(alphaoverline{beta}(1-i))&=operatorname{Re}(r_{alpha}e^{itheta_{alpha}}r_{beta}e^{-itheta_{beta}}sqrt{2},e^{-ipi/4}) \
&=sqrt{2},r_{alpha}r_{beta}operatorname{Re}left[e^{i(theta_{alpha}-theta_{beta}-pi/4)}right] \
&=sqrt{2},r_{alpha}r_{beta}cos(theta_{alpha}-theta_{beta}-pi/4).
end{align*}
As the $r$'s show up symmetrically, choose $r_{alpha}=r_{beta}=1/sqrt{2},$ and choose
begin{align*}
theta_{alpha}-theta_{beta}-frac{pi}{4}&=0 \
theta_{alpha}-theta_{beta}&=frac{pi}{4}
end{align*}
to maximize the cosine.
One final point: once you do all this, you should definitely back-calculate the actual probability, and make sure it isn't bigger than $1.$ If it is, you can always dial back something (the angles would be the easiest) to make it smaller.
A quick calculation:
$$frac12+frac{2}{sqrt{2},pi}(sqrt{2}(1/2))=frac12+frac{1}{pi}<1,$$
so you're safe.
answered Jan 30 at 21:55
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
For problem 3), we need to maximize $operatorname{Re}(alphaoverline{beta}(1-i))$ subject to $|alpha|^2+|beta|^2=1$. Let
begin{align*}
alpha&=r_{alpha}e^{itheta_{alpha}} \
beta&=r_{beta}e^{itheta_{beta}}, \
end{align*}
which implies
$|alpha|^2+|beta|^2=r_{alpha}^2+r_{beta}^2=1,$ and we're trying to maximize
begin{align*}
operatorname{Re}(alphaoverline{beta}(1-i))&=operatorname{Re}(r_{alpha}e^{itheta_{alpha}}r_{beta}e^{-itheta_{beta}}sqrt{2},e^{-ipi/4}) \
&=sqrt{2},r_{alpha}r_{beta}operatorname{Re}left[e^{i(theta_{alpha}-theta_{beta}-pi/4)}right] \
&=sqrt{2},r_{alpha}r_{beta}cos(theta_{alpha}-theta_{beta}-pi/4).
end{align*}
As the $r$'s show up symmetrically, choose $r_{alpha}=r_{beta}=1/sqrt{2},$ and choose
begin{align*}
theta_{alpha}-theta_{beta}-frac{pi}{4}&=0 \
theta_{alpha}-theta_{beta}&=frac{pi}{4}
end{align*}
to maximize the cosine.
One final point: once you do all this, you should definitely back-calculate the actual probability, and make sure it isn't bigger than $1.$ If it is, you can always dial back something (the angles would be the easiest) to make it smaller.
A quick calculation:
$$frac12+frac{2}{sqrt{2},pi}(sqrt{2}(1/2))=frac12+frac{1}{pi}<1,$$
so you're safe.
answered Jan 30 at 21:55
Adrian KeisterAdrian Keister
5,26971933
answered Jan 30 at 21:55
Adrian KeisterAdrian Keister
5,26971933
answered Jan 30 at 21:55
Adrian KeisterAdrian Keister
5,26971933
answered Jan 30 at 21:55
Adrian KeisterAdrian Keister
5,26971933
5,26971933
$begingroup$
Wow, thank you. Quick question: Is it possible to choose, say $r_alpha=1/2$, so that we have $r_beta=1-1/4 = 3/4$? (I do not understand the symmetry part.)
$endgroup$
– UnknownW
Jan 30 at 23:43
$begingroup$
Well, you could, but the final result would be smaller, and hence not the maximum. The symmetry argument has to do with how the variables are showing up: you could swap them, and you couldn't tell the difference. Often in situations like that, you'll find that the variables being equal is the extremum.
$endgroup$
– Adrian Keister
Jan 31 at 0:45
add a comment |
$begingroup$
Wow, thank you. Quick question: Is it possible to choose, say $r_alpha=1/2$, so that we have $r_beta=1-1/4 = 3/4$? (I do not understand the symmetry part.)
$endgroup$
– UnknownW
Jan 30 at 23:43
$begingroup$
Well, you could, but the final result would be smaller, and hence not the maximum. The symmetry argument has to do with how the variables are showing up: you could swap them, and you couldn't tell the difference. Often in situations like that, you'll find that the variables being equal is the extremum.
$endgroup$
– Adrian Keister
Jan 31 at 0:45
$begingroup$
Wow, thank you. Quick question: Is it possible to choose, say $r_alpha=1/2$, so that we have $r_beta=1-1/4 = 3/4$? (I do not understand the symmetry part.)
$endgroup$
– UnknownW
Jan 30 at 23:43
$begingroup$
Wow, thank you. Quick question: Is it possible to choose, say $r_alpha=1/2$, so that we have $r_beta=1-1/4 = 3/4$? (I do not understand the symmetry part.)
$endgroup$
– UnknownW
Jan 30 at 23:43
$begingroup$
Well, you could, but the final result would be smaller, and hence not the maximum. The symmetry argument has to do with how the variables are showing up: you could swap them, and you couldn't tell the difference. Often in situations like that, you'll find that the variables being equal is the extremum.
$endgroup$
– Adrian Keister
Jan 31 at 0:45
$begingroup$
Well, you could, but the final result would be smaller, and hence not the maximum. The symmetry argument has to do with how the variables are showing up: you could swap them, and you couldn't tell the difference. Often in situations like that, you'll find that the variables being equal is the extremum.
$endgroup$
– Adrian Keister
Jan 31 at 0:45
add a comment |
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$begingroup$
It looks right to me!
$endgroup$
– Adrian Keister
Jan 30 at 20:51
$begingroup$
@AdrianKeister I'm glad to hear it... If you have time, could you please show me a way for the last problem? The hint doesn't seem helpful to me.
$endgroup$
– UnknownW
Jan 30 at 21:24
$begingroup$
Hmm. Well, the maximization boils down to maximizing $operatorname{Re}(alphaoverline{beta}(1-i)),$ subject to $|alpha|^2+|beta|^2=1.$
$endgroup$
– Adrian Keister
Jan 30 at 21:35
$begingroup$
@AdrianKeister That's true. I could write $|beta|=sqrt{1-|alpha|^2}$, which implies $beta=sqrt{1-|alpha|^2}e^{ieta}$ for some number $eta$. Should I then calculate the real part of $alpha sqrt{1-|alpha|^2}e^{-ieta}(1-i)$ and then finding the maximum of it with respect to $ alpha$, as long as I assume $alpha$ is a real number?
$endgroup$
– UnknownW
Jan 30 at 21:53