Maximum position expectation value for 1D harmonic oscillator

[AwesomeTrains]

Hey, I'm stuck halfway through the solution it seems. I could use some tips on how to continue.

1. Homework Statement
I have to determine a linear combination of the states |0\rangle, |1\rangle, of a one dimensional harmonic oscillator, so that the expectation value \langle x \rangle is a maximum.

Homework Equations

x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^\dagger)

The Attempt at a Solution

I set |\alpha \rangle\equiv c_1|1\rangle + c_2|0\rangle
Then I calculate \langle \alpha | x | \alpha \rangle = \sqrt{\frac{\hbar}{2m\omega}} (c_1^*\langle1| + c_2^*\langle 0|)(a+a^\dagger)(c_1|1\rangle + c_2|0\rangle)=\sqrt{\frac{\hbar}{2m\omega}}[c_1^*(c_1+c_2)+c_2^*c_2] = \sqrt{\frac{\hbar}{2m\omega}}[1+c_1^*c_2]
I get the last equation because of the normalization condition:
c_1^*c_1+c_2^*c_2=1
This is where I don't know how to continue. Is this even the right approach?
(I'm not allowed to use the wave functions.)

Anything is appreciated.

Kind regards
Alex

 

[vela]

You should recheck your calculation of $\langle \alpha \lvert x \rvert \alpha \rangle$. The $c_i^*c_i$ terms shouldn't be there.

 

[blue_leaf77]

As for the maximization problem, the Lagrange multiplier method may be helpful.

 

[AwesomeTrains]

Thanks for the fast reply.
I've corrected my calculation, is this right?:
First I calculate (a+a^\dagger)|\alpha\rangle
a(c_1|1\rangle+c_2|0\rangle)=c_1|0\rangle+0c_2|0\rangle=c_1|0\rangle
and
a^\dagger(c_1|1\rangle+c_2|0\rangle)=c_1\sqrt{2}|2\rangle+c_2|1\rangle

Then because |\alpha\rangle^\dagger = c_1^*\langle1|+c_2^*\langle0|
\langle\alpha|(a+a^\dagger)|\alpha\rangle=c_1^*\langle1|(c_1\sqrt{2}|2\rangle+c_2|1\rangle+c_1|0\rangle)+c_2^*\langle0|(c_1\sqrt{2}|2\rangle+c_2|1\rangle+c_1|0\rangle)= c_1^*c_2+c_2^*c_1
Do I use the lagrange multiplier method here? Or how do I continue?

 

[blue_leaf77]

The Lagrange multiplier is used for finding the maximum value of <x>.

 

[AwesomeTrains]

So It's like a function of 4 variables f(c_1,c_1^*, c_2,c_2^*)=c_1^*c_2+c_2^*c_1 which I have to maximize under the condition c_1^*c_1+c_2^*c_2=1
Or is there an easier solution?

 

[Mikhail_MR]

Hello! I Have the same exercise in my homework. Using the Lagrange multiplier I get $c_{1}=c_{2}=1/\sqrt{2}$. Is that right?

 

[blue_leaf77]

So It's like a function of 4 variables f(c_1,c_1^*, c_2,c_2^*)=c_1^*c_2+c_2^*c_1 which I have to maximize under the condition c_1^*c_1+c_2^*c_2=1
Or is there an easier solution?

Yes, it is.

Hello! I Have the same exercise in my homework. Using the Lagrange multiplier I get $c_{1}=c_{2}=1/\sqrt{2}$. Is that right?

Whether it's true or not, you can do some kind of examination. For example, you can write $\langle \hat{x} \rangle \propto |c_1||c_2| \cos{(\phi_1-\phi_2)}$ where the $\phi$'s are the angle of each coefficients, since we want to maximize our expectation value, it's necessary that $\phi_1=\phi_2$.

 

[Mikhail_MR]

Thanks for advise, blue_leaf77. I have made my calculation new.
I get 4 different sets of values.
$c_{1}=c_{2} = i/\sqrt{2}$
$c_{1}=c_{2} = -i/\sqrt{2}$
$c_{1}=i/\sqrt{2}; c_{2} = -i/\sqrt{2}$
$c_{1}=-i/\sqrt{2}; c_{2} = i/\sqrt{2}$

Using your advise how to examine solutions only the first and the second solutions are of interest. I can get this result on another way just inserting values in $f$ and finding a maximum. My question is: how i decide which set is right (the first or the second)?

 

[blue_leaf77]

Thanks for advise, blue_leaf77. I have made my calculation new.
I get 4 different sets of values.
$c_{1}=c_{2} = i/\sqrt{2}$
$c_{1}=c_{2} = -i/\sqrt{2}$
$c_{1}=i/\sqrt{2}; c_{2} = -i/\sqrt{2}$
$c_{1}=-i/\sqrt{2}; c_{2} = i/\sqrt{2}$

Well as long as $\phi_1 = \phi_2$ is satisfied, any $\phi_1$ is possible.

how i decide which set is right (the first or the second)?

In the original question posted by the OP, he is asked to calculate the maximum of $\langle \hat{x} \rangle$. In your case, I don't know what you are asked to do in your own homework.

 

[Mikhail_MR]

I have to calculate the maximum of $\langle \hat{x} \rangle$ too

 

[blue_leaf77]

Well as long as $\phi_1 = \phi_2$ is satisfied, any $\phi_1$ is possible.

You have this freedom.

 

[vela]

Thanks for advise, blue_leaf77. I have made my calculation new.
I get 4 different sets of values.
$c_{1}=c_{2} = i/\sqrt{2}$
$c_{1}=c_{2} = -i/\sqrt{2}$

Using your advise how to examine solutions only the first and the second solutions are of interest. I can get this result on another way just inserting values in $f$ and finding a maximum. My question is: how i decide which set is right (the first or the second)?

Your first solution corresponds to the state
$$\lvert \alpha_1 \rangle = i\left(\frac {1}{\sqrt{2}} \lvert 0 \rangle + \frac{1}{\sqrt{2}} \lvert 1 \rangle\right) = e^{i\frac {\pi}{2}}\left(\frac {1}{\sqrt{2}} \lvert 0 \rangle + \frac{1}{\sqrt{2}} \lvert 1 \rangle\right),$$ and your second solution to
$$\lvert \alpha_2 \rangle = -i\left(\frac {1}{\sqrt{2}} \lvert 0 \rangle + \frac{1}{\sqrt{2}} \lvert 1 \rangle\right) = e^{-i\frac {\pi}{2}}\left(\frac {1}{\sqrt{2}} \lvert 0 \rangle + \frac{1}{\sqrt{2}} \lvert 1 \rangle\right).$$ The two states differ only by an overall phase factor, so…