# Homework Help: Maximum position expectation value for 1D harmonic oscillator

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1. Jun 18, 2015

### AwesomeTrains

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

1. The problem statement, all variables and given/known data

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.

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

3. 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

2. Jun 18, 2015

### vela

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

3. Jun 18, 2015

### blue_leaf77

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

4. Jun 18, 2015

### AwesomeTrains

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?

5. Jun 18, 2015

### blue_leaf77

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

6. Jun 18, 2015

### 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?

7. Jun 18, 2015

### 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?

8. Jun 18, 2015

### blue_leaf77

Yes, it is.
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 $\phi_1=\phi_2$.

Last edited: Jun 18, 2015
9. Jun 19, 2015

### Mikhail_MR

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)?

10. Jun 19, 2015

### blue_leaf77

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

11. Jun 19, 2015

### Mikhail_MR

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

12. Jun 19, 2015

### blue_leaf77

You have this freedom.

13. Jun 19, 2015

### vela

Staff Emeritus
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…