# Maximum position expectation value for 1D harmonic oscillator

• AwesomeTrains
In summary, the conversation discusses a homework problem involving determining a linear combination of states for a one-dimensional harmonic oscillator to maximize the expectation value of x. The Lagrange multiplier method is suggested as a possible approach, and the conversation then delves into examining various solutions and how to determine which is correct. Ultimately, it is determined that any solution where the phases of the coefficients are equal is valid.
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

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.

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?

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

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?

blue_leaf77
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?

AwesomeTrains said:
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.
Mikhail_MR said:
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##.

Last edited:
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)?

Mikhail_MR said:
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.
Mikhail_MR said:
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.

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

blue_leaf77 said:
Well as long as ##\phi_1 = \phi_2## is satisfied, any ##\phi_1## is possible.
You have this freedom.

Mikhail_MR said:
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…

Mikhail_MR

## What is the maximum position expectation value for a 1D harmonic oscillator?

The maximum position expectation value for a 1D harmonic oscillator is at the equilibrium position, where the potential energy is at its minimum. This corresponds to the point where the force acting on the oscillator is zero, and it has the highest probability of being found at that position.

## How is the maximum position expectation value related to the energy of the oscillator?

The maximum position expectation value is directly related to the energy of the oscillator. As the energy of the oscillator increases, the maximum position expectation value also increases. This is because a higher energy state allows the oscillator to explore a larger region of space, increasing the chances of finding it at a higher position.

## What is the significance of the maximum position expectation value in the context of the harmonic oscillator?

The maximum position expectation value is a key parameter in understanding the behavior of the harmonic oscillator. It helps us calculate the average position of the oscillator, as well as the spread of its position. It also gives us insights into the energy states and behavior of the oscillator.

## How does the maximum position expectation value change with time in a harmonic oscillator?

In a harmonic oscillator, the maximum position expectation value remains constant with time. This is because the oscillator is in a stable equilibrium state, and the probability of finding it at any position remains the same. However, the position of the oscillator may fluctuate around this maximum value due to its quantum nature.

## What factors can affect the maximum position expectation value in a harmonic oscillator?

The maximum position expectation value can be affected by factors such as the energy of the oscillator, the potential energy function, and the initial conditions of the system. Changes in these factors can lead to a shift in the equilibrium position and consequently, the maximum position expectation value of the oscillator.

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