Harmonic Oscillator and Ladder Operators

[MFAHH]

Homework Statement

Consider a linear harmonic oscillator with the solution defined by the ladder operators a and a^†. Use the number basis |n⟩ to do the following.

a) Construct a linear combination of |0⟩ and |1⟩ to form a state |ψ⟩ such that ⟨ψ|X|ψ⟩ is as large as
possible.

b) Suppose that the oscillator is in the state constructed in a) at time t = 0. Write an expression to describe the time dependence of this state state for t > 0. Evaluate the expectation value ⟨ψ(t)|X|ψ(t)⟩ as a function of time for t > 0.

c) Defining (∆x)² = ⟨ψ(t)|X²|ψ(t)⟩ − ⟨ψ(t)|X|ψ(t)⟩². Calculate (∆x)².

Homework Equations

The Attempt at a Solution

I have a hunch that for part a) I need to consider some |ψ⟩ = A|0⟩ + B|1⟩, then differentiate the expression ⟨ψ|X|ψ⟩. However I'm not sure how to implement that, any hints?

Many thanks

 

[PeroK]

Why not just go ahead and calculate <X> for arbitrary $A, B$ as your hunch suggests?

 

[MFAHH]

Why not just go ahead and calculate <X> for arbitrary $A, B$ as your hunch suggests?

Thanks for the reply.

Ok so I substituted |ψ⟩ = A|0⟩ + B|1⟩ into ⟨ψ|X|ψ⟩, and from simple manipulation I have:

<X> = A*A⟨0|X|0⟩ + A*B⟨0|X|1⟩ + B*A ⟨1|X|0⟩ + B*B ⟨1|X|1⟩

Is this correct? I wonder what can be done to simplify it, should I substitute X in terms of a and a†?

 

[PeroK]

Thanks for the reply.

Ok so I substituted |ψ⟩ = A|0⟩ + B|1⟩ into ⟨ψ|X|ψ⟩, and from simple manipulation I have:

<X> = A*A⟨0|X|0⟩ + A*B⟨0|X|1⟩ + B*A ⟨1|X|0⟩ + B*B ⟨1|X|1⟩

Is this correct? I wonder what can be done to simplify it, should I substitute X in terms of a and a†?

Yes, it's correct. You need to know (or calculate) $<0|X|0>$ and $<1|X|1>$. You have to calculate $<0|X|1>$ as well. The other term is related to this (hint: think about complex numbers).

You can use the ladder operator or just integrate (if you have a list of standard Gaussian type integrals).

 

[MFAHH]

Yes, it's correct. You need to know (or calculate) $<0|X|0>$ and $<1|X|1>$. You have to calculate $<0|X|1>$ as well. The other term is related to this (hint: think about complex numbers).

You can use the ladder operator or just integrate (if you have a list of standard Gaussian type integrals).

I've ended up with:

<X> = (h/2mw)² [A*B + B*A]

Now to obtain |ψ⟩ which maximizes <X>, we set <X> = 0 and solve for A and B. This gives: A*B = -B*A. How would one solve for the two unknowns then?

 

[PeroK]

I've ended up with:

<X> = (h/2mw)² [A*B + B*A]

Now to obtain |ψ⟩ which maximizes <X>, we set <X> = 0 and solve for A and B. This gives: A*B = -B*A. How would one solve for the two unknowns then?

I'd double check that factor you've got.

I'll give you that $A^*B + B^*A = A^*B + (A^*B)^* = 2Re(A^*B)$. It's useful not to forget that.

To maximise this, you know that $|A|^2 + |B|^2 = 1$. Try expressing these in polar form with magnitudes $a, b$ and note that:

If $a, b$ are real and $a^2 + b^2 = 1$ then $a = cos(\alpha)$ and $b = sin(\alpha)$ for some $\alpha$. It's useful to remember that as well.

It's not too hard to maximise $Re(A^*B)$ using this.

 

[MFAHH]

I'd double check that factor you've got.

I'll give you that $A^*B + B^*A = A^*B + (A^*B)^* = 2Re(A^*B)$. It's useful not to forget that.

To maximise this, you know that $|A|^2 + |B|^2 = 1$. Try expressing these in polar form with magnitudes $a, b$ and note that:

If $a, b$ are real and $a^2 + b^2 = 1$ then $a = cos(\alpha)$ and $b = sin(\alpha)$ for some $\alpha$. It's useful to remember that as well.

It's not too hard to maximise $Re(A^*B)$ using this.

Ah oops, is it: <X> = (h/2mw)^1/2 [A*B + B*A]

I ended up with α = π/4.

And so |ψ⟩ = A|0⟩ + B|1⟩ = cos(π/4) |0⟩ + sin(π/4) |1⟩

Not sure about that part as we've solved for the real parts of A and B, but not A and B themselves.

 

[PeroK]

Ah oops, is it: <X> = (h/2mw)^1/2 [A*B + B*A]

I ended up with α = π/4.

And so |ψ⟩ = A|0⟩ + B|1⟩ = cos(π/4) |0⟩ + sin(π/4) |1⟩

Not sure about that part as we've solved for the real parts of A and B, but not A and B themselves.

I'd move on with with $A = B = \frac{1}{\sqrt{2}}$

$Re(A^*B) = Re(ae^{-i\theta_A} be^{i \theta_B}) = abcos(\theta_B - \theta_A)$ was what you missed out.

 

[MFAHH]

I'd move on with with $A = B = \frac{1}{\sqrt{2}}$

$Re(A^*B) = Re(ae^{-i\theta_A} be^{i \theta_B}) = abcos(\theta_B - \theta_A)$ was what you missed out.

Ah I see now, so it's the case that:
$|ψ⟩ = \frac{1}{\sqrt{2}}|0⟩ + \frac{1}{\sqrt{2}}|1⟩$

 

[PeroK]

Ah I see now, so it's the case that:
$|ψ⟩ = \frac{1}{\sqrt{2}}|0⟩ + \frac{1}{\sqrt{2}}|1⟩$

That's the simplest. I.e. taking $\theta_A = \theta_B = 0$. Although, any common $\theta$ would have done.

 

[MFAHH]

That's the simplest. I.e. taking $\theta_A = \theta_B = 0$. Although, any common $\theta$ would have done.

Is it worth putting $\theta$ in or considering the simplest case for the following parts do you think?

 

[PeroK]

Is it worth putting $\theta$ in or considering the simplest case for the following parts do you think?

Definitely not! It says "construct a state ...". Always take the easiest option.

 

[MFAHH]

Definitely not! It says "construct a state ...". Always take the easiest option.

Good point :).

For the first part of b) I've got the following, just want to check it's right.

$|ψ⟩ = e^{\frac{-iE_0t}{h}}\frac{1}{\sqrt{2}}|0⟩ + e^{\frac{-iE_1t}{h}}\frac{1}{\sqrt{2}}|1⟩$ where $E_n$ is the nth energy eigenvalue given by $E_n = (n+\frac{1}{2})hw$

 

[PeroK]

Good point :).

For the first part of b) I've got the following, just want to check it's right.

$|ψ⟩ = e^{\frac{-iE_0t}{h}}\frac{1}{\sqrt{2}}|0⟩ + e^{\frac{-iE_1t}{h}}\frac{1}{\sqrt{2}}|1⟩$ where $E_n$ is the nth energy eigenvalue given by $E_n = (n+\frac{1}{2})hw$

Yes, that's right.

 

[MFAHH]

Yes, that's right.

Great! So I've calculated $⟨ψ|X|ψ⟩$ and obtained:

$⟨ψ|X|ψ⟩= \sqrt{\frac{h}{8mw}} (e^{\frac{it(E_0 - E_1)}{h}} + e^{\frac{it(E_1 - E_0)}{h}})$

As for part c, for $⟨ψ|X|ψ⟩^2$ I square the expression above to get:

$⟨ψ|X|ψ⟩^2 = \frac{h}{8mw} (e^{\frac{2it(E_0 - E_1)}{h}} + e^{\frac{2it(E_1 - E_0)}{h}} + 2)$

Is that correct so far.

 

[PeroK]

Great! So I've calculated $⟨ψ|X|ψ⟩$ and obtained:

$⟨ψ|X|ψ⟩= \sqrt{\frac{h}{8mw}} (e^{\frac{it(E_0 - E_1)}{h}} + e^{\frac{it(E_1 - E_0)}{h}})$

As for part c, for $⟨ψ|X|ψ⟩^2$ I square the expression above to get:

$⟨ψ|X|ψ⟩^2 = \frac{h}{8mw} (e^{\frac{2it(E_0 - E_1)}{h}} + e^{\frac{2it(E_1 - E_0)}{h}} + 2)$

Is that correct so far.

You are going to find QM difficult if you keep forgetting about complex conjugates! You need to simplify those expressions.

 

[MFAHH]

You are going to find QM difficult if you keep forgetting about complex conjugates! You need to simplify those expressions.

Yes, I should've clicked on. It would simplify to:

$⟨ψ|X|ψ⟩= \sqrt{\frac{h}{2mw}} cos(\frac{t}{h} (E_0 - E_1))$

And so:

$⟨ψ|X|ψ⟩^2= \frac{h}{2mw} cos^2(\frac{t}{h} (E_0 - E_1))$

 

[PeroK]

Yes, I should've clicked on. It would simplify to:

$⟨ψ|X|ψ⟩= \sqrt{\frac{h}{2mw}} cos(\frac{t}{h} (E_0 - E_1))$

And so:

$⟨ψ|X|ψ⟩^2= \frac{h}{2mw} cos^2(\frac{t}{h} (E_0 - E_1))$

You can do something with $E_0 - E_1$ as well. And, remember that $cos$ is an even function.

 

[MFAHH]

You can do something with $E_0 - E_1$ as well. And, remember that $cos$ is an even function.

Aha!

$⟨ψ|X|ψ⟩= \sqrt{\frac{h}{2mw}} cos(wt)$

And so:

$⟨ψ|X|ψ⟩^2= \frac{h}{2mw} cos^2(wt)$

Now it looks splendid.

 

[MFAHH]

Final thing, for $⟨ψ|X^2|ψ⟩$ I got:

$⟨ψ|X^2|ψ⟩= \frac{h}{mw}$, is that correct?

 

[PeroK]

Final thing, for $⟨ψ|X^2|ψ⟩$ I got:

$⟨ψ|X^2|ψ⟩= \frac{h}{mw}$, is that correct?

That's what I got.

 

[MFAHH]

Perfect! Many thanks for your time and help. I think I've got this down now :)