Griffiths Problem 3.35. Harmonic Oscillator, Bra-ket notation

[Irishdoug]

Homework Statement

Among the stationary states of the harmonic oscillator |n> = $\phi_{n}(x)$ only n = 0 hits the uncertainty limit $\frac{\hbar}{2}$. But certain linear combinations also minimise the uncertainty product. They are eigenfunctions of the lowering operator. Calculate $<x>$, $<x^{2}>$.

Relevant Equations

$x = \sqrt{\frac{\hbar}{2m\omega}}(a_{+} + a_{-})$

$x^{2} = \frac{\hbar}{2m\omega}$ ($a_{+}^{2} + 2a_{+}a_{-} +1 + a_{-}^{2}$)

The hermitian conjugate of $a_{+} = a_{-}$

Firstly, apologies for the latex as the preview option is not working for me. I will fix mistakes after posting.

So for $<x>$ = ($\sqrt{\frac{\hbar}{2m\omega}}$) $(< \alpha | a_{+} + a_{-}| \alpha >)$ = ($\sqrt{\frac{\hbar}{2m\omega}}$) $< a_{-} \alpha | \alpha> + <\alpha | a_{-} \alpha >$.

The final answer is then ($\sqrt{\frac{\hbar}{2m\omega}}$) $(\alpha + \alpha^{*})$. I am unsure as to why this is i.e. why does $< a_{-} \alpha | = 1$?

Likewise for $<x^{2}>$:

I can get as far as ($\frac{\hbar}{2m\omega}$)($<a_{-}^{2} \alpha | \alpha> + 2<a_{-} \alpha | a_{-} \alpha > + 1 + <\alpha |a_{-}^{2} \alpha>$

The answer is ($\frac{\hbar}{2m\omega}$) ($(\alpha^{*})^{2}) + 2\alpha^{*}\alpha + 1 + (\alpha^{*})^{2})$).

Again, I am unsure as to why the lowering operator squared leads to $(\alpha^{*})^{2}$ and $(\alpha)^{2}$?

Can I have some help please.

 

[PeroK]

Presumably $|\alpha \rangle$ is a normalised eigenstate of the lowering operator $a_{-}$ with eigenvalue $\alpha$. All follows from that:
$$a_{-} |\alpha \rangle = \alpha |\alpha \rangle$$

 

[ProxyMelon]

Hi, a coherent state is an eigenstate of the lowering operator as you said, so you have to work with:
$\eta|\beta\rangle=\beta|\beta\rangle\Rightarrow\langle\beta|\eta^{\dagger}=\langle\beta|\beta^*$

Now, I use a slightly different notation (mean value do not change as long as you keep the canonical commutation relations ok), so i write x as:
$x=-i\sqrt{\frac{\hbar}{2m\omega}}(\eta^{\dagger}-\eta)$
(your notation, which is more conventional, should work the same), so:
$\langle x \rangle = \langle \beta|x|\beta\rangle=-i\sqrt{\frac{\hbar}{2m\omega}}\langle \beta| (\eta^{\dagger}-\eta) |\beta\rangle=-i\sqrt{\frac{\hbar}{2m\omega}}\left(\langle \beta| \eta^{\dagger} |\beta\rangle-\langle \beta| \eta|\beta\rangle\right)$
Now you are supposed to apply $\eta^{\dagger}$ to the bra and $\eta$ to the ket, getting:
$-i\sqrt{\frac{\hbar}{2m\omega}}(\beta^*-\beta)$

(notice that once you apply the operator to the bra or ket, you are left with the eigenvalue and the braket $\langle\beta|\beta\rangle=1$).

You can easily check that with your definition of x you get the result you want (and with mine as well of course, a mean value is the result of a measurement and cannot depend on such choices).

You should be able to easily evaluate $\langle x^2\rangle$ in the same way, you just have to calculate it's value in terms of the rasing and lowering operator and then do the count, keep in mind that the two operator do not commute, so you have to treat them with care.

Tell me if everything is understandable.

 

[Irishdoug]

Thankyou for the response. SO am I correct in saying the following:

for clarity, I am going to call the eigenvalue $\beta$ and the eigenfunction $\alpha$ whilst the constant square root term C

(C)($<a_{-} \alpha | \alpha> + <\alpha | a_{-} \alpha> = \beta^{*} | \alpha > + <\alpha | \beta$)

This then equals (C)$(\beta^{*} + \beta) (<\alpha|\alpha> = (C)(\beta^{*} + \beta) (1) = <x>$

I feel that the jump from $|\alpha> + <\alpha|$ to $(<\alpha|\alpha>)$ is not correct but I am unsure.

 

[DrClaude]

(C)($<a_{-} \alpha | \alpha> + <\alpha | a_{-} \alpha> = \beta^{*} | \alpha > + <\alpha | \beta$)

The last equality doesn't hold. The first bra and the last ket cannot be replaced by scalars.

 

[ProxyMelon]

It is not correct, you are never left with single bras or ket, you get, step by step (for clarity, $\eta$ is the lowering operator, it's hermitiane conjugate $\eta^{\dagger}$ is the raising operator and $\gamma$ is the eigeinvalue associated to the coherent state $|\beta\rangle$):
$\langle\beta|\eta^{\dagger}|\beta\rangle=(\langle\beta|\eta^{\dagger})|\beta\rangle=\langle\beta|\gamma^*|\beta\rangle=\gamma^*\langle\beta|\beta\rangle=\gamma^*$
So to keep track of everything:
1) realize that $\eta|\beta\rangle=\gamma|\beta\rangle\leftrightarrow(\eta|\beta\rangle)^*=(\gamma|\beta\rangle)^*\leftrightarrow\langle\beta|\eta^{\dagger}=\langle\beta|\gamma^*$
2) understand that by applying the operator to an eigenstate you are left with the eigenvalue and the eigenstate, what you did is $\eta|\beta\rangle=\gamma$, which is clearly wrong, by applying an operator (that you can picture as a matrix) on a vector, you must get a vector back, and with eigenvectors you just have the added property of getting the exact same vector (multiplied by a constant, the eigenvalue)
3) exploit orthonormality $\langle\beta_n|\beta_m\rangle=\delta_{mn}$, so in this case $\langle\beta|\beta\rangle=1$
4) just to be clear, you are not supposed to sum ket and bras, giving a detailed explanaion of why not would be cumbersome, but just assume you will never have to do $|A\rangle+\langle B|$

 

[Irishdoug]

Ah right ok! That makes perfect sense. Thankyou both for your replies.

 

[PeroK]

Thankyou for the response. SO am I correct in saying the following:

for clarity, I am going to call the eigenvalue $\beta$ and the eigenfunction $\alpha$ whilst the constant square root term C

(C)($<a_{-} \alpha | \alpha> + <\alpha | a_{-} \alpha> = \beta^{*} | \alpha > + <\alpha | \beta$)

This then equals (C)$(\beta^{*} + \beta) (<\alpha|\alpha> = (C)(\beta^{*} + \beta) (1) = <x>$

I feel that the jump from $|\alpha> + <\alpha|$ to $(<\alpha|\alpha>)$ is not correct but I am unsure.

You need to decide whether you are using Dirac notation or not. This is a mixture of Dirac notation and more standard linear algebra. In Dirac notation we have (let's put hats on operators and use $\lambda$ for the eigenvalue):
$$\hat x = C(\hat a_+ + \hat a_-)$$
$$\langle x \rangle = \langle \alpha | \hat x | \alpha \rangle = \langle \alpha |C(\hat a_+ + \hat a_-) | \alpha \rangle = C(\langle \alpha |\hat a_+ | \alpha \rangle + \langle \alpha |\hat a_- | \alpha \rangle)$$
Now, in general, we have (this is an important result):
$$\langle \alpha | \hat A | \beta \rangle = \langle \beta | \hat A^{\dagger} | \alpha \rangle^*$$
Applying this and using $\hat a_+^{\dagger} = \hat a_-$ we have:
$$\langle x \rangle = C(\langle \alpha |\hat a_- | \alpha \rangle^* + \langle \alpha |\hat a_- | \alpha \rangle) = C(\langle \alpha |\lambda | \alpha \rangle^* + \langle \alpha |\lambda | \alpha \rangle) = C(\lambda^*\langle \alpha |\alpha \rangle^* + \lambda \langle \alpha |\alpha \rangle) = C(\lambda^* + \lambda)$$
You ought to check that you can justify every step. Then move on to the second case.

 

[Irishdoug]

Thankyou very much.