Expectation value of raising and lower operator

[renec112]

I am practicing old exams. I tried my best but looking at an old and a bit unreliable answer list, and i am not getting the same result.

Homework Statement

At time $t=0$ the nomralized harmonic oscialtor wavefunction is given by:
$\Psi(x,0) = \frac{1}{\sqrt{3}}(\psi_0(x) + \psi_1(x) + i \psi_2(x))$

$a_+$ and $a_-$ is respectively the raising and lower operator in the harmonic oscilatordefine $Y = i(a_+ - a_-)$
find the expectation value $\langle Y \rangle$

Homework Equations

$a_+ \psi_n = \sqrt{n+1} \psi_{n+1}$
$a_- \psi_n = \sqrt{n} \psi_{n-1}$

The Attempt at a Solution

Do a sandwich:
$\langle \Psi | Y | \Psi \rangle$
insert operator
$= \langle \Psi | i(a_+ - a_-) | \Psi \rangle$
Split inner product
$= \langle \Psi | i a_+ | \Psi \rangle - \langle \Psi | i a_- | \Psi \rangle$
take $i$ constant out
$=i \langle \Psi | a_+ | \Psi \rangle - i\langle \Psi | i a_-| \Psi \rangle$
Insert wavefunction the a's er operatring on
$= i\langle \Psi | a_+ | \frac{1}{\sqrt{3}}(\psi_0(x) + \psi_1(x) + i \psi_2(x)) \rangle - i \langle \Psi | a_- | \frac{1}{\sqrt{3}}(\psi_0(x) + \psi_1(x) + i \psi2(x))\rangle$
Operate with a's
$= i \langle \Psi | \frac{1}{\sqrt{3}}(\psi_1(x) + \sqrt{2} \psi_2(x) + \sqrt{3} i \psi_3(x)) \rangle - i \langle \Psi | \frac{1}{\sqrt{3}}( \psi_0(x) + i \sqrt{2} \psi_1(x))\rangle$
insert $\Psi$ on bra as well
$= i \langle \frac{1}{\sqrt{3}}(\psi_0(x) + \psi_1(x) + i \psi_2(x)) | \frac{1}{\sqrt{3}}(\psi_1(x) + \sqrt{2} \psi_2(x) + \sqrt{3} i \psi_3(x)) \rangle - i \langle \frac{1}{\sqrt{3}}(\psi_0(x) + \psi_1(x) + i \psi_2(x)) | \frac{1}{\sqrt{3}}( \psi_0(x) + i \sqrt{2} \psi_1(x))\rangle$
Take $1/\sqrt{3}$ out of inner product
$= i \frac{1}{3} \langle (\psi_0(x) + \psi_1(x) + i \psi_2(x)) | (\psi_1(x) + \sqrt{2} \psi_2(x) + \sqrt{3} i \psi_3(x)) \rangle - i \frac{1}{3} \langle (\psi_0(x) + \psi_1(x) + i \psi_2(x)) | ( \psi_0(x) + i \sqrt{2} \psi_1(x))\rangle$
Do inner product
$= i \frac{1}{3} (1 + i \sqrt{2}) - i \frac{1}{3} (1 + i \sqrt{2})$
Take minus inside parenthesis in the last term
$= i \frac{1}{3} (1 + i \sqrt{2}) + i \frac{1}{3} (-1 - i \sqrt{2})$
factorize
$= i \frac{1}{3} (1 + i \sqrt{2} -1 - i \sqrt{2})$
$= 0$

I got the same result last time. Maybe I'm doing the same mistakes? Would love you input.

 

[BvU]

insert $\Psi$ on bra as well

what happens to the $i$ in front of $\psi_2$ ?

 

[PeroK]

Do inner product
$= i \frac{1}{3} (1 + i \sqrt{2}) - i \frac{1}{3} (1 + i \sqrt{2})$
Take minus inside parenthesis in the last term
$= i \frac{1}{3} (1 + i \sqrt{2}) + i \frac{1}{3} (-1 - i \sqrt{2})$
factorize
$= i \frac{1}{3} (1 + i \sqrt{2} -1 - i \sqrt{2})$
$= 0$

I got the same result last time. Maybe I'm doing the same mistakes? Would love you input.

I think there is a mistake here (from the previous expression) manipulating $i$ and taking it out of the inner product.

PS: it's taking the $i$ out of the first term of the inner product: you need the complex conjugate.

 

[PeroK]

PS I would have done this slightly differently. First I would have calculated the expectation values of $a_+$ and $a_-$ separately. Then put them together to get the expectation value for $Y$:

$E(Y) = i(E(a_+) - E(a_-))$

You used this idea to some extent, but you could have broken the calculation down more.

 

[renec112]

what happens to the $i$ in front of $\psi_2$ ?

I checked, but i can't see if i dropped it somewhere?

Thank you PeroK. What are you thinking about more specifically about manipulating i?

 

[PeroK]

I checked, but i can't see if i dropped it somewhere?

Thank you PeroK. What are you thinking about more specifically about manipulating i?

$\langle i \psi | \psi \rangle = -i \langle \psi | \psi \rangle$

 

[PeroK]

For example, if you split the calculation up like this:

$a_- \psi = \frac{1}{\sqrt{3}}(\psi_0 + i\sqrt{2} \psi_1)$

$\langle a_- \rangle = \langle \psi | a_- \psi \rangle = \frac13 \langle \psi_0 + \psi_1 + i\psi_2 | \psi_0 + i\sqrt{2} \psi_1 \rangle$

$= \frac13( \langle \psi_0 | \psi_0 \rangle + \langle \psi_1 |i\sqrt{2} \psi_1 \rangle) = \frac13(1 + i\sqrt{2})$

Then it's a lot easier to check and isolate the mistake. You actually got the calculation of $\langle a_- \rangle$ correct and the mistake was calculating $\langle a_+ \rangle$.

 

[renec112]

$\langle i \psi | \psi \rangle = -i \langle \psi | \psi \rangle$

Oh right i see my mistake. So fixing $a_+$

$= i \frac{1}{3} \langle (\psi_0(x) + \psi_1(x) + i \psi_2(x)) | (\psi_1(x) + \sqrt{2} \psi_2(x) + \sqrt{3} i \psi_3(x)) \rangle$
Do inner product
$= i \frac{1}{3} (1 - i \sqrt{2})$

if i subract $a_-$ with my result from $a_+$
$= i \frac{1}{3} (1 - i \sqrt{2}) - i \frac{1}{3} (1 + i \sqrt{2})$
$= i \frac{1}{3} (1 - i \sqrt{2}) + i \frac{1}{3} (-1 - i \sqrt{2})$
$= i \frac{1}{3} (1 - i \sqrt{2} -1 - i \sqrt{2})$
$= i \frac{1}{3} (- 2 i \sqrt{2})$
Finally giving me
$= \frac{2 \sqrt{2} }{3}$

Now i have the right answer it seems. Thank you guys! I really appreciate your help