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 oscilator

define ## 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.

Homework Helper
insert ##\Psi## on bra as well
what happens to the ##i## in front of ##\psi_2## ?

renec112
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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.

Last edited:
renec112
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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
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?

Homework Helper
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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##

renec112
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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
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

BvU and PeroK