Expectation value of raising and lower operator

In summary, by calculating the expectation values of a+ and a- separately and then subtracting them, the expectation value for the operator Y was found to be 0. This was achieved by correctly manipulating the i's in the inner product calculations.
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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.
 
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  • #2
renec112 said:
insert ##\Psi## on bra as well
what happens to the ##i## in front of ##\psi_2## ?
 
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  • #3
renec112 said:
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.
 
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  • #4
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.
 
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  • #5
BvU said:
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?
 
  • #6
renec112 said:
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##
 
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  • #7
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##.
 
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  • #8
PeroK said:
##\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
 
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What is the expectation value of raising and lowering operators?

The expectation value of raising and lowering operators is a mathematical concept used in quantum mechanics to predict the average outcome of a measurement on a quantum system. It is calculated by taking the inner product of the state vector with the operator applied to the state vector.

How is the expectation value of raising and lowering operators related to quantum states?

The expectation value of raising and lowering operators is closely related to the quantum state of a system. It allows scientists to determine the average value of a physical quantity, such as energy or angular momentum, in a particular quantum state.

What are the mathematical expressions for raising and lowering operators?

The raising and lowering operators are represented by the symbols a+ and a-, respectively. They are defined as linear combinations of creation and annihilation operators, which are used to describe the creation and annihilation of particles in quantum systems.

How do raising and lowering operators act on quantum states?

Raising and lowering operators act on quantum states by changing the quantum numbers associated with the state. For example, the raising operator increases the energy of the state by one unit, while the lowering operator decreases it by one unit.

What is the significance of the expectation value of raising and lowering operators in quantum mechanics?

The expectation value of raising and lowering operators plays a crucial role in quantum mechanics, as it allows scientists to make predictions about the behavior of quantum systems. It is also used in many mathematical models and equations, such as the Schrödinger equation, to describe the evolution of quantum states over time.

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