Lowering operator for harmonic oscillator

[physica]

I have been told that for a ground state harmonic oscillator, if a lowering operator is placed on the extreme right no matter what operators follow the expectation value will be zero. I don't fully understand this. Is this because the right operator acts first and lowering the ground state will result in zero, once this is happened it cannot be raised by a raising operator? Can somebody be so kind as to post maths on this if possible.

 

On a further note I am completely clueless as to why unequal numbers of raising and lowering operators equate a ground state harmonic oscillator to a zero expectation value.

[Mordred]

If I'm not mistaken and I could be QM isn't my strong suit

 

The reason Is covered on this page where it discusses zero point energy and the Heisenburg uncertainty.

 

http://en.m.wikipedia.org/wiki/Quantum_harmonic_oscillator

 

 

"Third, the lowest achievable energy (the energy of the n = 0 state, called the ground state) is not equal to the minimum of the potential well, but ħω/2 above it; this is called zero-point energy. Because of the zero-point energy, the position and momentum of the oscillator in the ground state are not fixed (as they would be in a classical oscillator), but have a small range of variance, in accordance with the Heisenberg uncertainty principle. This zero-point energy further has important implications in quantum field theory and quantum gravity."

Edited December 12, 2014 by Mordred

[ajb]

Have a look at the idea of normal ordering, it may help you. I am sure a quick 'google' will give you the idea.

[physica]

I've had a look but I'm still a little stuck. I've googled normal ordering and I get the impression that if the ground state is acted on a lowering operator first then the the wave functions in the integral build on an orthonormal basis thus resulting in zero??? I'm not sure why this happens if it does.

[Mordred]

What you need to do is look at Wicks theorem and how it applies to the term normal ordering. This article specific to the harmonic oscillator shows the maths involved with the harmonic oscillator and wicks theorem. Also pay attention to the ordering of the creation and annihilation operators.

http://www.google.ca/url?sa=t&source=web&cd=2&ved=0CCMQFjAB&url=http%3A%2F%2Fweb.physik.rwth-aachen.de%2F~meden%2Fvielteilchenneu%2Fskriptka2.pdf&rct=j&q=normal%20ordering%20of%20quantum%20harmonic%20oscillator&ei=ul2SVOrhHMOvogTe34DgAg&usg=AFQjCNGbt0Ht-dYx3c0TG_nI5R-tGVyaSg&sig2=uO2WklOZPeDdKz2lY2yHVQ

 

As stated QM isn't my strong suit so I'm going off my limitted understanding.

In wicks theorem your annihilation operators are to the right of your creation operators. In normal ordering.

Here is wikis coverage of Wicks theorem

http://en.m.wikipedia.org/wiki/Wick's_theorem

As the first article also refers to the Bose Einstein and Fermi Dirac statistics these two articles may be of use.

 

http://arxiv.org/pdf/hep-th/0503203.pdf"Particle Physics and Inflationary Cosmology" by Andrei Linde

http://www.wiese.itp.unibe.ch/lectures/universe.pdf:"Particle Physics of the Early universe" by Uwe-Jens Wiese

 

The second article has excellent examples of how to use those two equations in chapter 3

 

Hope this helps Ajb would be better stepping you thru the math than I.

Edited December 18, 2014 by Mordred

[Mordred]

Found the paper on Wick calculus.

http://www.google.ca/url?sa=t&source=web&cd=3&ved=0CCEQFjAC&url=http%3A%2F%2Farxiv.org%2Fpdf%2Fphysics%2F0212061&rct=j&q=wick's%20theorem%20of%20quantum%20harmonic%20oscillator&ei=MEyTVLX-HMK2oQSg8oGICQ&usg=AFQjCNHvnUvsxkAGdH7-BqnSpM1-szd42A&sig2=QGuj7wjsPqzS1XmiZX2KYg&bvm=bv.82001339,d.cGU

[elfmotat]

Basically it just comes from the fact that [math]a |0 \rangle = 0[/math]. So if you have some operator [math]M[/math], then [math]Ma |0 \rangle = M \, (0) = 0.[/math].

 

Is this because the right operator acts first and lowering the ground state will result in zero, once this is happened it cannot be raised by a raising operator?

 

Exactly.

Edited December 18, 2014 by elfmotat

[Mordred]

Thanks Elfomat I recall where I've seen this before. I definetely need to review my Dirac notation lol. In particular the Brac and ket portions and the raising and lowering operators. It's been awhile since I last looked at it. Thankfully I still have the textbooks I need to review

For that matter I can't recall how to latex the raising operator notation. Could you post an example please. I'll use the + sign in its place.

 

[a,a+]=1. As stated I can't recall the latex symbol for the raising operator a+

[physica]

Thank you guys so much. If I have an unequal number of lowering operators to raising operators in a ground state would this also equal zero? If so is it because the wavefunctions will be orthogonal resulting in a zero value?

Edited December 19, 2014 by physica

[elfmotat]

Thank you guys so much. If I have an unequal number of lowering operators to raising operators in a ground state would this also equal zero? If so is it because the wavefunctions will be orthogonal resulting in a zero value?

 

Yes. For example: [math] \langle 0| a^{\dagger} \, a \, a^{\dagger} |0 \rangle = A \langle 0 | 1 \rangle = 0[/math], where A is some normalization constant.

Also, the operation [math]a |0 \rangle = 0[/math] is sometimes referred to as "killing/annihilating the vacuum/ground state." In lectures or notes you might see someone say "such and such operator will annihilate the vacuum, leaving such and such terms left over..."

[Mordred]

Thanks that was the latex command I forgot [latex]\dagger[/latex]