Register to reply 
Second quantizationby daudaudaudau
Tags: quantization 
Share this thread: 
#1
Dec309, 10:39 AM

P: 296

Hi. In second quantization (not QFT or anything advanced like that) we have the particle density [itex]\hat n(x)=\Psi^{\dagger}(x)\Psi(x)[/itex] using the usual field creation/annihilation operators. For a single particle we obtain for the expectation value in the state [itex]\psi\rangle[/itex]: [itex]\langle \psi  \Psi^{\dagger}(x)\Psi(x)  \psi\rangle=\psi(x)^2[/itex]. So does this mean that I should think of [itex]\Psi(x)[/itex] as the second quantized wave function? What is the significance of this formal similarity between the wave function and the creation/annihilation operators?
It seems that whenever you have something in first quantization written in terms of a wave function (i.e. the probability current density), you can replace the wave functions by the creation/annihilation operators and get the second quantized operator... 


#2
Dec409, 02:15 PM

P: 5

Yes, field operator [tex]\hat \Phi[/tex] and wave function [tex]\phi[/tex] are very similar objects. The formal reason is that [tex]\hat \Phi(x) = \sum_k \phi_k(x) \hat a_k [/tex]  it is a "superposition" of wave functions with coefficients being annihilation operators. That's why the technique is called "second quantization". In the "first quantization" we go from physical obserables (which are numbers) to operators. And now we go from wave functions (which are numbers) to operators. For example, if you write motion equations for [tex]\hat\Psi(x,t)[/tex] in the free particle theory then it will coincide with Schroedinger equation for [tex]\psi(x,t) [/tex]



#3
Dec409, 02:37 PM

P: 296

Thank you for the reply. If you have the particle density operator [itex]\hat n(x)=\Psi^{\dagger}(x)\Psi(x)[/itex], you get the particle density in a particular state by taking the expectation value of this operator. Is it possible somehow to get the wave function by taking the expectation value of an operator?



#4
Dec409, 02:43 PM

Sci Advisor
P: 779

Second quantization
To expand slightly on SnarkyFellow's response:
Think of a secondquantized field operator [tex]\hat{\Phi}(x)=\sum_k \phi_k(x)\hat{a}_k[/tex] as the "hammer that bangs the vacuum"  and everytime it bangs, it "creates" a state with wavefunction [itex]\phi_k(x)[/tex]. Thus, when you want to talk about multiple particle states in quantum mechanics, it is useful to use this "field operator" rather than an ordinary wavefunction. That is why you use "second quantized quantum mechanics" in "many body theory". And in fact, to dau^4's first sentence: there is no real difference between this and QFT! But that is just semantics. ;) 


#5
Dec409, 02:46 PM

Sci Advisor
P: 779

When [itex]\hat{n}(x)[/itex] bangs the state you're in (analogous to you taking the expectation value of the operator), you get [itex]\phi_k(x)^2[/itex] which is, of course, the particle density, so yes!



#6
Dec409, 02:53 PM

P: 1,160

How about the vacuum average of Ψ(x)a_{k}^{+} ?



#7
Dec409, 03:00 PM

Sci Advisor
P: 779

you mean, i left out the "negative frequency" terms? sure. that's there too! i was typing quickly. I should have written:
[tex]\hat{\Phi}(x)=\sum_k\left(\phi_k(x)a_k+\phi_k(x)^*a_k^\dag\right)[/tex] Then [itex]\langle 0\hat{\Phi}^\dag(x)\hat{\Phi}(x)0\rangle = \sum_k\phi_k(x)^2[/itex]. 


#8
Dec409, 03:14 PM

P: 296

Or maybe I'm looking for the second quantization equivalent of [itex]\langle x\psi\rangle=\psi(x)[/itex], i.e in second quantization the creation or annihilation operator would be on the right and something other on the left hand side. 


#9
Dec409, 03:23 PM

Sci Advisor
P: 779




#10
Dec409, 03:33 PM

P: 296




#11
Dec409, 03:36 PM

P: 296




#12
Dec409, 03:52 PM

P: 296

If someone can recommend a good book about this, that would be great.. I have read Ballentine's chapters on this topic.



#13
Dec409, 04:31 PM

Sci Advisor
P: 779

[tex]\psi\rangle=\int dx\psi(x)\psi\rangle[/tex] [itex]\psi[/itex] is not a position eigenstate, and the "wavefunction" is given by [itex]\langle x\psi\rangle[/itex]. My sum over k is analogous the integral over x in this past example. So if you want to have the wavefunction of a specific mode, you would write [tex]\langle k\hat{\Phi}(x)0\rangle=\phi^*_k(x)[/tex] (where the c.c. is b/c of my phase conventions). Is that better? 


#14
Dec409, 05:22 PM

P: 171

Moreover, the state with 0 particles is orthogonal to the one with a particle. Perhaps there is a confusion between different second quantized notation going on. As I understand it, the notation in the original post is the following: [itex]\hat{\Phi}^{\dagger}(x)[/itex] creates a particle at x. i.e, [itex]\hat{\Phi}^{\dagger}(x)0\rangle =  x\rangle[/itex] [itex]\hat{\Phi}(x)[/itex] destroys a particle at x. The fact that the symbol for the operators is [itex]\Phi[/itex] is neither here nor there, we could have chosen [itex]c^{\dagger}_x[/itex]. I was under the impression, that for a 1particle system, we have the wave function when the system is in state [itex]\psi\rangle[/itex] as [itex]\psi(x) = \langle x\psi\rangle[/itex] Inserting the the other expression for [itex]x\rangle[/itex] in second quantized language [tex] \psi(x) = \langle 0\Phi(x)\psi\rangle [/tex] The quantity [tex]\psi^2[/tex] is then [tex] \psi(x)^2 = \langle \psi\Phi^\dagger(x)0\rangle\langle 0\Phi(x)\psi\rangle [/tex] As long as [itex]\psi\rangle[/itex] is a one particle state (so our single particle wavefunction makes sense), the vacuum projection operator just acts like the resolution of the identity so we can ignore it to give [tex] \psi(x)^2 = \langle \psi\Phi^\dagger(x)\Phi(x)\psi\rangle [/tex] For a many particle state, the particle density is more complicated in the wavefunction picture, but the second quantised operator remains the same. As far as book recommendations go, if you are coming at this from the condensed matter perspective, the book by Negele and Orland is highly recommended (by me at least)  the first couple of chapters on second quantisation are very thorough. "Condensed matter field theory" by Altland and Simons has a good intro to second quantisation  you can get the flavour of the book by reading Ben Simon's lecture notes for his Quantum Condensed Matter Field Theory course. (google him). MIT Open Course Ware might well have some decent lecture notes available too, I haven't checked. Wen has a recent graduate level Condensed Matter Field Theory textbook too, which surely has a chapter on second quantization. 


#15
Dec409, 05:33 PM

Sci Advisor
P: 779

i think, as you say, this is notational. but if you look at my earlier post, i have the field operator as the SUM of creation and annihilation operators, so this works itself out. If you split it up as positive and negative frequency terms, then you are right.
I also used Negele and Orland, an excellent book, but it is quite advanced (at least I think so). 


#16
Dec409, 05:39 PM

P: 296

Thank you for clearing things up, Peter, and I will definitely have a look at those books. Could you maybe explain a little more about why the vacuum projector becomes the identity operator in the single particle case?



#17
Dec409, 06:24 PM

P: 171

A full identity operator (for Fock space) would have been
[tex] 0\rangle\langle 0 + \sum_kk\rangle\langle k + \sum_{k_1,k_2}k_1,k_2\rangle\langle k_1,k_2 + \ldots [/tex] I.e, the 0particle identity [tex]\oplus[/tex] the identity for a one particle state [tex]\oplus[/tex] the identity operator for two particle states etc. I've chosen 'k' to be the labels of a basis of single particle states. The creation (annihilation) operators move us from an nparticle state to an n+1 (n1) particle state. States with different numbers of particles are orthogonal. So consider: [tex]\langle\psi\Phi^{\dagger}(x)S\rangle[/tex] If [tex]\langle \psi[/tex] is a one particle state, then the only state S for which the above is nonzero is a zero particle state, and there is only one of those, the particle vacuum. So inserting a full resolution of the indentity would reduce to exactly the expression I gave. 


#18
Dec509, 01:29 AM

P: 5

As peteratcam has noticed, wavefunction function can be obtained as matrix element [tex] \langle 0\hat \Psi \psi_k \rangle [/tex]. We can't get a wavefunction as the expectation in a state with defined number of particles. It's due to the fact that [tex]\hat \Psi [/tex] decreases number of particles so in order to give nonzero result, there should be the product of equal number of creation and annihilation operators. But such a product is invariant under the gauge transformation [tex]\phi \rightarrow \phi e^{i\alpha}[/tex]. Therefore it can't give us the wavefunction.
But if we prepare somehow [tex] \alpha0\rangle+\beta\phi\rangle [/tex] state, then expectation of [tex] \hat \Psi [/tex] is [tex] \alpha^*\beta \phi [/tex]. It's rather curious fact, but I suppose it is not really useful in physics because [tex] \hat \Psi [/tex] isn't selfconjugate on Fock space. Therefore it is not observable and we can't get this expectation in an experiment. 


Register to reply 
Related Discussions  
Find the hamiltonian and lagrangian of a chained system  Advanced Physics Homework  0  
Quantization over quantization...  Quantum Physics  1  
Third Quantization?...  General Physics  2 