Quantum States and ladder operator

[kashokjayaram]

In any textbooks I have seen, vacuum states are defined as:
a |0>= 0
What is the difference between |0> and 0?

Again, what happens when a⁺ act on |0> and 0?

and Number Operator a⁺a act on |0> and 0?

 

[mfb]

0 is a real number, like 1 and $\pi$
Edit: Sorry not here, see post #5.

|0> is a state. The digit inside is just a convenient name. You could name it |myfavoritestate> and nothing would change (well, the formulas would look more complicated).

Operators "act on" states. If you write a⁺ 0 (or any other operator instead of a⁺), this is not an operator acting on a state, it is an operator multiplied with a real number - and a multiplication with 0 has the result 0.

Again, what happens when a+ act on |0>

You get another state, called |1>.

and Number Operator a+a act on |0>

You get the number of the state |0>, which is 0 (hence the name of the state).

 

[stevendaryl]

Really, you should first thoroughly understand the way raising and lowering operators work in the non-relativistic quantum mechanics of the Harmonic Oscillator. There, you start with observables x and p, and you make the switch to ladder operators

A = \sqrt{\frac{m \omega}{2 \hbar}} x + i \sqrt{\frac{1}{2 m \omega \hbar}} p
A^\dagger = \sqrt{\frac{m \omega}{2 \hbar}} x - i \sqrt{\frac{1}{2 m \omega \hbar}} p
In terms of A and A^\dagger, the corresponding states are:

|n\rangle where

A\ |n \rangle = \sqrt{n}\ |n-1\rangle
A^\dagger\ |n\rangle = \sqrt{n+1}\ |n+1\rangle

The state |0\rangle, when you translate back into x language, is something like C e^{-\lambda x^2} for appropriate constants \lambda and C. It's not zero, it's just that the operator A acting on it produces 0.

 

[dauto]

|0> represents a physical state. 0 is just a number.

 

[Bill_K]

0 is a real number, like 1 and $\pi$

|0> represents a physical state. 0 is just a number.

Not in this case. When you act on a state like |0> with an operator like a, you do not get a number, you get another element of the Hilbert space.

In the example given, a |0>= 0, the 0 on the right hand side is the zero element of the Hilbert space.

 

[mfb]

Ah, right. Sorry.

 

[kashokjayaram]

In the example given, a |0>= 0, the 0 on the right hand side is the zero element of the Hilbert space.

Zero element of the Hilbert space means what??
|0> is also the element of Hilbert space...! Isn't it?
Is it the "smallest element(norm of which is small)"..??
Or why is then a acts gives zero..??

Sorry I can't distinguish... that's why..!

 

[vanhees71]

No, |0 \rangle is the vacuum (ground) state. It's not the null vector of the Hilbert space. That's why often one better writes another symbol, e.g., |\Omega \rangle for the vacuum or ground state, which is a state of minimal energy.

 

[WannabeNewton]

Or why is then a acts gives zero..??

Because $|0\rangle$ is the lowest possible energy eigenstate of $\hat{H}$. You can't go any lower. $\hat{a}$ is a lowering operator that takes you from excited states to less excited states but if $|0 \rangle$ is the lowest possible mode excitation of the harmonic oscillator then $\hat{a}$ can't take you any lower than that so it must annihilate $|0 \rangle$ i.e. $\hat{a} |0 \rangle = 0$.

But $|0\rangle$ is not the zero vector. In fact if you solve for $\psi_0(x) = \langle x | 0 \rangle$ in the differential equation $\langle x | \hat{a} | 0 \rangle = x\psi_0 + x_0^2 \frac{d\psi_0}{dx} = 0$ you'll get a Gaussian.

 

[stevendaryl]

Zero element of the Hilbert space means what??
|0> is also the element of Hilbert space...! Isn't it?
Is it the "smallest element(norm of which is small)"..??
Or why is then a acts gives zero..??

Sorry I can't distinguish... that's why..!

If |\psi\rangle is any state, and \alpha is any number (real or complex), then there is another state |\psi'\rangle = \alpha |\psi\rangle. The "zero element" 0 is what you get when you choose \alpha = 0.

The state |0\rangle is not the zero element. It's a state with no particles in it--the vacuum. The operator N = a^\dagger\ a is the "number operator". The state |n\rangle is the eigenstate with n particles. It satisfies the equation:

N\ |n\rangle = n\ |n \rangle

So the state |0\rangle is the state satisfying N |0\rangle = 0 |0\rangle =0.

 

[kashokjayaram]

If |\psi\rangle is any state, and \alpha is any number (real or complex), then there is another state |\psi'\rangle = \alpha |\psi\rangle. The "zero element" 0 is what you get when you choose \alpha = 0.

The state |0\rangle is not the zero element. It's a state with no particles in it--the vacuum. The operator N = a^\dagger\ a is the "number operator". The state |n\rangle is the eigenstate with n particles. It satisfies the equation:

N\ |n\rangle = n\ |n \rangle

So the state |0\rangle is the state satisfying N |0\rangle = 0 |0\rangle =0.

Why is it another state...?? \alpha is a number which doesnot change the vectors in vector space. So it will be the same state. Am I right..?
What happens when a^\dagger act on 0??

 

[stevendaryl]

Why is it another state...?? \alpha is a number which doesnot change the vectors in vector space. So it will be the same state. Am I right..?
What happens when a^\dagger act on 0??

Well, there is an ambiguity about what "the same state" means. A lot of quantum mechanics uses normalized states, where you multiply by a constant so that \langle \Psi | \Psi \rangle = 1. But the zero vector cannot be normalized. It's not equal to any other state.

And operating on the zero vector with any operator just returns the zero vector again.

 

[vanhees71]

In standard quantum theory the pure states are equivalently described by either rays in Hilbert space or the special class of statistical operators that are projection operators.

A ray in Hilbert space is given by an equivalence class of non-zero vectors. Two non-zero vectors |\psi_1 \rangle and \psi_2 \rangle belong to the same class if and only if there is a non-zero complex number \lambda such that |\psi_2 \rangle=\lambda |\psi_1 \rangle. Sometimes (and again equivalently) for convenience one restricts oneself to normalized state vectors only. Then two state vectors are in the same ray if they differ by a phase factor only, i.e., a complex number of modulus 1.<br /> <br /> A Statistical operator is any self-adjoint positive semidefinite operator \hat{R} with \mathrm{Tr} \hat{R}=1. It represents a pure state if and only if it is a projection operator, i.e.,<br /> if additionally \hat{R}^2=\hat{R}.<br /> <br /> It&#039;s clear that both definitions of a pure state are equivalent. For each representant |\psi \rangle \neq 0 the operator<br /> \hat{R}=\frac{1}{\|\psi \|^2} |\psi \rangle \langle \psi|<br /> fulfills the requirements of a Statistical operator, representing a pure state.<br /> <br /> If, on the other hand, \hat{R} represents a pure state in the above given sense, then due to \hat{R}^2=\hat{R} it has only eigenvalues 0 and 1. Since at the same time \mathrm{Tr} \hat{R}=1 there is one and only one eigenvector with eigenvalue 1. Choosing it normalized \langle \psi|\psi \rangle this implies that the statistical operator must be the projection operator<br /> \hat{R}=|\psi \rangle \langle \psi|.<br /> <br /> It is important to keep in mind that a pure state is represented not exactly by a (normalized) Hilbert-space vector but by an entire ray (normalized Hilbert-space vector modulo an arbitrary phase factor), because it implies important subtle points: e.g., it admits that also half-integer eigenvalues make sense for angular momenta and thus that particles with half-integer spins exist. Among others these are the electrons, protons, and neutrons, making up all everyday matter!

 

[stevendaryl]

It is important to keep in mind that a pure state is represented not exactly by a (normalized) Hilbert-space vector but by an entire ray (normalized Hilbert-space vector modulo an arbitrary phase factor), because it implies important subtle points: e.g., it admits that also half-integer eigenvalues make sense for angular momenta and thus that particles with half-integer spins exist. Among others these are the electrons, protons, and neutrons, making up all everyday matter!

That's a leap that's too big for me to follow. Why does the use of rays, rather than vectors, imply anything about half-integer spins? Oh, maybe because you don't require that a 360 degree rotation is the identity, but only that it brings you another vector in the same ray?