# Quantum States and ladder operator

1. Mar 2, 2014

### 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?

2. Mar 2, 2014

### Staff: Mentor

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.

You get another state, called |1>.

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

Last edited: Mar 2, 2014
3. Mar 2, 2014

### stevendaryl

Staff Emeritus
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.

Last edited: Mar 2, 2014
4. Mar 2, 2014

### dauto

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

5. Mar 2, 2014

### Bill_K

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.

6. Mar 2, 2014

### Staff: Mentor

Ah, right. Sorry.

7. Mar 3, 2014

### kashokjayaram

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 cant distinguish... thats why..!!

8. Mar 3, 2014

### 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.

9. Mar 3, 2014

### WannabeNewton

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.

10. Mar 3, 2014

### stevendaryl

Staff Emeritus
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.

11. Mar 3, 2014

### kashokjayaram

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??

12. Mar 3, 2014

### stevendaryl

Staff Emeritus
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.

13. Mar 3, 2014

### 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 $|\psi_1 \rangle and [itex]\psi_2 \rangle$ 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.

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.,
if additionally $\hat{R}^2=\hat{R}$.

It's clear that both definitions of a pure state are equivalent. For each representant $|\psi \rangle \neq 0$ the operator
$$\hat{R}=\frac{1}{\|\psi \|^2} |\psi \rangle \langle \psi|$$
fulfills the requirements of a Statistical operator, representing a pure state.

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
$$\hat{R}=|\psi \rangle \langle \psi|.$$

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!

14. Mar 3, 2014

### stevendaryl

Staff Emeritus
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?