### Re: Quantum mech question regarding raising and lowering oporators.

Date: Tue Jun 23 10:58:57 2009
Posted By: Michael Wohlgenannt, PostDoc
Area of science: Physics
ID: 1245197570.Ph
Message:

Dear Geoff,

let us denote the ground state of the harmonic oscillator with |0>, and the excited states with |1>,|2>, .... respectively. By a^+ = b we denote the creation (raising) operator, and by a the annihilation (lowering) operator. They act in the following way on the oscillator states:

```
b|n > = c |n+1 >,

a|n > = d |n-1 >,  a|0> = 0

```
An expection value of some operator A is given by the matrix element
```< m| A |n>.
```
Therefore, for an expectation value in the ground state, we have
```
< 0| A |0 >

```
and we see that whenever A contains an lowering operator a on the right, i.e. A = B a, the expectation value will vanish, because a|0 > = 0. The oscillator states have another important property: they build an orthonormal basis. Every oscillator state function can be written as
```|f > = sum_{a=0}^{infiniy} c_a |a >,
```
with coefficients c_a, AND
```< m|n > = 0, for all m =/= n,
< m|m > = 1, for all m.
```
This also the reason why the expectation value in the ground state of a product of an unequal number of lowering and raising operators always vanishes. E.g. consider the following example:
```< 0| a b^2 |0 > = const < 1 | 2 > = 0.
```

best regards,
Michael

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