What are the properties of Dirac notation and operators?

[Ronin2004]

Homework Statement

[A^{+}A]=1
A|a>=\sqrt{a}|a-1>
A^{+}|a>=\sqrt{a+1}|a+1>
<a'|a>=\delta_{a&#039;}_{a}

Homework Equations

what is
1 <a|A|a+1>
4. <a+1|A^{+}|a>
3. <a|A^{+}A|a>
4. <a|AA^{+}|a>

The Attempt at a Solution

1. <a|A|a+1> =<a|\sqrt{a+1}|a+1-1>=\sqrt{a+1}<a|a>
since a=a and since its the Knocker delta it equals 1 so it equals
\sqrt{a+1}

2.<a+1|A^{+}|a>=<a+1|\sqrt{a+1}|a+1>=
\sqrt{a+1}<a+1|a+1>
since a+1=a+1 and since its the Knocker delta it equals 1 so it equals
\sqrt{a+1}

<a|A^{+}A|a> this is where i am having problems
(<a|A^{+})*(A|a>)= A^{+}A<a|a>=1 is this right

the same is for part 4

 

[fzero]

(<a|A^{+})*(A|a>)= A^{+}A<a|a>=1 is this right

The left-hand side is a good idea, but the right-hand side makes no sense. You can't just pull the operators out of the inner product. However, if you can compute the action of the operator in terms of an eigenvalue, you can pull the ordinary number out. In this case, note that

(\langle a | A^\dagger) = (A | a\rangle)^\dagger,

so that you can use the expression for A|a\rangle to compute(\langle a|A^\dagger)(A|a\rangle) directly.

 

[Ronin2004]

(\langle a | A^\dagger) = (A | a\rangle)^\dagger,

so that you can use the expression for A|a\rangle to compute(\langle a|A^\dagger)(A|a\rangle) directly.

Ok so in this case (\langle a | A^\dagger) = (A | a\rangle)^\dagger, would that make it equal to \sqrt{A}|(a+1) and then how do you take the adjoint of that expresion

 

[fzero]

Ok so in this case (\langle a | A^\dagger) = (A | a\rangle)^\dagger, would that make it equal to \sqrt{A}|(a+1) and then how do you take the adjoint of that expresion

You want to be a bit more careful typing things out here. I realize it can be hard to figure out all of the codes to type formulas here, but you should stick to the same notation you've been using. So instead of \sqrt{A}|(a+1), you probably mean \sqrt{a}|(a+1), since we were using capital A for the operator and lowercase a for the eigenvalue.

In any case, A is a lowering operator, so it lowers the index of the state it acts on. From your original post, you were given the correct result A|a\rangle = \sqrt{a}|a-1\rangle.

As for taking adjoints, the results you need are that

\langle \psi | = ( |\psi \rangle )^\dagger.

( c | \psi \rangle )^\dagger = \langle \psi | c^* ,

where c is a complex number and c^* is its complex conjugate.

 

[Ronin2004]

In any case, A is a lowering operator, so it lowers the index of the state it acts on. From your original post, you were given the correct result A|a\rangle = \sqrt{a}|a-1\rangle.

As for taking adjoints, the results you need are that

\langle \psi | = ( |\psi \rangle )^\dagger.

( c | \psi \rangle )^\dagger = \langle \psi | c^* ,

where c is a complex number and c^* is its complex conjugate.

Since there is no complex number involved with the would it just be that ( A | a \rangle )^\dagger = \langle a | A^* , is equal to \langle a-1 | \sqrt{a} ,

and the answer be \langle a-1 | \sqrt{a} \sqrt{a}|a-1\rangle.
or a*<a-1|a-1>

sorry just having a hard time seeing this its hard to picture
from my understanding the whole thing with bra ket notation is that is is supposed to represent vector space <a| are the columns and |a> are the rows So you have the operator acting upon the rows in such kind of like a scalar on an vector.


\sqrt{a}|a-1\rangle \sqrt{a}|a-1\rangle[/itex].
So in the above is the \sqrt{a} the eigenvalue acting on the vector |a-1\rangle

 

[fzero]

Since there is no complex number involved with the would it just be that ( A | a \rangle )^\dagger = \langle a | A^* , is equal to \langle a-1 | \sqrt{a} ,

The second expression is fine because you just have a number multiplying the state, but when you have an operator in the first expression, you should really use the adjoint A^\dagger. The reason is that, if you are working with vectors and matrices, like you mention below, the adjoint of a matrix operator is obtained by taking the complex conjugate of the matrix elements combined with the transpose of the matrix.

and the answer be \langle a-1 | \sqrt{a} \sqrt{a}|a-1\rangle.
or a*<a-1|a-1>

Yes and you can simplify this a bit more by using orthonormality like you did earlier.

sorry just having a hard time seeing this its hard to picture
from my understanding the whole thing with bra ket notation is that is is supposed to represent vector space <a| are the columns and |a> are the rows So you have the operator acting upon the rows in such kind of like a scalar on an vector.


\sqrt{a}|a-1\rangle \sqrt{a}|a-1\rangle[/itex].
So in the above is the \sqrt{a} the eigenvalue acting on the vector |a-1\rangle

That's correct.

 

[Ronin2004]

Thank you so much for the help. You are a lifesaver

 

[maalpu]

Is the commutator [A⁺ A] = 1 in the givens correct ?

I get [A⁺ A] = -1 based on the definitions that follow.

And therefore [A A⁺] = 1, consistent with my getting answers a and a+1 for equations 3 and 4, regardless of whether I apply both operators in turn to the ket, or apply the first to the bra and the second to the ket.