Ladder operators in quantum mechanics

[bigevil]

Homework Statement

This is problem 2.11 from Griffith's QM textbook under the harmonic oscillator section.

Show that the lowering operator cannot generate a state of infinite norm, ie, \int | a_{-} \psi |^2 < \infty

Homework Equations

This isn't so hard, except that I consistently get the wrong "sign". I have worked based on Griffith's a- operator

a_{-} = \frac{1}{\sqrt{2m}} (\frac{h}{i} \frac{d}{dx} - im\omega x).

As suggested by the question, I am to find that \int (a_{-}\psi)^{"*"} (a_{-}\psi) dx = \int \psi^{"*"} (a_{+} a_{-} \psi) dx. It's easy to get the answer after getting this intermediate step, but I get the wrong sign, ie, - \int \psi^{"*"} (a_{+} a_{-} \psi) dx instead of with a positive sign.

\int (a_{-}\psi)^{"*"} (a_{-}\psi) dx = \int (\frac{1}{\sqrt{2m}}\frac{h}{i} \frac{\partial \psi^{"*"}}{\partial x} - im\omega x \psi^{"*"}) (a_{-} \psi) dx.

Once I multiply out the two brackets (which is ok because there is only one operator involved here right?) I get

\int \frac{h}{i} \frac{1}{\sqrt{2m}} (a_{-}\psi) \frac{\partial \psi^{"*"}}{\partial x} - I am \omega x a_{-} \psi \psi^{"*"} dx

If I apply integration by parts to the first term, I get an expression along the lines of -\int \psi^{"*"} \frac{\partial \psi}{\partial x} dx. This would transform the initial expression to -a_{+} and leave a minus sign in the final expression!

I do know that due to the way Griffiths derives the operators, there is an alternative operator, ie a_{-} = - \frac{h}{i} \frac{d}{dx} + m\omega x. This would turn out the correct answer I think.

But am I missing something here? Why wouldn't both methods check out equally as well? Surely the sign is important in these types of derivations.

 

[Nick89]

Maybe you have a different edition than me, but in my book he states:
a_{\pm} = \frac{1}{\sqrt{2\hbar m \omega}}\left(\mp i p + m \omega x\right)

This would lead to:
a_- = \frac{1}{\sqrt{2\hbar m \omega}}\left(+ ip + m\omega x\right)

Furthermore, it is much easier to show this by noting that a_- and a_+ are hermitian conjugates. So you can write:
\int \left| a_- \psi \right|^2 \, dx = \int (a_- \psi)^*(a_- \psi) \, dx = \int \psi^*(a_+a_- \psi) \, dx
Then use equation 2.54:
a_+a_- = \frac{1}{\hbar \omega} H - \frac{1}{2}
And you should be on your way to get the answer:
\int \left| a_- \psi \right|^2 \, dx = \frac{1}{\hbar \omega} E - \frac{1}{2} < \infty

 

[bigevil]

I've got an older edition (ca 1995!) Nick. It's weird because Griffiths does a few examples (ie, expand a+ a-, etc) with the operator that I stated at the beginning.

I didn't know about the Hermitian part. But this particular question, which I don't think is in newer editions (I checked), requires me to do out the whole thing with integration by parts.

 

[turin]

Nick's method is the way that I would show this to myself, but, perhaps the exercise is to show this using the expressions for a_± in terms of x and p.

You have to be very careful about which operators are acting on which functions.
You also need to be careful about the operator definitions.

\int (a_{-}\psi)^{"*"} (a_{-}\psi) dx = \int (\frac{1}{\sqrt{2m}}\frac{h}{i} \frac{\partial \psi^{"*"}}{\partial x} - im\omega x \psi^{"*"}) (a_{-} \psi) dx

You need to be careful how you are doing the complex conjugation.

\int \frac{h}{i} \frac{1}{\sqrt{2m}} (a_{-}\psi) \frac{\partial \psi^{"*"}}{\partial x} - I am \omega x a_{-} \psi \psi^{"*"} dx

You need to be careful about what the a_- is operating on (i.e. keeping parenthesis would be a good idea).

If I apply integration by parts to the first term, I get an expression along the lines of -\int \psi^{"*"} \frac{\partial \psi}{\partial x} dx.

Momentum (the partial derivative operator) does not commute with a_-.

I do know that due to the way Griffiths derives the operators, there is an alternative operator, ie a_{-} = - \frac{h}{i} \frac{d}{dx} + m\omega x.

I don't have Griffiths QM text, but I suspect that this is a typo. This gives a Hermitian operator, but a_- should not be Hermitian if it is a lowering operator.

 

[dolphus]

There's not a typo in Griffiths. Remember to negate the first trm of the lowering operator that contains an i, as well as the second when you take the complex conjugate. I got hung up at first trying to go too far with the math. For the steps I finally used to get it correct see:

http://copaseticflow.blogspot.com/2011/08/its-obvious-not-knowing-when-not-to-do.html"

Hamilton