# Homework Help: Linear operators & the uncertainty principle

1. Jan 27, 2006

### Ahmes

Hi,
I try to understand the proof for the uncertainty principle for two Hermitian operators A and B in a Hilbert space. My questions are rather general so you don't need to know the specific proof.

The first thing I couldn't get into my head was the definition of uncertainty
$$(\Delta A)^2=\langle\psi|(A- \langle A \rangle)^2|\psi\rangle$$
Using the properties of the inner product I can be convinced this is equivalent to the ordinary definition of $$(\Delta A)^2=\langle A^2 \rangle-{\langle A \rangle}^2$$ BUT $$A$$ is an operator and $$\langle A \rangle$$ is a number! How can $$(A- \langle A \rangle)^2$$ even be defined?

Second - somewhere else in the proof a new vector is defined $$|\varphi \rangle=C| \psi \rangle$$ OK, they can do it, but then they say that if so, then also $$\langle \varphi |= \langle \psi | C^\dagger$$ and I just can't see why... Why the dagger I mean

Can anyone help?

Last edited: Jan 27, 2006
2. Jan 27, 2006

### Galileo

There's nothing wrong with adding a scalar to an operator, since you can interpret every scalar itself as an operator. It just multiplies the vector it acts upon by the scalar. If it helps, you can also imagine the identity operator next to it.

By the way, the ordinary definition of $(\Delta A)^2$ is either $\langle (A-\langle A \rangle)^2\rangle$, or $\langle A^2 \rangle-{\langle A \rangle}^2$. They are identical ofcourse, but from the former you can easily see that
$$(\Delta A)^2=\langle\psi|(A- \langle A \rangle)^2|\psi\rangle$$

since, for any operator $\langle Q \rangle = \angle \psi|Q|\psi \rangle$.

The bra associated with the ket $C|\psi \rangle$ is $\langle \psi|C^\dagger$, I`m sure that's explained in your book somewhere.
$$\langle \psi |C|\psi\rangle^*=\langle \psi |C^\dagger|\psi\rangle$$

3. Jan 28, 2006

### Ahmes

Thanks.
It took me quite a while to convince myself about the latter...