# Quantum operators and trasformation under rotations

## Main Question or Discussion Point

Good morning!
I have a problem in understanding the steps from vectors to operators.
Imagine you are given a vectorial observable.
In classical mechanics, after rotating the system it transform with a rotation matrix R.
If we go to quantum mechanics, this observable becomes an operator that is not obvious to transform with the same law.
Imagine for example the momentum: in classical mechanics it is clearly a vector, but when we go to quantum mechanics it becomes a gradient, so it will transform differently when rotating the system of reference.

How is this possible?

• strangerep

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Imagine for example the momentum: in classical mechanics it is clearly a vector, but when we go to quantum mechanics it becomes a gradient, so it will transform differently when rotating the system of reference.
How is this possible? It's not. Momentum is still a vector. It is no more "a gradient" in QM than it was "multiplication" classically. Do not confuse an observable with with an expression that calculates it.

PeroK
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Gold Member
Good morning!
I have a problem in understanding the steps from vectors to operators.
Imagine you are given a vectorial observable.
In classical mechanics, after rotating the system it transform with a rotation matrix R.
If we go to quantum mechanics, this observable becomes an operator that is not obvious to transform with the same law.
Imagine for example the momentum: in classical mechanics it is clearly a vector, but when we go to quantum mechanics it becomes a gradient, so it will transform differently when rotating the system of reference.

How is this possible?
If you have studied linear algebra, you should be familiar with how vectors and matrices transform under unitary transformations.

In QM, if we have an operator ##\hat O## and a state vector ##|\alpha \rangle##, then the expected value of the observable associated with ##\hat O## is given by:
$$\langle O \rangle = \langle \alpha | \hat O | \alpha \rangle$$
If we have a transformation that transforms the state to ##|\alpha ' \rangle = \hat U |\alpha \rangle##, then we require the transformed operator to satisfy:
$$\langle \alpha' | \hat O' | \alpha' \rangle = \langle O \rangle = \langle \alpha | \hat O | \alpha \rangle$$
Because both these expressions represent the expected value of the same physical observable. Also, expanding ##|\alpha' \rangle## gives:
$$\langle \alpha' | \hat O' | \alpha' \rangle = \langle \alpha| \hat U^{\dagger} \hat O' \hat U| \alpha \rangle$$
Hence:
$$\hat U^{\dagger} \hat O' \hat U = \hat O$$
And, finally, the transformation rule for operators:
$$\hat O' = \hat U \hat O \hat U^{\dagger}$$

• vanhees71
Good morning!
I have a problem in understanding the steps from vectors to operators.
Imagine you are given a vectorial observable.
In classical mechanics, after rotating the system it transform with a rotation matrix R.
If we go to quantum mechanics, this observable becomes an operator that is not obvious to transform with the same law.
Imagine for example the momentum: in classical mechanics it is clearly a vector, but when we go to quantum mechanics it becomes a gradient, so it will transform differently when rotating the system of reference.

How is this possible?
I recommend you to take a look at the book "quantum mechanics" by Messiah, chapter 13 and section 3. The rotation of scalar, vector and other tensor operators is described in full detail.

• vanhees71