Why must the state transformation be unitary?

Hi there,

In QM it is said that state transformations must be implemented via unitary operators. The reason is that, if ##\left| \Psi\right\rangle## is a normalized state and ##U## is a unitary operator, than ##U\left|\Psi\right\rangle## is also normalized.

But why do I need ##U\left|\Psi\right\rangle## to be normalized? A state in QM is actually a ray in a Hilbert space, so if ##\left|\Psi\right\rangle## represents my state, ##\alpha\left|\Psi\right\rangle## also represents my state. Therefore, I can apply a non-unitary operator and then divide the resulting state to make it normalized.

What am I missing?

eoghan said:

TL;DR Summary: Why can't we have non-unitary transformations followed by a normalization of the transformed state, instead of considering only unitary transformations that lead to normalized states?

Hi there,

In QM it is said that state transformations must be implemented via unitary operators. The reason is that, if ##\left| \Psi\right\rangle## is a normalized state and ##U## is a unitary operator, than ##U\left|\Psi\right\rangle## is also normalized.

But why do I need ##U\left|\Psi\right\rangle## to be normalized? A state in QM is actually a ray in a Hilbert space, so if ##\left|\Psi\right\rangle## represents my state, ##\alpha\left|\Psi\right\rangle## also represents my state. Therefore, I can apply a non-unitary operator and then divide the resulting state to make it normalized.

What am I missing?

Is that a well-defined process? The scale factor ##\alpha## will depend on the initial state. That's no longer a well-defined operator. Consider ##Tu, Tv, T(au + bv)##, where ##T## is a non-unitary operator.

eoghan said:

What am I missing?

If you got a probability of 7, would this be sensible?

eoghan said:

TL;DR Summary: Why can't we have non-unitary transformations followed by a normalization of the transformed state, instead of considering only unitary transformations that lead to normalized states?

Hi there,

In QM it is said that state transformations must be implemented via unitary operators. The reason is that, if ##\left| \Psi\right\rangle## is a normalized state and ##U## is a unitary operator, than ##U\left|\Psi\right\rangle## is also normalized.

But why do I need ##U\left|\Psi\right\rangle## to be normalized? A state in QM is actually a ray in a Hilbert space, so if ##\left|\Psi\right\rangle## represents my state, ##\alpha\left|\Psi\right\rangle## also represents my state. Therefore, I can apply a non-unitary operator and then divide the resulting state to make it normalized.

What am I missing?

It's not only that you have to preserve the norm of a state ket but the entire system of equations of QT. This means a transformation that preserves one to one all equations of QT you have to think of what's the essence of the entire formalism.

You have quantum states, which are described in the most general case by a statistical operator ##\hat{R}##, which is a positive semidefinite adjoint operator with trace 1. Then you have a set of self-adjoint operators, which build the observable algebra with the commutator as the algebraic operation (building a Lie algebra). Finally the physical meaning of the state is that for a complete set of compatible observables, ##A_1,\ldots,A_n##, the probability to find the eigenvalues ##a_1,\ldots,a_n## of their representing operators ##\hat{A}_1,\ldots,\hat{A}_n##,
$$P(a_1,\ldots,a_n)=\langle a_1,\ldots,a_n|\hat{R}|a_1,\ldots,a_n \rangle.$$
It turns out that the most general type of transformations of this entire mathematical system are unitary (or antinitary) transformations.

PeroK said:

Is that a well-defined process? The scale factor ##\alpha## will depend on the initial state. That's no longer a well-defined operator. Consider ##Tu, Tv, T(au + bv)##, where ##T## is a non-unitary operator.

Vanadium 50 said:

If you got a probability of 7, would this be sensible?

My idea goes like this. Let's suppose there is an observable ##A## with eigenvalues ##a_1##,...,##a_n## and eigenvectors ##|a_1\rangle##,...,##|a_n\rangle##. Our state is described by a state vector ##|\Psi\rangle##. The probability to measure the value ##a_1## for ##A## is then given by

$$
P(a_1) = \frac{\langle a_1|\Psi\rangle\langle\Psi|a_1\rangle}{\langle\Psi|\Psi\rangle}
$$

Similarly, under a non-unitary transformation ##\hat R##, we have

$$
P(a_1) = \frac{\langle a_1|\hat R|\Psi\rangle\langle\Psi|\hat R^\dagger |a_1\rangle}{\langle\Psi|\hat R^\dagger \hat R|\Psi\rangle}
$$

Therefore, the probability cannot be, e.g., 7. Nor it doesn't matter if the normalization factor depends on the initial state. The denominator would take care of both cases.

vanhees71 said:

It turns out that the most general type of transformations of this entire mathematical system are unitary (or antinitary) transformations.

Interesting. Can you please elaborate on this? Or, do you know some textbook / reference about this?

eoghan said:

My idea goes like this. Let's suppose there is an observable ##A## with eigenvalues ##a_1##,...,##a_n## and eigenvectors ##|a_1\rangle##,...,##|a_n\rangle##. Our state is described by a state vector ##|\Psi\rangle##. The probability to measure the value ##a_1## for ##A## is then given by

$$
P(a_1) = \frac{\langle a_1|\Psi\rangle\langle\Psi|a_1\rangle}{\langle\Psi|\Psi\rangle}
$$

Okay.

eoghan said:

Similarly, under a non-unitary transformation ##\hat R##, we have

$$
P(a_1) = \frac{\langle a_1|\hat R|\Psi\rangle\langle\Psi|\hat R^\dagger |a_1\rangle}{\langle\Psi|\hat R^\dagger \hat R|\Psi\rangle}
$$

What is your physical interpretation of that calculation?

eoghan said:

My idea goes like this. Let's suppose there is an observable ##A## with eigenvalues ##a_1##,...,##a_n## and eigenvectors ##|a_1\rangle##,...,##|a_n\rangle##. Our state is described by a state vector ##|\Psi\rangle##. The probability to measure the value ##a_1## for ##A## is then given by

$$
P(a_1) = \frac{\langle a_1|\Psi\rangle\langle\Psi|a_1\rangle}{\langle\Psi|\Psi\rangle}
$$

Similarly, under a non-unitary transformation ##\hat R##, we have

$$
P(a_1) = \frac{\langle a_1|\hat R|\Psi\rangle\langle\Psi|\hat R^\dagger |a_1\rangle}{\langle\Psi|\hat R^\dagger \hat R|\Psi\rangle}
$$

Therefore, the probability cannot be, e.g., 7. Nor it doesn't matter if the normalization factor depends on the initial state. The denominator would take care of both cases.Interesting. Can you please elaborate on this? Or, do you know some textbook / reference about this?

This is known as Wigner's theorem. A very concise derivation can be found in the (older) textbook by Gottfried:

Kurt Gottfried, Quantum mechanics, CRC (1974)

eoghan said:

Similarly, under a non-unitary transformation ##\hat R##, we have

$$
P(a_1) = \frac{\langle a_1|\hat R|\Psi\rangle\langle\Psi|\hat R^\dagger |a_1\rangle}{\langle\Psi|\hat R^\dagger \hat R|\Psi\rangle}
$$

Therefore, the probability cannot be, e.g., 7. Nor it doesn't matter if the normalization factor depends on the initial state. The denominator would take care of both cases.

For a general transformation ##\hat R##, that won't necessarily be a real number. In fact, I suspect it's only a real number for all states ##\Psi## if ##\hat R## is a scalar multiple of a unitary transformation.