# Is there a deterministic observable that has only single eigenvalue?

Is there an observable in quantum mechanics which has only one eigenvalue and an eigenspace associated with that single eigenvalue? This observable is deterministic in the sense that it gives same measurement value all the time. But the final state would be any of the wave functions living in its eigenspace corresponding to the single eigenvector, with different probabilities.

What would that mean practically, to quantum mechanics?

• I suppose every operator of that type would be the identity operator (multiplied by a real number). – yuggib Jan 7 '15 at 7:39
• @RajeshD I don't understand your comment, sorry. The identity operator $Id$ is the operator such that, for any $\psi\in\mathscr{H}$, $Id\psi=\psi$ so it obviously preserves the norm. – yuggib Jan 7 '15 at 7:46
• The identity operator is always supposed to exist in every theoretical formulation of quantum theories I know. My interpretation (but I don't know if everyone agrees) is that "measuring" the identity operator is making the trivial measurement, i.e. no measurement at all (leaving the system unchanged). This is "deterministic" because it would give you a sure output: it would leave the system as it is. – yuggib Jan 7 '15 at 8:02
• It is usually supposed that an eigenstate is left unchanged after a measurement of the corresponding operator; since all (pure) states are eigenstates for the identity operator, they are left unchanged by the measurement process. – yuggib Jan 7 '15 at 8:57
• It is usually assumed that given an initial state, the measurement process is such that it gives you as a value one of the eigenvalues, and the resulting state is the projection of the original one on the eigensubspace corresponding to the eigenvalue. Therefore, since the eigensubspace of the identity operator with eigenvalue one is the whole space, and the initial state is in the space, the result of the projection would be trivially the state itself, i.e. the system is left unchanged. – yuggib Jan 7 '15 at 10:03

If I understand the question and the comments correctly, what is needed is an everywhere defined operator that preserves norms and has only a single point in the spectrum. The first condition forces the operator to be a partial isometry, while the second forces it to be a multiple of the identity. The intersection is then any operator $zI$, where $z$ is a complex number of norm one and $I$ is the identity operator.

• I guess I found a deterministic (single eigenvalue) operator which is not an identity operator (the one that takes $\psi$ and results in $\psi$ itself even after measurement). It takes $\psi$ and results in something else after measurement. I want to show that and test its validity by posting it here, but before that I want to enlive my discovery before it is refuted. So I want to know what are the implications in QM in case such an operator exists. – Rajesh Dachiraju Jan 7 '15 at 13:02
• It can't possibly be self-adjoint, so it is not an observable. – Phoenix87 Jan 7 '15 at 13:18

My contention is that there is an observable other than the identity operator, which is deterministic (meaning single eigenvalue). It is given as follows.

The Hilbert transform acts on functions of the form $f:\mathbb{R}\to\mathbb{R}$, to give out functions in the same domain, while preserving the norm. Let $f^h$ denote the Hilbert transform of $f$.

Let us stick to $1$-space dimension. Let all the wave functions live in a Hilbert space and be of unit norm.

We define the operator $i\mathcal{H}$ as follows. Let $\psi(x) = \psi_R(x) + i\psi_I(x)$, where $\psi_R, \psi_I$ are real avlued functions. Lets assume $||\psi|| = 1$. The operator $i\mathcal{H}$ is defined as follows.

$$i\mathcal{H}\psi = i(\psi_R^h + i\psi_I^h)$$

Properties of this operator $i\mathcal{H}$,

1. Linear and self-adjoint, and hence an observable.

2. Single eigenvalue $\lambda= 1$. The corresponding eigenspace consists of all the unit norm wave functions of the form $\psi = f+if^h$, where $f:\mathbb{R}\to \mathbb{R}$ and $||f|| = 0.5$

Edit : After comment from @Phoenix87, it has another eigenvalue $\lambda = -1$ and the corresponding eigen space is wave functions of the form $f-if^h$.

Note : $\lambda = 1$ corresponds to particle moving in positive $x$-direction, and $\lambda = -1$ corresponds to particle moving in negative $x$-direction. Wonder what I should name this operator!

1. The final state after the measurement is $$\frac{1}{2}{(\psi_R + \psi_I^h) +/- i\frac{1}{2}(\psi_I - \psi_R^h)}$$

2. Commutes with operators like translations in space and also with the momentum operator.

There may be more properties but these are the ones I could think of.

PS : The properties 1,2,3,4 can be verified using the fact that the Hilbert transform of $f^h = -f$ and few other properties of the Hilbert transform.

• @Phoenix87 : Wish if you could verify this and comment. – Rajesh Dachiraju Jan 7 '15 at 13:46
• @Qmechanic : Request your help in verification and your valuable comments. – Rajesh Dachiraju Jan 7 '15 at 14:04
• According to en.wikipedia.org/wiki/Hilbert_transform#Boundedness, for $p=2$ the Hilbert transform is a contraction, but not necessarily an isometry, i.e. it is not true in general that $\Vert f^h\Vert = \Vert f\Vert$, but just $\Vert f^h\Vert\leq\Vert f\Vert$. – Phoenix87 Jan 7 '15 at 14:29
• @Phoenix87 : In the same page, se $H(H(u)) = -u$, which indicates its not a contraction. Hilbert transform is a Fourier multiplier and hence preserves the norm. – Rajesh Dachiraju Jan 7 '15 at 14:40
• that is ${f^h}^h = -f$, i.e. that the Hilbert transform is an anti-involution – Phoenix87 Jan 7 '15 at 14:41