Let's say I have an operator $\hat{A}$ and a state $|\psi\rangle$. What exactly is the state $\hat{A}|\psi\rangle$? Is it just another different state that I am describing using my $\hat{A}$ and $|\psi\rangle$? For example, if

$$\hat{A} \doteqdot \text{put chocolate syrup}$$

Then is $\hat{A}$ just a tool to describe a state, like:

$$|\text{vanilla ice cream with chocolate syrup}\rangle = \hat{A}|\text{vanilla ice cream}\rangle$$

But on the other hand, we have something like

$$\hat{H} |\psi\rangle = E|\psi\rangle$$ I interpret this equation as "If you apply the (time-independent) Hamiltonian to a state, the result is proportional to your original state". But $E|\psi\rangle$ itself cannot be a new state, because it's in general not normalized. So here the operator $\hat{H}$ is just used to establish a mathematical property of $|\psi\rangle$, not to describe another state. You can't say that $\hat{H}$ is a machine that takes a state and returns another state, not in the same way $\hat{A}$ takes an ice cream and puts syrup on it. Or can you?

And what does taking a measure mean? If you measure an observable, it returns an eigenvalue and the state collapses into an eigenstate. Is this resulting eigenstate the one you obtain when you apply the operator to the state?


I think you may be misguided by the concept that we associate $\textbf{observables}$ to self-adjoint operators. They do operate on the Hilbert space, but to see them as entities that transform states or prepare them is a little bit tricky. I will describe here self-adjoint operators and preparation of states.

1) The true (physical) power of self-adjoint operators for describing observables lies in the spectral theorem, and not in its $\psi \mapsto A\psi$ action. Physically, what it means? There is a set called spectrum of an observable, and it is the set of possible outcomes on its measure for given states. For example, a spin observable $S$ on a 1/2-spin system has spectrum $\sigma(S) = \{-1/2,+1/2\}$, and decomposes as a sum of its spectral projections, $S = +1/2 P_{+} -1/2 P_{-}$. In general, there is a spectral resolution $E$, that is, a bunch of projection related to the spectrum, such that the operator can be written as $A = \int_{\sigma(A)}\lambda dE(\lambda)$.

And what are the spectral projections? Those are again (self-adjoint) operators, but the whole collection of spectral projections will give you a probability measure when coupled with a state. In the spin system example, if you take a state $\psi$, then $\langle\psi, P_+\psi\rangle$ would give you the probability of measuring a +1/2 spin, and likewise for -1/2.

Now suppose you had a 1/2 spin system with prepared state $\psi$, and you measure the spin, and get +1/2. After the measurement, your state collapses to a $|+1/2\rangle$ state.

In a more detailed formalism, suppose you have prepared a state $\psi$ and you are going to make a measurement of an observable expressed as $A = \int_{\sigma(A)} \lambda dE(\lambda)$ (where the $E$ is the spectral resolution of your operator, just think of the 1/2-spin example intuitively). Then suppose your measurement is on a subset $\Lambda \subset \sigma(A)$ (you may think of the set $\{+1/2\} \subset \{-1/2,+1/2\}$. Your state $\psi$ then collapses to the following state $\phi$:

$\psi \rightarrow \phi = \frac{E(\Lambda)\psi}{\|E(\Lambda)\psi\|}. $

(notice that $\phi$ is normalized and well defined, since $E(\Lambda)\psi=0$ then the probability of the outcome being in $\Lambda$ would be zero to start).

Summing up, you do not simply apply a self-adjoint operator on a state, since, as you have seen, it doesn't have much meaning. This is a point most introductory QM books do not stress as much as I would want. What happens with measurements and collapses and whatsoever uses, as I tried to point out, the spectral projections more than the operator itself. So, as you said about your Hamiltonian operator, it does not act like your syrup machine, which we will try to cover up next.

2) Now what you describe as "tools", in your example, the putting syrup, is not a measurement per se, it is a preparation of states, which would grab a state without syrup and put syrup in it. The modeling of such procedure is usually ignored, at least to my knowledge.

One choice would be just saying "my state now is syrup", end of discussion.

Other option is using unitary operators ($U$ such that $UU^* = U^*U = 1$). Those transform state vectors in state vectors.

If you would like more sophisticated examples, it starts to get tricky, and I will shut up before I say something very wrong about it. But rest assured this is not easy at all, and your question is really nice. Hope to see some other inspiring aswers.

  • $\begingroup$ Ah, that makes things clearer for me. Yeah, I was confused because my textbook just says "if you apply the Hamiltonian to a state, you get the same state times E" which I found weird because I was interpreting as the operator physically modifying the state. $\endgroup$ – Ignacio Nov 16 '12 at 18:59

It seems that OP's question arises because he assumes that a state $|\psi\rangle$ is normalized $\langle\psi |\psi\rangle=1$ at all stages of developing the quantum mechanical language.

Let $H$ be a Hilbert space. Note that the set $$\{|\psi\rangle\in H \mid \langle\psi |\psi\rangle=1\}$$ of normalized states is not a vector space, and therefore not a Hilbert space.

It is better to only assume that a state $|\psi\rangle$ is just normalizable


with the implicit assumption that when one wants a probabilistic interpretation, then one should normalize $|\psi\rangle$ via the standard procedure:

$$ |\psi\rangle ~\longrightarrow ~ |\psi^{\prime}\rangle~:=~\frac{|\psi\rangle}{\sqrt{\langle\psi|\psi\rangle}} , $$

so that $$\langle\psi^{\prime} \mid\psi^{\prime}\rangle~=~1.$$

So to answer OP's question, in the elementary version$^1$ of quantum mechanics, a state is a (ket) element $|\psi\rangle$ of the Hilbert space $H$. In particular, it is a normalizable element. An Observable is a linear Hermitian operators $\hat{A}:H\to H$ that takes states to states. The expectation value $\langle\hat{A}\rangle$ of the observable $\hat{A}$ in the state $|\psi\rangle$ is then

$$\langle\hat{A}\rangle ~=~ \frac{\langle\psi|\hat{A}|\psi\rangle}{\langle\psi|\psi\rangle}. $$

Concerning the subquestion about measurements in quantum mechanics and the collapse of the wave-function, I suggest to first check out Wikipedia, and if needed, then ask a more specific question.


$^1$The kind of version that ignores un-normalizable states and unbounded linear operators.


Your attempt to summarize a state vector by attaching to it a daily life meaning is probably a reason why you can't understand the more abstract situation prevailing in QM.

$A|\psi\rangle$ is simply the image of the vector $|\psi\rangle$ under the operator $A$. There is no need that either $|\psi\rangle$ or its image is a state in the sense of being normalized. They are just elements of the Hilbert space (or, sometimes, unnormalizable weak limits of such states.)

There is also no necessary relation to measurement. Interpreting realistic measurement is a quite complex matter, and the textbook recipe (Born's rule) is applicable only to the simplest or very idealized situations.


A really cool example came up when a friend of mine and me discussed a spin-1/2 system. I had some trouble and it turns out that exactly the same misunderstanding that started this topic lead me to really strange results:

I'll write $S_x, S_y$ and $S_z$ for the canonical components of the spin operator and $|S_i; \pm\rangle$ for its eigenvectors, i.e. $S_y \lvert S_y; -\rangle = -\frac{\hbar}{2}\lvert S_y; -\rangle$. For brevity we write $\lvert \pm \rangle = \lvert S_z; \pm\rangle$

For future reference we recall

$$ \lvert S_x; \pm \rangle = \frac{1}{\sqrt 2} \lvert + \rangle \pm \frac{1}{\sqrt 2} \lvert - \rangle$$ and also $$ \lvert \pm \rangle = \frac{1}{\sqrt 2} \lvert S_x;+ \rangle \pm \frac{1}{\sqrt 2} \lvert S_x;- \rangle$$

OK, here's what I thought:

Assume we have an electron that's definitely in state $\lvert + \rangle$ (perhaps we got it out of a Stern-Gerlach experiment). Then we send it through another Stern-Gerlach experiment, oriented in the $x$-direction. I thought "that means applying the $S_x$ operator, right?", so we got $$ S_x \lvert + \rangle = S_x \left(\frac{1}{\sqrt 2} \lvert S_x;+ \rangle + \frac{1}{\sqrt 2} \lvert S_x;- \rangle \right) = \frac{\hbar}{2\sqrt 2} \left( \lvert S_x;+\rangle - \lvert S_x;- \rangle \right) $$

In the $z$-basis the last term is equivalent to

$$ S_x \lvert +\rangle = \frac{\hbar}{2\sqrt 2} \left(\frac{1}{\sqrt 2} \lvert + \rangle + \frac{1}{\sqrt 2} \lvert - \rangle - \left[\frac{1}{\sqrt 2} \lvert + \rangle - \frac{1}{\sqrt 2} \lvert - \rangle \right]\right) = \frac{\hbar}{2} \lvert - \rangle$$

Hence, if we confronted this outcome with a third Stern-Gerlach experiment, this time again in $z$-direction, we will definitely get a spin down. This means that the sequence of subsequent SG experiments will swap the initial spin up to spin down.

Of course, this is in direct contradiction to both theory and experimental evidence. The point is exactly what has been discussed in this thread: Quoting Yul from above:

The true (physical) power of self-adjoint operators for describing observables lies in the spectral theorem, and not in its $\psi \mapsto A \psi$ action

It's a mistake to think of $S_z$ as the application of the measurement process. The outcome of SG is actually defined by the spectrum.

Fixing this thought, we get the following: Our initial $\lvert + \rangle$ can be written as $\lvert + \rangle = \frac{1}{\sqrt 2} \lvert S_x;+ \rangle + \frac{1}{\sqrt 2} \lvert S_x;- \rangle$. Measuring $S_x$ will then choose one of "spin up" and "spin down" with equal probability, let's say $\lvert S_x;- \rangle$. This again can be represented in $S_z$ base kets so we get another 50-50 chance for spin-up and spin-down with respect to $z$.

I hope that my mistake will help other students as well. Thanks Pascal for revealing the crucial point to me (also by referring to this thread).

  • $\begingroup$ Thank you o wise one to the eternity and back...I started learning QFT and with exactly the same ideas about measurement process I was having troubles geting around some ideas...this is what was missing. Tnx! $\endgroup$ – Žarko Tomičić May 7 at 14:44

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