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David Bohm in Section (4.5) of his wonderful monograph Quantum Theory after defining the usual density probability function $P(x,t)=\psi^{*} \psi$ for the Schrödinger equation for the free particle in one dimension: \begin{equation} i \hbar \frac{\partial \psi}{\partial t}= - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}, \end{equation} states that $P(x,t)$ is the unique function of $\psi(x,t)$ and the partial derivatives of $\psi$ with respect to $x$ all computed in $(x,t)$ which satisfies the following properties:

  1. P is never negative;

  2. the probability is large when $|\psi|$ is large and small when $|\psi|$ is small;

  3. the significance of $P$ does not depend in a critical way on any quantity which is known on general physical grounds to be irrelevant: in particular this implies (since we are dealing with a nonrelativistic theory) that $P$ must not depend on where the zero of energy is chosen;

  4. $\int P(x,t) dx$ is conserved over time, so that by eventually normalizing $P$ we can choose $\int P(x,t) dx=1$ for all $t$.

Bohm gives no mathematical argument at all and actually the statement seems completely unjustified to me.

Does someone know some reason why it should be true?

NOTE (1). Since the Schrödinger equation is a first-order equation, the time evolution of $\psi$ is fixed given the initial state $\psi(x,0)$: this is the reason why we require that the probability $P(x,t)$ depends on the state at time $t$, that is $\psi(x,t)$, and its spatial derivatives. To be explicit, the requirement that $P(x,t)$ is a function only of $\psi(x,t)$ and the partial derivatives of $\psi$ with respect to $x$ all computed in $(x,t)$ means that there exists a function $p$ such that $P(x,t)=p\left(\psi(x,t),\frac{ \partial \psi}{\partial x}(x,t),...,\frac{\partial^m \psi}{\partial x^m}(x,t)\right)$.

NOTE (2). The one given above is not a mathematically rigorous formulation of the problem, but the one originally given by Bohm. So we can feel free to attach a rigorous mathematical meaning to the different properties. In particular, as for property (iv) we can formulate it in a different (and not equivalent) mathematical form, by requiring that a local conservation law holds, in the sense that there exists a function $\mathbf{j}$, such that, if we put $\mathbf{J}(x,t)=\mathbf{j}\left(\psi(x,t),\frac{ \partial \psi}{\partial x}(x,t),...,\frac{\partial^m \psi}{\partial x^m}(x,t)\right)$, we get \begin{equation} \frac{\partial P}{\partial t} + \nabla \cdot \mathbf{J} = 0. \end{equation}

NOTE (3). Similar questions are raised in the posts Nonexistence of a Probability for Real Wave Equations and Nonexistence of a Probability for the Klein-Gordon Equation. Presumably Bohm had in mind the same kind of mathematical argument to tackle these three problems, so the real question is: what mathematical tool did he envisage to use? Maybe some concept from classical field theory or the theory of partial differential equations?

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  • $\begingroup$ Reason for which statement? $\endgroup$ – Avantgarde Oct 9 '18 at 18:06
  • $\begingroup$ The statement by Bohm is that $P$ is the unique probability that you can define by using $\psi$ and its partial derivatives which satisfies the properties listed in the post. Obviously, these properties required are not defined in a very rigorous mathematical way, but this is how Bohm deals with the issue. $\endgroup$ – Maurizio Barbato Oct 9 '18 at 19:36
  • $\begingroup$ You can find justification for these kind of statements in an introductory textbook on QM(e.g., Griffiths'). $\endgroup$ – Omar Nagib Oct 9 '18 at 19:43
  • $\begingroup$ If P(x) is the probability of finding a particle at x -then the integral says that the total probability must be 1 as the particle must be found between -infinity to +infinity value of x. $\endgroup$ – drvrm Oct 9 '18 at 19:50
  • $\begingroup$ I asked a related question here. Doesn't have exactly what you want, but it may have some useful pointers. $\endgroup$ – knzhou Oct 25 '18 at 20:16
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Bohm's assumptions are not mathematically precise, so you must attach a mathematical interpretation to them (especially statements $2$ and $3$). Since you have not done so yourself, I will try to interpret them in a way that I find reasonable.

Definition for $P$: We will require that the probability density $P_\psi(x,t)$ of any smooth function $\psi$ be a local function of its partial derivatives at $(x,t)$.

More formally, let $\psi(x,t)$ be a smooth function and let $j^\infty_{x,t}\psi$ denote the infinite jet prolongation of $\psi$ at $(x,t)$, i.e., its formal Taylor series expansion about $(x,t)$. Then we can write $$P_\psi(x,t) = p(j_{x,t}^\infty\psi),$$ for some function $p$ defined on the jet bundle. In terms of regularity, we will require $p$ to be continuous.

This is essentially the definition you proposed for $P$. In fact, this is already problematic since we will necessarily have to work with wavefunctions which are not smooth. Therefore it is already problematic to require that $P$ depend on higher derivatives since they are not guaranteed to even exist. Nevertheless, we will allow arbitrary dependence on higher partials and apply the consistency conditions for $P$ evaluated on smooth functions. We can then recover $P$ uniquely for arbitrary $L^2$ functions by continuity.

Assumption 1: The function $p$ is a real-valued and non-negative.

Assumption 2: The probability density $P_\psi(x,t)$ is a non-decreasing function of $|\psi(x,t)|$, i.e., $$P_{\psi_1}(x,t) \ge P_{\psi_2}(x,t) \iff |\psi_1(x,t)| \ge |\psi_2(x,t)|.$$

Assumption 3: The function $p$ is invariant under global phase, i.e., $$p(e^{i\theta} j^\infty_{x,t}\psi) = p(j^\infty_{x,t}\psi).$$

Assumption 4: If $\psi(x,t)$ is a normalized function, then $P_\psi(x,t)$ is likewise a normalized function.

Let us go through the assumptions one by one. Assumption $1$ is relatively straightforward as probability densities must be real and non-negative.

Property $2$ is, in my view, the most difficult to properly interpret. The way I have interpreted the property in Assumption $2$ is to say that the magnitude of the probability density at a point is directly reflective of the magnitude of the wavefunction at that point. This is what I feel to be the most direct transcription of Bohm's second property.

This assumption is in fact extremely strong, and it necessarily implies that $p$ is independent of all derivatives of $\psi$. This is essentially because the value of a smooth function and all of its derivatives can be independent prescribed at any point. This was already pointed out by @Kostas.

Lemma: Suppose that $p(j^\infty_{x,t}\psi)$ is a continuous non-decreasing function of $|\psi(x,t)|$. Then $p$ is independent of all derivatives of $\psi$, i.e., $$p(j^\infty_{x,t}\psi) = p(j^0_{x,t}\psi) = p(\psi(x,t)).$$

Proof: By Borel's theorem, given any complex sequence $(a_{n,m})_{n,m=0}^\infty,$ and any point $(x,t)$, there exists a smooth function $\psi$ such that $$\frac{\partial^{n+m}}{\partial x^n \partial t^m}\psi(x,t) = a_{n,m}.$$ Therefore we are able vary the individual entries of the Taylor series completely independently.

Suppose that $p$ is not constant on some partial $\partial_i \psi$. Then by Borel's theorem we can find smooth functions $\psi_1$ and $\psi_2$ such that all Taylor coefficients of $\psi_1$ and $\psi_2$ agree at $(x,t)$ except for $\partial_i$. Then $$p(j_{x,t}^\infty\psi_1) \neq p(j_{x,t}^\infty\psi_2),$$ and without loss of generality we may assume that $$p(j_{x,t}^\infty\psi_1) > p(j_{x,t}^\infty\psi_2).$$ Next, we can find some other smooth function $\psi_3$ which agrees with $\psi_2$ for all Taylor coefficients at $(x,t)$ except for the constant term, $\psi_3(x,t)\neq \psi_2(x,t)$. By continuity, we can choose $\psi_3(x,t)$ slightly larger than $\psi_2(x,t)$ but still such that $$p(j_{x,t}^\infty\psi_1) > p(j_{x,t}^\infty\psi_3).$$ Therefore we have $$|\psi_1(x,t)| = |\psi_2(x,t)| < |\psi_3(x,t)|,\ \ \ \ \text{and}\ \ \ \ \ p(j_{x,t}^\infty\psi_1) > p(j_{x,t}^\infty\psi_3),$$ in contradiction to the monotonicity assumption. $\square$

Therefore we will assume that $P_\psi(x,t) = p(\psi(x,t))$ from now on. Note that at this point, we no longer need the assumption that $\psi$ is smooth.

Assumption $3$ is also essentially just a direct transcription of Property 3. If we change the energy, we effectively change the wavefunction by a global phase factor $e^{iEt}$. Since we can always make the wavefunction real and positive at any given point $(x,t)$ by an appropriate choice of a global phase factor, it follows that $p$ is independent of the phase of $\psi(x,t)$, i.e., $$p(\psi(x,t)) = p(|\psi(x,t)|).$$

Note that this assumption is actually completely unnecessary. We could've deduced the above equation by a slightly modification of our lemma using Assumption $2$. I will keep it just for the sake of completeness.

Finally, we come to Assumption $4$. Bohm's statement for his Property $4$ is that the probabilities should be normalized at all times, namely for all $t$ we should have $$1 = \int_\mathbb{R} P(x,t)\ dx.$$

This has certain ambiguities however. Which time-evolution should we use? Naively, any self-adjoint operator $H$ with a spectrum which is bounded below (so that there is a lowest energy level) should be able to serve as a valid Hamiltonian. If we require that the assignment $\psi \mapsto P_\psi$ be be universally valid, i.e., Hamiltonian independent, then we must require that $P(x,t)$ be normalized with respect to the unitary evolution generated by any Hamiltonian.

It can be shown that given any unitary $U$, there exists some admissible (self-adjoint, bounded below) Hamiltonian $H$ such that $U = e^{iH}$. In fact, we do not even need to consider the set of all admissible Hamiltonians, but rather just the set of all bounded Hamiltonians due to the following theorem.

Theorem: Let $\mathcal{H}$ be a Hilbert space and let $U$ be any unitary operator on $\mathcal{H}$. Then there exists a bounded self-adjoint operator $A$ (with norm at most $\pi$) such that $$U = e^{iA}.$$

Proof: This is a simple consequence of the Borel functional calculus for bounded operators applied to the principal branch of the logarithm. See here for a complete proof. $\square$

Now, let $\psi_1(x)$ be some normalized wavefunction. Let us assume without loss of generality that $P$ is normalized so that $$1 = \int_\mathbb{R} P_{\psi_1}(x)\ dx.$$ Let $\psi_2(x)$ be some other arbitrary normalized wavefunction. Let $U$ be any unitary such that $\psi_2 = U\psi_1$. Then there exists some bounded Hamiltonian such that the time-evolution brings the initial state $\psi(x,t=0) = \psi_1(x)$ to $\psi(x,t=1) = \psi_2(x)$. This means that we must have $$1 = \int_\mathbb{R} P_{\psi_1}(x)\ dx = \int_{\mathbb{R}} P_{\psi_2}(x)\ dx.$$ Since $\psi_2$ was an arbitrary normalized function, it follows that $P_\psi$ is normalized for all normalized $\psi$. We take this as our Assumption $4$.

Physically, this assumption is essentially saying that we should be able to vary the potential of the Hamiltonian so as to drive any normalized wavefunction arbitrarily close to any other normalized wavefunction. Since $P$ is conserved under this evolution, it must be normalized given any normalized wavefunction.

Note that this implies that we must have $p(0) = 0$. Otherwise $p(0) > 0$ will give a divergent integral for any normalized compactly supported function $\psi$.

Now let $y>0$. Define $\psi_y(x,t)$ to be equal to $1/\sqrt{y}$ for $x\in (0,y)$ and zero elsewhere. Then we have $$\int_\mathbb{R} |\psi_{y}(x,t)|^2\ dx = 1 = \int_\mathbb{R} p(|\psi_y(x,t)|)\ dx = \int_0^y p(1/\sqrt{y})\ dx = yp(1/\sqrt{y}).$$ Therefore we must have $$p(|\psi(x,t)|) = |\psi(x,t)|^2.$$

This is the desired claim. Of course, you might disagree with how I've interpreted some of Bohm's statements. But as you've said yourself in the question, some rigorous definitions must be assigned to these physical properties. These are simply what I felt to be the most faithful.

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    $\begingroup$ I have carefully thought upon your assumption 4, and even though it is not completely unjustified, for sure it is not an adequate interpretation of property 4 required by Bohm, which only says that $\int P(x,t) dx$ must be constant over time. Your assumption 4 makes the problem trivial, while Bohm's statement is not trivial at all. $\endgroup$ – Maurizio Barbato Oct 27 '18 at 21:05
  • $\begingroup$ @MaurizioBarbato You say that Property 4 only requires that $\int P\ dx$ be constant. But constant with respect to what? Whatever probability assignment we make, it should not be dependent on the Hamiltonian itself. Granted that $P$ is universal, $\int P\ dx$ must be constant with respect to all possible Hamiltonian evolutions. The unitaries generated by the set of all admissible (i.e., self-adjoint, bounded from below) Hamiltonians will be dense in the full unitary group of the Hilbert space. $\endgroup$ – EuYu Oct 28 '18 at 14:29
  • $\begingroup$ This means that given any two normalized wavefunctions $\psi_1(x)$ and $\psi_2(x)$, there will exist some Hamiltonian $H$ such that $U_H(1)\psi_1 \approx \psi_2$ to arbitrary precision. By continuity, this means that we must have $\int P_{\psi_1}\ dx = \int P_{\psi_2}\ dx$. Normalizing $P$ for any single normalized $\psi$ means that $P$ will be normalized for all possible normalized $\psi$. This is precisely the assumption I've made. If you don't think that this is an adequate interpretation, then you should clearly state in the question what you believe to be adequate. $\endgroup$ – EuYu Oct 28 '18 at 14:30
  • $\begingroup$ Now I understand your point, and this seems to me a very brilliant idea! Sorry for not having grasped it at the first time. If I do not overindulge, I would like to have some reference for your statement "the unitaries generated by the set of all admissible Hamiltonians is dense in the fulll unitary group of the Hilbert space". I think this has something to do with Stone's Theorem, but I don't know what you exactly mean, since I never met the notion of "bounded from below" operator. Thank you very very ... much in advance. $\endgroup$ – Maurizio Barbato Oct 28 '18 at 19:39
  • $\begingroup$ @MaurizioBarbato I've updated the answer so that it contains our discussion so far, as well as a reference to the proof you wanted. $\endgroup$ – EuYu Oct 29 '18 at 3:40
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(i) Clearly $P$ is non-negative since $a^*a=|a|^2$ is non-negative for all complex numbers $a$.

(ii) The probability to find the particle in an infinitesimal interval between $x$ and $x+dx$ is given by $Pdx=|\psi(x,t)|^2dx$ where this probability clearly is large for large $|\psi|$ and vice versa.

(iii)Indeed it can be proven that shifting the zero of energy by a constant(i.e., $V(x) \to V(x)+V_o$, where $V_o$ is constant) changes the overall phase of the wavefuntion so that $\psi(x,t) \to \psi(x,t)\exp(-iV_ot/\hbar)$, but this does not affect $P$, since $P_{\text{new}}=\psi_{\text{new}}^*\psi_{\text{new}}=[\psi^*\exp(+iV_ot/\hbar)][\psi\exp(-iV_ot/\hbar)]=\psi^*\psi=P$.

(iv) There're a lot of ways to prove this; One way is to show $\dfrac{d}{dt}{\displaystyle \int_{-\infty}^{+\infty}\psi^*\psi dx }=0$, You can go about doing this by pulling the derivative inside the integral and applying the product rule on $\dfrac{d[\psi^*\psi]}{dt}$ and using Schrodinger equation and integration by parts to prove the desired result.

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    $\begingroup$ Dear Omar, thank you very much for your answer, but my post is asking something completely different. You proved that $P=\psi^{*} \psi$ satifies the required property, and this is standard staff. But my question is about the $\textbf{uniqueness}$ of $P$. Bohm asserts that $P=\psi^{*} \psi$ is essentially the unique function of $\psi$ and its partial derivatives which satifies the required properties. Do you some idea to prove this statement? $\endgroup$ – Maurizio Barbato Oct 10 '18 at 7:37
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OP is asking for a mathematical proof of uniqueness of some or all of the properties of P stated by Bohm. It is easy to see that the i), ii), and iii) can be violated separately and even together. Here is a trivial example: $P = (Ψ^* Ψ + |\partial Ψ|^2)^2$ It doesnt satisfy iv) of course. Prooving that no such expression satisifes iv) is hard. One way to do that is to assume P is a polynomial function of Ψ and its derivatives (it is in the example) and proove that all coefficients of the polynomial vanish except for the coefficient of the term with $Ψ^* Ψ$. It is a silly business, but it may help you if you are trying to understand why the Schroedinger equation is the only possible equation!

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  • $\begingroup$ I am not going to rewrite my answer, but the second part of ii) says P is "small when |ψ| is small" to me this sounds like partial derivatives are simply not allowed. Then the job of prooving iv) would be much easier. $\endgroup$ – Kostas Oct 12 '18 at 11:10
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After a long search in the literature, I must admit that the issue of establishing the uniqueness of the probability density for the Schrödinger equation as discussed by Bohm has attracted very little interest. Actually the only work which explicitly deals with this issue is Uniqueness of Conserved Currents in Quantum Mechanics, even though the uniqueness problem had been previously discussed in the context of de Broglie-Bohm pilot wave theory (see the papers there quoted).

Holland uses relativistic considerations to show the uniqueness of the conserved current for the Klein-Gordon equation, from which he deduces in the non-relativistic limit the analogous uniqueness result for the Schrödinger equation.

Anyway, as the same author suggested to me in a private communication, a more direct proof could be given if we start directly from the Schrödinger equation in the case of a potential $V$: \begin{equation} i \hbar \frac{\partial \psi}{\partial t}(\mathbf{x},t) = -\frac{\hbar^2}{2m} \Delta \psi(\mathbf{x},t) + V(\mathbf{x}) \psi(\mathbf{x},t), \end{equation} and use the method he introduces in his paper to derive all the conserved currents $(P,\mathbf{J})$ which are functions only of $\psi$, the first-order partial derivatives of $\psi$ and eventually $V$, by studying their transformation properties under the Galilei transformations and under possible changes of $V$. Let us note that we have to be careful in applying this procedure since $\psi$ is not invariant under Galilei transformations, but it changes appropriately: see e.g. Commins, Quantum Mechanics or Galilei Invariance of the Schrodinger Equation.

This will give a general set of conserved currents, and we could investigate whether the additional assumption that $P$ must depend only on $\psi$ and not on its first-order partial derivatives (which is a particular consequence of property (ii) required by Bohm) eventually implies a unique expression for $P$ or not. In the last case, some other condition (maybe also derived by property (ii) which is quite vague) should be added in order to obtain the uniqueness.

Anyway, this approach would in any case suffer of a lack of generality, since you must assume from the very beginning that $P$ and $\mathbf{J}$ do not depend on partial derivatives of $\psi$ of order greater than one, an assumption which seems completely absent in Bohm's discussion, even though a very plausible assumption on physical grounds (see the remark made on this point by Holland in his work). For this reason, I am pretty sure that this approach was not the one Bohm had in mind when he made his uniqueness statement, even though I have no idea about the kind of argument he could have envisaged.

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